Topic Page No.Theory 01 - 29Exercise - 1 30 - 38Exercise - 2 39 - 44Exercise - 3 45- 49Exercise - 4 50 - 55Exercise - 5 56 - 57Answer Key 58 - 59ContentsTHERMODYNAMICS&THERMOCHEMISTRYSyllabusTHERMODYNAMICS&THERMOCHEMISTRYFirstlawofthermodynamics;Internalenergy,workandheat,pressure-volumework;Enthalpy,Hess'slaw;Heatofreaction,fusionandvapourization;Secondlawofthermodynamics;Entropy;Freeenergy;Criterionofspontaneity.Name : ____________________________ Contact No. __________________ETOOS ACADEMY Pvt.LtdF-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor, BSNL Lane,Jhalawar Road, Kota, Rajasthan (324005)Tel. : +91-744-242-5022, 92-14-233303ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 1THERMODYNAMICS&THERMOCHEMISTRYIntroduction :Thermodynamics : The branch of science which deals with different forms of energy & their interconversion. THERMODYNAMICSFirst law of ThermodynamicsThermochemistry (application of I law in chemical reactions)Second law of ThermodynamicsApplication of thermodynamics :In chemistry using thermodynamics! Wecanpredictfeasibilityofthereactionthatisiftwosubstancesaremixedthenthereactionbetween them will takes place or not.! If reaction does take place then what are the energy changes involvedduring the reaction.! Ifinachemicalreaction,equilibriumisgoingtogetattainedthenwhatwillbetheequilibriumconcentrations of different reactants & products, can be calculated with thermodynamics.Limitations of Thermodynamics :! Laws of thermodynamics are applicable to matter in bulk or on system as a whole, these can not beapplied on individual particles(temperature, pressure, enthalpy etc have meanings only for systemas a whole).! Using thermodynamics we cannot calculate the time taken for completion of a reaction or for attainmentof chemical equilibrium.Terms to be used in Thermodynamics :! System : Part of the universe which is under study for energy changes.Ex. Air in a room, water in a bottle, any living body.! Surrounding : Rest of the universe.! Universe :Universe = System + Surroundings! Boundary : Anything which separates system & surroundings is called boundary.! Boundary can be real or imaginary.! Boundary can be flexible or rigide.g. - air in a flexible balloon (flexible boundary) while air in a room (fixed boundary).! Boundary can be adiabatic(non-conducting) or diathermic(conducting).Types of system :! Open system : System which can exchange energy & matter both with the surroundings.e.g. : Living systems( any living organism) are open systems, air in an open room! Closed system : System which can exchange only energy but cannot exchange matter with thesurroundings is called closed system.e.g. :any matter in a closed container.! Isolated system : System which cannot exchange energy and matter both with the surroundings.e.g. :Water in thermos flask.(Though not a perfectly isolated system but can be taken as, forsmall interval of time as the energy exchanges are negligible). ! Whole of universe is a perfect isolated system.ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 2. Open SystemClose System Isolated system ! State of a system :! It means the condition in which the system is present.! It can be specified/defined by measuring/ specifying some observable/measurable properties of thesystem like pressure, volume, temperature,amount of substance, elasticity, heat capacity etc.e.g.Foranidealgaseoussystemstateofthesystemcanbedefinedbyspecifyingvolume,temperature and pressure.! We may have to specify more properties of the system depending on the complexity of the system.State function :! Property of a system which is dependent only on the state of the system i.e. it is a point function! It is independent of the path adopted to attain a particular state.e.g.In Mechanics, Displacement of any object will a state function but distance travelled by theobject will be a path function.For any thermodynamic system,Temperature, Pressure, Volume, Total internal energy (E or U), Enthapy(H), Gibbs free energy (G),Entropy (S) are all state functions.e.g.In the above example the final temperature, pressure, and the volume will be same in both the above waysbut the work involved and the heat exchanged during the processes will be different.! Path function :! Quantities which are dependent on the path/way the system has achieved a particular state.e.g. Heat, work, Heat capacities(Molar heat capacities, specific heat capacities etc.).! These quantities are define when there is a process going on.! These can not have any definite (particular) value in any particular state of the system.! TypesofprepertiesExtensive properties :! Functions or properties of the system which are dependent on mass or on size of the system arecalled Extensive Properties .! Extensive functions are additive in nature( The addition of the volumes of the two parts equals thevolume of the whole of the room.)e.g.Volume, Mass, Total heat capacity, Total internal energy (E), Enthalpy(H), Gibbs FreeEnergy(G), Entropy(S).ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 3Intensive properties :! Functions or properties which are not mass dependentor size dependentare called intensivefunction.! Intensive properties are not additive in nature.eg. Temperature, pressure, molar heat capacity, specific heat capacity, density,concentration, vapourpressure.How to identify extensive or intensive propertiesIf a system in a particular state is divided into two equal or unequal parts, the properties which have valueequal to the original value of that property for the whole of the system is called an Intensive property. Whilethe properties which have values different from the values for whole of the system are called ExtensiveProperties.! For example consider air in a room at temp of 300K, 1 atm pressure. Now, if the room is divided bysome boundary( imaginary or real) into two parts( equal or unequal) then in these two parts :# The temperature, pressure, density of the gas, concentration of gaseous molecules etc. will havethe same value as that of for whole of the system. (intensive)# While the volume of two parts, mass of gas in two parts, total energy of the gaseous moleculesin the two parts, entropy the two parts etc. will be different from the values of these properties as forthe whole of the system initially. (extensive) ! Thermodynamic equilibrium :! When there is no change in any observable or measurable property of a system with time then thesystem is said to be in thermodynamic equilibrium.! Thermodynamic equilibrium consist of three types of equilibrium.(a) Mechanical equilibrium(b) Thermal equilibrium(c) Chemical equilirbrium! Mechanical equilibrium :There should not be any pressure gradient with time or with space (for anyideal gaseous system, for a liquid system there can be pressure gradientwith space as pressure at the bottom of the container in which a liquid isfilled will be greater than the pressure at the surface of the liquid.) in thesystem.! Thermal equilibrium : There should not be any temperature gradient (difference).Temperature may have different values at different places/locations in a system but it should remainconstant with time.! Chemical equilibrium : There should not be any concentration gradient of any of the species inthesystem. ! Types of thermodynamic process on basis of state/conditionsThermodynamic process : Any method/process by which system can change its state from one state ofthermodynamic equilibrium to another state of thermodynamic equilibrium.There can be infinite type of thermodynamic processes, out of these the following are important ones:ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 41. Isothermal process : T = constantdT = 0!T = 02. Isochoric process : V = constantdV = 0!V = 03. Isobaric process : P = constantdP = 0!P = 04. Adiabatic process : q = constantor heat exchange with the surrounding = 0(zero) ! Types of thermodynamics processes on basis of the way the processes are carried out :! Reversible process :The process that can be reversed by a very small change is known as reversibleprocess.! If a process is carried out in such a manner so that the system is always in thermodynamic equilibriumat every stage of the process. ! Ifthe process is carried out such that the difference in driving force and opposing force is infinitesimallysmall so that process takes place at infinitesimally slow rate.Fdriving Fopposing = dF and dF " 0! An ideal reversible process will take infinite time to get completed.! It is carried out infinitesimally slowly.! Strictly speaking there is no ideal reversible process in universe.To get an idea of a reversible process we can consider the following system.An ideal gas is enclosed in a container and a massless piston is put onthe gas on which a pile of sand is placed having particles of negligiblemass. To carry out a reversible expansion we will slowing decrease themass of the sand lets say by removing the particles one by one, so theexpansion of the gas will take place at infinitesimally small rate and wecan always assume the system to in thermodynamic equilibrium. So, theexpansion will be of reversible type. FBD of piston For piston to be in equilibrium :Pgas = Patm + Mg/A!Irreversible process : The process can not be reversed by a small change is known as irreversible.! If a process is carried out in such a manner so that the system is in thermodynamic equilibrium (I)Only at initial & final state of the process but not at the intermediate stages.(II) System may be in thermodynamic equilibrium state at some finite number of intermediate stagesonly - for example - n step irreversible expansion of a gas! If during the process there is a finite difference in driving force and opposing force so that processtakes place with a finite rate.! Irreversible processes will get completed in finite time.! Atintermediatestagesoftheirreversibleprocess,differentstatefunctionsuchasPressure,temperature etc. are not defined.! All real process are irreversible. ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 5Consider the above system. If the stopper placed over the piston is removed, then the piston will move withalmost infinite acceleration and will keep moving to a position where the pressure of the gas becomes equalto the external pressure. Since the process will get completed in finite time and there was a finite differencebetween the driving force and the opposing force so, process is irreversible.During the process, the pressureof the gas can not be defined as it will be having different values at different locations. ! Modes of energy exchange :These are two ways by which a system can interact or can exchange energy with its surroundings.(i) Heat & (ii) WorkHeat & Work both are forms of energy.Heat:Whentheenergytransfertakesplacebecauseoftemperaturedifferencebetweensystem&surroundings. It is known as heat.Work : Energy transfer which is not heat or which is not because of temperature difference is called work.Work can be of many types: Mechanical work, Electrical work, Magnetic work, Gravitational work etc.# The same energy transfer can be called work or can also be called heat depending on choiceof the system.To understand this, consider a system shown below in which water is taken in a closed container at 25C,the surroundings is also at temperature of 25C and there is a heater coil in the dipped in the water which isconnected to a battery through a switch S.Heater coil is also at 25C initially.

Now, there are two ways in which system can be chosen!System : All contents of the container (water + Heater coil). When switchis turned on there will be increment in the temperature of the system. Since the temperatureof the surroundings was equal to temperature of the system so, heat can not flow but still there is incrementin the energy of the system and hence, there is temperature increment. This must be because of electricalwork done by the battery on the system not because of the heat transfer as initially temperatures were equal.!!System : Water only is our system. Heater coil will be part of the surroundings.In this case when switch is turned on the temperature of the heater coil will increase first so there will be atemperature difference between system & surroundings. Hence, this energy transfer will be called heat. ! IUPAC Sign convention about Heat and WorkAny energy given to system is taken positive so heat given to system = positiveheat given out from system / taken out from system = NegativeWork done on the system= PositiveWork done by the system= NegativeThe sign convention is different from physics, but the meaning always comes out tobe same only in equation we have to use a different sign convention for work.So if in any problem, w = 10 JIt means system has done work of 10 Joule on surroundings.According to Chemistry :!w!Q

!w!Q

## $ In Physics!U = !Q + !W !U = !Q !WETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 6IUPAC convention of Heat1. Heat given to the system = +ve2. Heat coming out of the system = ve3. Work done on the system = +ve4. Work done by the system = veFirstLaw of Thermodynamics (FLOT) :Law of energy conservation :Energy of total universe is always conserved.orTotal energy of an isolated system is always conserved.orHence absolute value of E can never be calculated only change in value of E can be calculated for a particularprocess.Mathematical form of First Law of thermodynamics.If a system is intially in a particular state in which its total internal energy is E1. Now q amount of heat is givento it and w amount of work is done on it so that in new state its total internal energy becomes E2. Thenaccording to 1st Law of thermodynamics.E2 = E1 + q + wso !E = (E2 E1) = q + wApplication of First Law!U=!Q+!W Since !W= P !V& !U=!QP!V! Calculation ofdifferent quantities in FirstLaw of Thermodynamics (FLOT) :Calculation of "E :Thermodynamic definition of an ideal gas :! If for a gas the internal energy is directly proportional to its absolute temperature then the gas istermed as an ideal gas.so TVE'()*+,-- = 0 , TPE'()*+,-- = 0InternalEnergy(E, also denoted by U) :! Every system having some quantity of matter is associated with a definite amount of energy,called internal energy.! It is the sum of all forms of energies present in the system.E = ETranslational + ERotational + EVibrational + Ebonding + .....!E =EFinal .EInitial.!E = qv, heat supplied to a gas at constant volume, since all the heat supplied goes to increase theinternal energy of the gas .! Itisanextensiveproperty &astatefunction . Itisexclusively a function of temperature.If!T = 0 ;!E = 0as well.! With change in temperature only kinetic energy changes.! Degree of freedom # The total no of modes on which a molecule of an ideal gas can exchange energyduring collisons is known as its degrees of freedom.Translational degree of freedom = 3 - for all type of gases.Rotational degree of freedom= 0 monoatomic gases= 2 diatomic or linear polyatomic gases= 3 non-linear polyatomic gases.If f is initial degree of freedom for that gas.f = 3 for monoatomic = 5 for diatomic or linear polyatomic = 6 for non - linear polyatomicETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 7 ! Law of equipartion of energy :Energy equal to 21kT is associated with each degree of freedom per ideal gas moleculeWhere k is Boltzmann constantU/molecule = f x 21kT/ U/mole = 2f RT(k NA = R)& For n moles,U = 2fnRT only for ideal gas./!U = 2fnR!T ! Calculation of Heat (q)$ Heat is a path function and is generally calculated indirectly using IstLaw of thermodynamics$ First calculate !E and W & then q or heat can be calculated if heat capacity of any process isgiven to us.$ Heat capacity is a path function and different type of heat capacities are definedRemember heat capacity ofa substance is not fixed itis dependent on type of processwhich is beingperformed on that substance!Total Heat Capacity (CT)Heat required to raise the temperature of system by 1C under the given process is known as totalheat capacity.Mathematically, CT = dTdqTq0!! J/C$ It is extensive properties and path function. So, dq = CTdTon integrating q = 1dT CT! Molar heat capacity (C)Heat required to raise temperature of 1 mole of a subtance by 1CMathematically, C = ndTdqT nq0!! J mole1 K1So, dq = nCdTq = T nC nCdT ! 01C isintensive path function.Cp is molar heat capacity at constant pressureCV is molar heat capacity at constant volumeCp and CV are intensive but not a path function!Specific heat capacity (s) :Heat required to raise temperature of unit mass (generally 1 g) of a substance by 1C.S = mdTdqT mq0!! Jg1 K1So, dq = msdTq =T ms dT ms dq ! 0 01 1S is intensive path functionSP is specific heat capacity at constant pressureSV is specific heat capacity at constant volumeSP & SVare intensive but not a path functionETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 8Total heat capacity, molar heat capacity & specific heat capacity of a process on a substance are related asCT = nC = ms & C = MsWhere m weight of substanceM molar mass of substancen no. of moles of the substanceFor isothermal process C = 2For isobaric process C = CpFor isochoric process C = CvFor adiabatic process C = 0Heat capacity can have value from 2 to +2depending on the process.WORK DONE(w) :Energy that is transmitted from one system to another in such a way that difference of temperature isnot directly involved is known as work. It is a path function.Thisdefinitionis consistent with our understanding of work asdw = Fdx. The force F can arise fromelectrical, magnetic, gravitational & other sources.Units :Heat & work both are forms of energy . Hence, their units are units of energy. i.e.S3 system: Joules (J).Much data is available in the old units of calories (cal) as well.P V = (litre. atmosphere) term which has unit of energy . It is useful to remember the conversion1 litre. atm= 101.3 Joules = 24.206 cal$ For irreversible processes, state parameters such as P,Tetc cannot be defined. Hence, workcannot be estimated using Pgas. But by the work energy theoremWgas = Wext + !KpisstonWhen the piston comes to rest again !Kpiston = 0& Wgas = Wext = dv Pext1as the external pressure is always defined hence, for all processes work can be calculated usingWext =dv . Pext1Work : dW = Fext dxP-V Work :Assuming that under an external force Fx, the piston moves down by a distance dx.Fextdx) dx A (AFdWext0 / dW = Pext (dV)/ Wext = 1dV Pext & Wgas = Wext = 1dV PextAs the work done in the above case by the external agent is +ve and as the expression conveys otherwise,hence a -ve sign is introduced.& dW = Pext. dV! Calculation of work for different type of process on an ideal gas.(A) Isothermal expansion : There are many ways in which a gas can be expanded isothermally.(a) Isothermal reversible expansion :Pext = P0 + Amg = Pgas = P (always)In reversible process,Pext= Pgas (thermodynamic equilibrium always)Since process is isothermalETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 9P = VnRTW = 1fiVVextdV P = 1fiVVdV P /W=1fiVVdVVnRT W = nRT ln ''()**+,ifVV PV diagram RepresentationIn expansion work is done by system on the surroundings and Vf > ViW = veWork = Area under the PV diagram(b) Irreversible isothermal expansion :(i) Single step isothermal expansionWe are assuming expansion against atmospheric pressure which need not be the case in a given problem.A mass equal to m0 is placed on piston initially to maintain equilibrium.Initially,Pi = Pgas= Patm + m0 g/APV diagram RepresentationForexpansion totake place,m0massis suddenlyremovedsogasexpands againstconstantexternalpressure of PatmIn this case, the pressure of the gas will not be defined as the sudden expansion has taken place so all themolecules of sample will not get the information of expansion simultaneously, there will be a time gap andhence, there will be a state of turbulence.From some intermediate state of volume V, the work done is slight expansion fromV#" #(V + dV)dw = Pext . dV (IUPAC sign convention)So, W = 1 1. 0fiVVextdv . P dwW = Pext (Vf Vi)ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 10Only initial and final states can be located (as at intermediate stages pressure of the gas is not defined)(ii)Two step isothermal expansion :Mass M0 is divided into two mass (may be equal or unequal)) m m ( M2 1 04 0Now, if m1 only is removed, then the expansion of gas will take place against constant external pressure1extP = (Patm + m2g/A)and this expansion will take place only upto volume V1 such that1extP. V1 = Pi Vi (isothermal)Now, if second mass m2 is also removed then expansionV1 #" #Vf will take place against constant pressureSo,2extP = Patmwork done is expansion Vi#" # V1W1 = (Patm +M2g/A) (Vi V1)& work done is expansion V1#" # VfW2 = Patm (Vf Vi)Total work = W1 + W2PatmVVfM0ViM1 PV Diagram representation! Work done in this irreversible expansion is greater than work done by gas during the single stage expansionof gas and so on for three step expansion we divide the mass m0 into three masses m1, m2and m3 andremove these step by step and so on.(iii) For n step expansion and n#" # 2Irreversible process becomes PV Diagram representationETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 11(B) isothermal compression of an ideal gas :(a) Reversible isothermal compression of an ideal gasThis can be achieved by placing particles of sand one by one at a very slow take in the assembly whichkeeps the temperature of gas constant in this case the expression of work done will be exactly similar to asobtained in case of reversible expansion of gasW = nRT ln(Vf/Vi)This will automatically come out to be +ve as Vf < ViPatmViVi(P + Mg/A) = Patm0 fVi VfPatm(b) Irreversible isothermal compression of an ideal gas(i) Single step compression :To compress gas a mass m0 is suddenly placed on massless pistondw = Pext. dv= (Patm + m0g/A) dvso. to calculate total work done on the gasW = 1 1. 0fiVVextdv . P dw; W = Pext (Vf Vi)(ii) Two step compression :Place mass m0 in two fragments (m1 + m2) the graphical representation will make the calculation ofwork done.If m1 is placed first, then the first compression has taken place aganist external pressure of (Patm + m1g/A)So, W1 = (Patm + m1 g/A) (V1 Vi)Simlarly, W2 = (Patm + m0 g/A) (Vf Vi)Note : If process takes place in n steps and n " 2then process will be like reversible compressionETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 12Conclusion :Whenever work is done on the gas then it will be minimum in case of reversible process.Thats why different machines/engines are designd to work reversibly so maximum output can be obtainedbut minimum input is given to it output work done by engine/machine/systeminput work done by system us a surrounding# If expansion/compression takes place against constant external pressure then it is irreversible.# If there is sudden change then the process is irreversible.Reversible and irreversible isothermal process.! Except the infinite stage compression/expansion, all are irreversible.! We can redefine reversible and irreversible as follows :Reverssible process : If a process operates is such a fashion that when it is reversed BAback both the system as well as surroundings are restored to their initial position w.r.t. both work and heat,is known as reversible process.If for the process A " Bwork = w, heat = Q then if for the process B " A, work = w, heat = Q then theprocess is reversible.! If the external pressure is constant in isothermally process, process is irreversibleCalculation of Cp and CV(a) Constant volume process (Isochoric)dU = dq + dwdU = dq pdV& dU = (dq)vHeat given at constant volume = change in internal energydu = (dq)& dU = (nCdT)vdU = nCvdTCv is Specific molar heat capacity at constant volume.Cv =dTdU.n1 0 dT2 / ) fnRT ( dn1 = 2fR(b) Constant pressure process (Isobaric) :dU = dQ + dWdU = dQ PdV/ dQ = dU + PdV ...............(i)Defining a new thermodynamic functionH5Enthalpy! It is a state function and extensive property! It is mathematically defined as :H = U + PVas dH = dU + d (Pv)as P = constantdH = dU + PdV ..............(ii)from equation (i) & (ii)& dH = (dq)ponly at constant pressure.Heat given at constant pressure = Change in enthalpydH = (nCdT)pdH = nCp dT ! Relation between Cp ad Cv for an ideal gasH = U + PV& dH = dU + d(PV)for an ideal gasPV = nRTd(PV) = d(nRT) = nRdTnCp dT = nCvdT + nRdT/ Cp Cv = R only for ideal gas/ Mayers RelationshipETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 13ADIABATIC PROCESS :dQ = 0 (no heat changed b/w system and surroundingdU = dQ + dW/ nCV dT = PdV /1 1. 0 dv .VnRTdT ncV/ 121TTvTdT . C= 121VVdT .VR/ Cv ln 12TT= P ln 12VVln 12TT= ln vc / R12VV.''()**+,/12112VVTT. 6''()**+,0''()**+,T2 V2 61 =T1 V1 61 or T V61 = constantPV6 = constant! This is only valid when the quantity PV6 orTV61 is constant only for a quasi-static or reversibleprocess.! For irreversible adiabatic process these equations are not applicable.Operation of adiabatic process(a) Reversible Adiabatic! Operation wise adiabatic process and isothermal process are similar hence all the criteria that isused for judging an isothermal irreversible processes are applicable to adiabatic process.! Also, volume in case of isothermal volume is more than that of adiabatic at constant pressure and noof moles, V7 Tw = 1dv P. ext, but Pext = Pint = 6VK& w = 1 6dv .VK, / W = K 8 9: ; 6 ..4 6 . 4 6 .1V V1112= 1V . V P V . V P11 1 112 2 2. 6 .6 . 6 6 . 6/work done =1V P V P1 1 2 2. 6 .(as K = P2V26 = P1V16 )ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 14(b) IrreversibleAdiabatic! Adiabatic irreversible expansion W = dv P. ext1.= Pext (V2 V1) , and 1 10 dw du& W = !uW = nCv (T2 T1) = 1) T T ( nR1 2. 6. = 1) V P V P1 1 2 2. 6.Note: If two states A and B are connected by a reversible path then they can never be connected by anirreversible path. If the two states are linked by an adiabatic reversible and irreversible path thenwrev. = !urev.But as u is a state function& !urev.= !uirrev./ wirrev. = wrev.as work is a path function.If we assume thatwirrev. = wrev./ It implies that!urev. < !uirrev.. which again is a contradiction as U is a state function.! Two states A and B can never lie both on a reversible as will as irreversible adiabatic path.! There lies only one unique adiabatic path linkage between two states A and B. ! Comparison of Adiabatic Expansion (single stage Vs Infinite stage)Single stage means irreversible processInfinite stage means reversible processIn adiabatic expansion process,(Wgas)rev >(Wgas)irrev/ !urev. > !uirrev.& (T2)rev.< (T2) irrev(P2)rev.< (P2) irrev(If volume change are same)(V2)rev.< (V2) irrev(If pressure change are same) ! Comparison of Adiabatic Expansion (single stage Vs Infinite stage)Single stage means irreversible processInfinite stage means reversible processIn adiabatic compression process,(Wgas)rev work done inone stage by the gas. So, !Utwo stage > !Usingle stage&Tfintwostagedecrease in internal energy in one stage. ! Adiabatic Irreversible process (calculation of state parameters)State A # #" #irrevState BP1,V1,T1 P2,V2,T2W = 1V P V P1) T T ( nR1 1 2 2 1 2. 6 .0. 6. = Pext. (V2 V1) , 11 1TV P = 22 2TV P! Free expansion Always going to be irrerversible and since Pext = 0So, dW = Pext . dV= 0If no heat is supplied q = 0 then !E = 0 So, !T = 0.! Calculation of "H, "U, work, heat etc.Case - I For an ideal gas undergoing a process.the formula to be used aredU = nCvdT = 2fnRdTdH = nCpdT ='()*+,412f nRdTW = 1dV . PextdH = dU + d (PV)!H = !U + nR !Tdu = dQ + dwETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 16Case - II For solids and liquid system :dU= nCvdT 0 / process is non spontaneous(iii) If !G system= 0 / system is at equilibriumSo, at every temparature !G < 0!H system!S system!G system = !H system T!S system+ Ve Ve + Ve + Ve + Ve + Ve Ve at high temprature Ve Ve + Ve at high temprature"G = standrad free energy change :When the reactants under standrad conditions gets converted into products which is also under standeredcondition, then the free energy change is known as !G (it is a constant) for 1 mole at 1 bar.ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 24At standared conditions :!G = !H T!S = constant for a GIVEN REACTIOND A4 EB #" # 6C +F D!G = (Gbbs energy of product) (Gibbs energy of reactant)!G = 60C , mG + 0D , mG F 0A , mG D 0B , mG E ,where 0mG cannotbecalculatedas 0mH cannotbecalculated.hence, we can convert this relation in to Gibbs energy of formation of substance.!G = 0C , fG ! 6+ 0D , fG ! F 0A , fG ! D 0B , fG ! E0fG !: standered Gibbs energy of formation.0fG !(elements in their standard states) = 0!G = 0product , fG ! 0ts tan reac , fG !For a reaction in progress!G = !G +RT !n QQ is reaction quotient , !G > 0 backward is feasible , !G < 0 forward is feasibleAt equilibrium !G = 0 & Q = K & !G = RT !n K at equilibriumTHERMOCHEMISTRY ! Enthalpy of a substance :! Every substance has a fixed value of enthalpy under any particular state. Though, its exact value cannot becalculated but it has some finite fixed value.! The enthalpy of one mole of a substance called molar enthalpy in any particular state can be assigned symbol! Hm(substance) :For example molar enthalpy of water vapours at 398 K and 1 atm pressure may be representedas Hm (H2O, g, 398 K, 1 atm). In very simple works, enthalpy can be considered as heat constant (amount)of substance, and during reaction this heat is being released or absorbed.! Molar enthalpy of substance under standard conditions is called standard molar enthalpy of a substance.Standard state of any substance means.! For a GAS standard state means ideal gas at 1 bar partial pressure at any give temperature.! For a LIQUID pure liquid at one bar pressure at 1 bar pressure at any given temperature.! For a PURE CRYSTALLINE SOLID pure crystalline solid at 1 bar pressure and at any given temperature! For any SUBSTANCE orIONIN SOLUTION the species should be in unit molality (can also be takenas 1M concentration), at one bar pressure and at any given temperature.! Molar standard enthalpy of water vapours at 398 K will be represented as H(H2O, g , 398 K) and molarstandard enthalpy of liquid water at 398 K will be represented as Hm (H2O, l, 398 K)(It is hypothetical but can be calculated).! We cannot exactly calculate enthalpy content of a substance only the change in enthalpy can be calculatedwhen substance is taken from one state to other.Forexample: Let enthalpy contant initially be 01 , mH& finally enthalpy content be 02 , mHThen, !H = 02 , mH 01 , mH= heat added at constant pressure to change temperature from 25C to 50C.= CP!T = (18 cal/mole C) (25C) = 450 calETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 25 ! Enthalpy of formation :It is not possible to determine the absolute value of the molar enthalpy of a substance. However, based onthe following convention, the relative values of standard molar enthalpies of formation of various substancescan be built.!The standard enthalpy of formation of every element in its stable state of aggregation at one bar pressureand at specified temperature is assigned a zero value.The specified temperature is usually taken as 25 C.A few exmaples are !Hf (O2 , g) = 0!Hf (C, graphite) = 0 !Hf (C, diamond) < 0!Hf (Br2 , !) = 0!Hf (S, rhombic) = 0 !Hf (S, monoclinic) < 0!Hf (P, white) = 0 !Hf (P, black) < 0 ! The standard enthalpy of formation of a compound is the change in the standard enthalpy when one moleof the compound is formed starting from the requisite amounts of elements in their stable state of aggregation.The chemical equations corresponding to enthalpy of formation of few substances are given below.Enthalpy of formation of HBr(g) : 21 H2 (g) + 21 Br2 (l) " HBr(g)!Hf(HBr, g) = C GB 0mH(B) =0mH (HBr, g) 210mH (H2, g) 210mH(Br2, 1) ...(1)Enthalpy of formation of SO2 (g) : S (rhombic) + O2 (g) " SO2 (g)!Hf (SO2, g) = 0mH(SO2, g) 0mH (S, rhombic) 0mH (O2, g) ...(2)But above equations cannot be for calculation of enthalpy of reaction as the molar enthalpies of differentspecies can not be exactly known.!Enthalpy of Reaction from Enthalpies of Formation:The enthalpy of reaction can be calculated by!Hr = C GB !Hf,products C GB !Hf,reactants GB is the stoichiometric coefficientabove equation holds good for any reaction asthe same reference state is used for reactantsand products (shown in figure).Hesss Law of constant heat summation :! The heat absorbed or evolved in a given chemical equation is the same whether the process occurs in onestep or several steps.! The chemical equation can be treated as ordinary algebraic expressions and can be added or subtracted toyield the required equation. The corresponding enthalpies of reactions are also manipulated in the same wayso as to give the enthalpy of reaction for the desired chemical equation.! Since !rH stands for the change of enthalpy when reactants (substances on the left hand side of the arrow)are converted into products (substances on the right hand side of the arrow) at the same temperature andpressure, if the reaction is reversed (i.e., products are written on the left hand side and reactants on the righthand side), then the numerical value of !rH remains the same, but its sign changes.! The utility of Hesss law is considerable. In almost all the thermochemical numericals, Hesss law is used.! One of the important applications of Hesss law is to determine enthalpy of reaction which is difficult todetermine experimentally. For example, the value !rH for the reactionC(graphite) + 21O2 (g)#" #CO(g)which is difficult to determine experimentally, can be estimated from the following two reactions for which !rHcan be determined experimentally.C(graphite) + O2(g)#" #CO2(g) !rH1CO(g) + 21O2(g)#" #CO2(g) !rH2Substracting the latter from the former, we getC(graphite) + 21O2(g) #" #CO(g)Consequently, !rH = !rH1 !rH2ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 26Enthalpy of Combustion :! It is the enthalpy change when one mole of a compound combines with the requisite amount of oxygen togive products in their stable forms.For example, the standard enthalpy of combustion of methane at 298 K is 890 kJ mol1. This implies thefollowing reaction :CH4 (g) + 2O2(g)#" #CO2(g) + 2H2O (!) !H = 890 kJ mol1The standard enthalpy of combustion of methane at 298 K may be written as!Hc (CH4, g, 298 K) = 890 kJ mol1! The data on the enthalpy of combustion can be determined experimentally.! With the help of such data, we can determine the enthalpy of formation of a compound, which otherwise isdifficult or impossible to determine experimentally. Consider for example, the enthalpy of formation of CH4(g):C(graphite) + 2H2(g)#" #CH4(g)First of all, the combination of carbon and hydrogen does not occur readily. Secondly, if the reaction is evencompleted, the end product would not be pure methane. Therefore, the enthalpy of formation of methane canbe determined indirectly through the enthalpy of combustion of methane :CH4(g) + 2O2(g)#" #CO2(g) + 2H2O(!)!HC(CH4, g) = !Hf(CO2, g) + 2!Hf(H2O , !) !Hf (CH4, g)therefore!Hf(CH4, g) = !Hf(CO2, g) + 2!Hf(H2O , !) !HC (CH4, g)! The enthalpies of formation of CO2 and H2O can be determined experimentally by the combustion of carbon(graphite) and hydrogen. Thus, knowing the mesured value of !HC(CH4, g), the enthalpy of formation of CH4can be calculated. The value is!Hf(CH4, g) = !Hf(CO2, g) + 2 !Hf(H2O , !) !HC (CH4, g)= [ 393 + 2 (285) (890)] kJ mol1 = 73 kJ mol1or, equivalently, we may add the following three chemical equations.C(graphite) + O2 (g)#" #CO2 (g) !CH = 393 kJ mol12 [H2(g) + 21O2 (g)#" #H2O(!)] !rH = 2( 285) kJ mol1 [CH4(g) + 2O2 (g)#" #CO2 (g) + 2H2O(!)] !rH = ( 890) kJ mol1 C(graphite) + 2H2(g)#" #CH4 (g) !fH = 73 kJ mol1Measurement of Enthalpy of Combustion :Enthalpy of combustion are usually measured by placing a known mass of the compound in a closed steelcontainer (known as bomb calorimeter) which is filled with oxygen at about 30 bar pressure.

The calorimeter is surrounded by a knownmass of water. The entire apparatus is kept in aninsulated jacket to prevent heat entering into orleaving from the container, as shown in figure.Thesample is ignited electrically to bring aboutthecombustion reaction. The heat evolved is usedinraisingthetemperatureofwaterandthecalorimeter.If total heat capacity of calorimeter and all of itscontents = C, rise in temperature = !Tthenheatreleased=q=C!Tofthisheatisbecauseof mass m of substance thendue to 1 mole, heat released ='()*+,mMq = !EC(constant volume reaction).Now, !HC can be calculated by using !HC = !EC + !ng RT.Where !ng is the change in stoichiometric number of gaseous species in the balanced chemical equationrepresenting the combustion process.ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 27Integral enthalpy of solution :The integral enthalpy of solution at the given concentration is the enthalpy change when one mole ofthe solute is dissolved in a definite quantity of solvent to produce a solution of a desired concentration.While recording integral enthalpies of solution it is a general practice to state the amount of thesolvent in which 1 mole of solute is dissolved ; ThusHCl(g) + 10H2O(!)#" #HCl (10H2O) !H1 = 69.5 kJ mol1indicates that when 1 mole of hydrogen chloride gas is dissolved in 10 mol of water, there is anevolution of 69.5 kJ of heat. Other values are(i) HCl(g) + 25 H2O(!)#" #HCl (25H2O) !H2 = 72.3 kJ mol1(ii) HCl(g) + 40 H2O(!)#" #HCl (40H2O) !H3 = 73.0 kJ mol1(iii) HCl(g) + 200 H2O(!)#" #HCl (200H2O) !H4 = 74.2 kJ mol1(iv) HCl(g) + aq#" #HCl (aq) !H5 = 75.0 kJ mol1Whenever amount of solvent is not specified then take its amount to be very large just likein equation no. (iv).Enthalpy of Hydration :Enthalpy of hydration is used in following to ways.! Enthalpy of hydration of anhydrous or partially hydrated salts :Enthalpy of hydration of a given anhydrous or partially hydrated salt is the enthalpy change when it combineswith the requisite amount of water to form a new hydrated stable salt.For example, the hydration of anhydrous cupric sulphate is represented byCuSO4 (s) + 5H2O (!) " CuSO4 . 5 H2O(s)There is a almost invariably a liberation of heat in such reactions, i.e. the value of !H is negative.CuSO4(s) + 800 H2O (!) " CuSO4 (800 H2O) !Hr = 68 kJ mol1CuSO4 . 5H2O (s) + 795 H2O (!) " CuSO4 (800 H2O) !Hr = + 10 kJ mol1by subtraction, we get CuSO4(s) + 5H2O (!) " CuSO4 . 5 H2O(s)!Hr = 78 kJ mol1! Enthalpy of hydration of gaseous ions.Enthalpy of hydration of any gaseous ion is the enthalpy change when 1 mole of the gaseous ion is hydratedin large amount of water to form aqeous ion.By convention, the standard enthalpy of formation of H+(aq) is taken to be zero.Enthalpy of hydration of Cl gaseous ions will be represented by :Cl(g) + aq.#" #Cl(aq) !Hr = !Hf (Cl, aq)Enthalpy of Neutralization :! The amount of heat released when one gram equivalent of an acid is neutralised by one gram equivalent of abase.or! The amount of heat released in formation of one mole of water when an acid is neutralised by a base.or! Enthalpy of neutralization is defined as the enthalpy change when one mole of H+ in dilute solution combineswith one mole of OH to give rise to undissociated water, i.e.H+(aq) + OH(aq)#" #H2O(!) !H= 57.1 kJ/mole = 13.7 kcal/molRemember :! For Strong Acid + Strong Base, heat of neutralisation is always equal to 13.7 kcal/mole or 57.1 kJ/mole.! For any other combination of acid and base this heat is less than 13.7 kcal/mole or 57.1 kJ/mole.Enthalpy of Ionization :! Whenever a weak acid (or base) reacts with a strong base (or acid), the release of heat is less than 57.1 kJmol1.! It is because of the fact that these acids or bases are not completely ionized in solution. Some of the heat isconsumed in ionizing there acids and bases this heat is known as enthalpy of ionization.Examples are :HCN + Na+ OH " Na+ + CN + H2O !rH = 12 kJ mol1CH3COOH + Na+OH" Na+ + CH3COO + H2O !rH = 49kJ mol1The enthalpy of ionization can be calculated as follows. The neutralization of a weak acid, say HCN, may berepresented in two steps, namely,Ionization HCN#" #H+ + CN !H1 = xNeutralization H+ + OH#" #H2O !H2 = 57.1 kJ/moleETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 28The complete reaction is obtained by adding the above two steps. ThusHCN + OH#" #H2O + CN !H = 12 kJ/moleObviously, !H = !H1 + !H2!H1 =!H . !H2 = [12 (57.1)] = 45.1 kJ/mole! Greater the enthalpy of ionization of any weak acid or weak base, weaker will be the acid or base.Enthalpy of Transition :Enthalpy of transition is the enthalpy change when one mole of one allotropic form changes to another.For example : C(graphite) " C(diamond) !Htrs0 = 1.90 kJ mol1so if C(graphite) + O2(g) " CO2(g) !HC0 = 393.51 kJ mol-1and C(diamond) + O2(g) " CO2(g) !HC0 = 395.41 kJ mol1Subtracting, we have, C(graphite) " C(diamond) !Htrs0 = 1.90 kJ mol1Enthalpy of Precipitation :Enthalpy of precipitation is the enthalpy change when one mole of a precipitate is formed.For example : BaCl2(aq.) + Na2SO4(aq) " BaSO4(s) + 2NaCl(aq) !rH0=24.27 kJ mol1Enthalpy of Formation of ions :The enthalpy change when one mole of hydrated ions is obtained from element in its standardstate as. Cl2 (g) + aq#" #Cl (aq)!Hr = !Hf (Cl, aq)By convention, the standard enthlpy of formation of H+(aq) is taken to be zero.We have seen that H+ (aq) + OH(aq) " H2O(l ) !rH0 = 57.1 kJ mol1For this reaction, !Hr0 = !Hf0 (H2O,l ) {!Hf0 (H+, aq) + !Hf0 (OH,aq)}Hence, at 25C, we get !Hf0 (H+, aq) + !Hf0 (OH ,aq) = !Hf0 (H2O, l ) !Hr0so !Hf0 (OH ,aq) = { 286.1 ( 57.1)} kJ mol1 = 229.00 kJ mol1! With the enthalpies of formation of these two ions, the enthalpy of formation of any other ion can be foundfrom the enthalpies of formation and solution of its pure compound with H+ or OH. for example, the enthalpyof formation of Na+ can be calculated from the enthalpy of formation and enthalpy of infinite dilute solution ofNaOH. The two values are :! The chemical equation for the formation of infinite dilute solution of NaOH(s) isNaOH(s) + nH2O(!)#" #Na+(aq) + OH(aq) !aqH(NaOH, s) = 44.50 kJ mol1Since there are equal amounts of water on both sides of the above equation, the two enthalpies give no neteffect and thus!aqH(NaOH, s) = !fH(Na+, aq) + !fH(OH, aq) !fH(NaOH, s)or !fH (Na+, aq) = !aqH (NaOH, s) !fH(OH, aq) + !fH(NaOH, s)= [44.50 (229.99) + (425.61)] kJ mol1 = 240.12 kJ mol1! Similarly, from NaCl(aq) or HCl(aq), the enthalpy of formation of Cl(aq) can be determined, and so on. Thechanges in enthalpy of any ionic reaction can then be found from these ionic enthalpies of formation and theusual enthalpies of formation of compounds.Bond Enthalpies :The bond enthalpy is the average of enthalpies required to dissociate the said bond present in differentgaseous compounds into free atoms or radicals in the gaseous state. While bond dissociation enthalpy isthe enthalpy required to dissociate a given bond of some specific compound.for example the enthalpy ofdissociation of the OH bond depends on the natureof molecular species from which the H atom is beingseparated. For example, in the water molecule.H2O(g) " H(g) + OH(g) !Hr0 = 501.87 kJ mol1However, to break the OH bond in the hydroxyl radical required a different quantity of heat :OH(g) " O(g) + H(g) !Hr0 = 423.38 kJ mol1The bond enthalpy, HOH, is defined as the average of these two values, that is :HOH = 2kJmol 38 . 423 mol 87 . 5011 1 . .4 = 462.625 kJ mol1Inthecaseofdiatomicmolecules,suchasH2,thebondenthalpyandbonddissociationenthalpyareidentical because each refers to the reaction.H2(g) " 2H(g) HH H = !Hr0 = 435.93 kJ mol1Thus,the bond enthalpy given for any particular pair of atoms is the average value of the dissociationenthaplies of the bond for a number of molecules in which the pair of atoms appears.ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 29Estimation of Enthalpy of a reaction from bond Enthalpies :Let the enthalpy change for the gaseous reactionC2H4(g) + HCl(g) " C2H5Cl(g) (g)be required from the bond enthalpy data. This may be calculated as follows :!H ='()*+,.'()*+,atoms gasesous the fromproducts form to released Enthalpyatomsgasesous intoreactants break to required Enthalpy= [4HCH + HC=C + HHCl] + [5HCH HCC HCCl]= (HC=C + HHCl) (HCH + HCC + HCCl)Resonance Energy :Difference between energy of resonance hybrid and resonating structure in whichresonancehybrid have lower energy because stabilised by resonance.!Hresonance = !Hf, experimental !Hf, calclulated= !Hcombustion, calclulated!Hcombustion, experimentalRelation between energy and enthalpy of a reaction :"rH = "rU + ("vg)RT where !vg is the change in the stoichiocmetric number of gaseous species in goingfrom reactants to products.It should be noted that while computing !vg of a reaction, only the stoichiometric numbers of gaseous iscounted and those of liquids and solids are completely ignored.Kirchoffs equation(Variation of !H with temprature)Since the enthalpy (or standard enthalpy) of a substance is dependent on state of the substance, value ofenthalpyofasubstancechangeswithtemperatureandhencetheenthalpychangeofreactionisalsodependent on temperaturte at which the reaction is being carried out. This change is enthalpy change (or !E,of reaction is carried out at constant volume) is represented by Kirchoffs Equations.DP + EQ #" # 6R + FSat temperature T1 let the standard enthalpy of reaction be !H1 , then!H1 = 6Hm (R,T1) + FHm (S, T1) DHm (P, T1) EHm (Q, T1)If the same reaction is carried out at temperature T2, then!H2 = 6Hm (R,T2) + FHm (S, T2) DHm (P, T2) EHm (Q, T2)Then, the change in enthalpy (or difference in enthalpy at these two temperatures)!H = !H2 !H1 = 6 {Hm (R,T2) Hm (R,T1)} +{Hm (F,T2) Hm (F,T1)} D {Hm (P,T2) Hm (P,T1)} - E {Hm (Q,T2) Hm (Q,T1)}Hm (R, T2) Hm (R,T1) = CP, R (T2 T1) = Heat required at constant pressure to increase temperature of onemole of R from T1 to T2SimilarlyHm (S,T2) Hm (S,T1) = CP,S (T2 T1)Hm (P,T2) Hm (P,T1) = CP,P (T2 T1) andHm (Q,T2) Hm (Q,T1) = CP,Q (T2 T1)so !H = !H2 !H1 = 6 CP,R (T2 T1) + FCP,S(T2 T1) D CP,P (T2 T1) ECP,Q(T2 T1)= [6 CP,R + FCP,S D CP,P ECP,Q] (T2 T1)= !CP (T2 T1)!CP = 6CP, R+ FCP,S DCP,P ECP,Q = Difference in molar heat capacities of products and reactants.so !H2 = !H1 + 1! dT . CPfor example for the reactionN2 (g) + 3H2 (g) #" # 2NH3 (g)!H2 = !H1 + !CP (T2 T1)where !CP = 2 2 3H , P N , P NH , PC 3 C C 2 . .* for a constant volume reaction1! 4 ! 0 ! dT . C E EV0102ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 30 *Marked Questions are having more than one correct option.1. Basic Terminology1.1 Warming ammonium chloride with sodium hydroxide in a test tube is an example of :(A) Closed system (B) Isolated system (C) Open system (D) None of these1.2 Out of boiling point (!), entropy (!!), pH (!!!) and e.m.f. of a cell (!V), intensive properties are :(A) !, !! (B) !, !!, !!! (C) !, !!!, !V (D) All of the above2.First law of thermodynamics, calculations of !E, w & q.2.1 What is "U for the process described by figure. Heat supplied during the process q = 100 kJ.(A) + 50 kJ (B) 50 kJ (C) 150 kJ (D) + 150 kJ2.2 Two mole of an ideal gas is heated at constant pressure of one atomosphere from 27C to 127C. If Cv,m = 20+ 102 T JK1 mol1, then q and "U for the process are respectively :(A) 6362.8 J, 4700 J (B) 3037.2 J, 4700 J (C) 7062.8, 5400 J (D) 3181.4 J, 2350 J2.3 One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27C. If thework done by the gas in the process is 3 kJ, the final temperature will be equal to (CV=20 J/K mol)(A) 100 K (B) 450 K (C) 150 K (D) 400 K2.4 What is the change in internal energy when a gas contracts from 377 ml to 177 ml under a constantpressure of 1520 torr, while at the same time being cooled by removing 124 J heat? [Take : (1 L atm) =100 J ](A) 24 J (B) 84 J (C) 164 J (D) 248 J2.5 A sample of liquid in a thermally insulated container (a calorimeter) is stirred for 2 hr. by a mechanical linkageto a motor in the surrounding, for this process :(A) w < 0; q = 0; "U = 0(B) w > 0; q > 0; "U > 0 (C) w < 0; q > 0; "U = 0 (D) w > 0; q = 0; "U > 02.6 An ideal gas is taken around the cycle ABCDA as shown in figure. The net work done during the cycleis equal to :B DCPP2P1V1V2V(A) zero (B) positive (C) negative (D) we cannot predict2.7 In the cyclic process shown in P-V diagram, the magnitude of the work done is : (A) 21 22P P#$%&'( )*(B) 21 22V V#$%&'( )*(C) 4* (P2 P1) (V2 V1) (D) * (V2 V1)2ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 312.8 10 mole of ideal gas expand isothermally and reversibly from a pressure of 10 atm to 1 atm at 300 K. Whatis the largest mass which can lifted through a height of 100 meter ?(A) 31842 kg (B) 58.55 kg (C) 342.58 kg (D) None of these2.9 A cyclic process ABCD is shown in PV diagram for an ideal gas. Which of the following diagram representsthe same process?(A)(B)(C)(D) 2.10 A gas expands adiabatically at constant pressure such that T + V1/2. The value of , (Cp,m/Cv,m) of the gas will be:(A) 1.30 (B) 1.50 (C) 1.70 (D) 22.11* P-V plot for two gases (assuming ideal) during adiabatic processes are given in the figure. Plot A and plot Bshould correspond respectively to :(A) He and H2(B) H2 and He (C) SO3and CO2(D) N2 and Ar2.12 A diatomic ideal gas initially at 273 K is given 100 cal heat due to which system did 209 J work.Molar heatcapacity (Cm) of gas for the process is :(A) 23R (B) 25R (C) 45R (D) 5 R2.13 One mole of an ideal monoatomic gas expanded irreversibly in two stage expansion.State-1 (8.0 bar, 4.0 litre, 300 K)State-2 (2.0 bar, 16 litre, 300 K)State-3 (1.0 bar, 32 litre, 300 K)Total heat absorbed by the gas in the process is :(A) 116 J (B) 40 J (C) 4000 J (D) None of these2.14 1 mole of NH3 gas at 27 C is expanded in reversible adiabatic condition to make volume 8 times (, = 1.33).Final temperature and work done respectively are :(A) 150 K, 900 cal (B) 150 K, 400 cal (C) 250 K, 1000 cal (D) 200 K, 800 cal2.15 For an ideal monoatomic gas during any process T = kV, find out the molar heat capacity of the gas duringthe process. (Assume vibrational degree of freedom to be active)(A) 25R (B) 3R (C) 27(D) 4 R2.16 1 mole of an idal gas A (C v,m = 3R) and 2 mole of an ideal gas B are #$%&'(. R23Cm , vtaken in acontainer andexpanded reversible and adiabatically from 1 litre to 4 litre starting from initial temperature of 320 K. "E or "Ufor the process is :(A) 240 R (B) 240 R (C) 480 R (D) 960 RETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 322.17 An insulated container of gas has two chambers separated by an insulating partition. One of the chambershas volume V1 and contains ideal gas at pressure P1 and temperature T1. The other chamber has volume V2and contains ideal gas at pressure P2 and temperature T2.If the partition is removed without doing any workon the gas, the final equilibrium temperature of the gas in the container will be(A) 1 2 2 2 1 12 2 1 1 2 1T V P T V P) V P V P ( T T/ /(B) 2 2 1 12 2 2 1 1 1V P V PT V P T V P//(C) 2 2 1 11 2 2 2 1 1V P V PT V P T V P//(D) 2 2 2 1 1 12 2 1 1 2 1T V P T V P) V P V P ( T T//2.18 An ideal gas is taken around the cycle ABCA as shown in P-V diagram.The net work done by the gas during the cycle is equal to :(A) 12P1V1(B) 6P1V1(C) 5P1V1(D) P1V12.19 What is the net work done (in calories) by 1 mole of monoatomic ideal gas in a process described by 1, 2,3, 4 in given VT graph.Use : R = 2cal / mole Kln 2 = 0.7(A) 600 cal (B) 660 cal (C) + 660 cal (D) + 600 cal2.20 A heat engine carries one mole of an ideal mono-atomic gas around thecycle as shown in the figure, the amount of heat added in the process ABand heat removed in the process CA are :(A) qAB = 450 R and qCA = 450 R(B) qAB = 450 R and qCA = 225 R(C) qAB = 450 R and qCA = 375 R(D) qAB = 375 R and qCA = 450 R2.21 Two moles of an ideal gas(CV = 25R) was compressed adiabatically against constant pressure of 2 atm.Which was initially at 350 K and 1 atm pressure. The work involve in the process is equal to(A) 250 R (B) 300 R (C) 400 R (D) 500 R2.22 Two moles of Helium gas undergo a reversible cyclic process as showin infigure. Assuming gas to be ideal, what is the net work involved in the cyclicprocess c?(A) 100 R!n4 (B) +100R!n4(C) +200R!n4 (D) 200R!n43. Calculation of !E, !H, w and q3.1 A gas (Cv,m = 25R) behaving ideally was allowed to expand reversibly and adiabatically from 1 litre to 32 litre.It's initial temperature was 327C. The molar enthalpy change (in J/mole) for the process is(A) 1125 R (B) 575 R (C) 1575 R (D) None of theseETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 333.2 One mole of an ideal gas #$%&'(. R25Cm , v at 300 K and 5 atm is expanded adiabatically to a final pressure of 2atm against a constant pressure of 2 atm. Final temperature of the gas is :(A) 270 K (B) 273 K (C)248.5 K (D) 200 K3.3 The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressureof 2 atm starting from intial pressure of 1 atm and initial temperature of 300 K (R = 2 cal/mol-degree)(A) 360 cal (B) 720 cal (C) 800 cal (D) 1000 cal3.4 One mole of non-ideal gas undergoes a change of state (1.0 atm, 3.0 L, 200 K) to (4.0 atm, 5.0 L, 250 K) witha change in internal energy ("U) = 40 L-atm. The change in enthalpy of the process in L-atm ;(A) 43 (B) 57 (C) 42 (D) None of these3.5* 0.5 mole each of two ideal gases A#$%&'(. R25Cm , vand B(Cv, m = 3R) are taken in a container and expandedreversibly and adiabatically, during this process temperature of gaseous mixture decreased from 350 K and250 K. Then for the process :(A) "U= 100 R (B) "U = 275 R (C) "H = 375 R (D) "H= 300 R3.6 50 L of a certain liquid is confined in a piston system at the external pressure 100 atm. This pressure issuddenly released and liquid is expanded against the constant atmospheric pressure, volume of the liquidincreases by 1 L and the final pressure on the liquid is 10 atm. Find the workdone.(A) 1L.atm (B) 5 L.atm (C) 500 L.atm(D) 50 L.atm3.7 A vessel contains 100 litres of a liquid X. Heat is supplied to the liquid in such a fashion that, Heat given =change in enthalpy. The volume of the liquid increases by 2 litres. If the external pressure is one atm, and202.6 Joules of heat were supplied then, [U - total internal energy](A) "U = 0 ,"H = 0 (B) "U = + 202. 6J , "H =+ 202.6 J(C) "U = 202.6J, "H = 202.6J (D) "U = 0,"H = + 202.6J3.8 For the real gases reaction 2CO (g) + O2 (g)01 02CO2 (g) ; "H = 560 kJ. In 10 litre rigid vessel at 500K, the initial pressure is 70 bar and after the reaction it becomes 40 bar. The change in internal energy is :(A) 557 kJ (B) 530 kJ (C) 563 kJ (D) None of these4.Thermochemistry4.1 For which of the following change "H 2 "E?(A) H2 (g) + I2 (g)01 2HI (g) (B) HCl (aq)+ NaOH(aq) 01 NaCl(aq) + H2O(l)(C) C(s) + O2(g) 01 CO2(g) (D) N2 (g)+ 3H2(g) 01 2NH3(g)4.2 In Haber's process of manufacturing of ammonia :N2(g) + 3H2(g)01 02NH3(g) ; 0C 25H3 = 92.2 kJMolecule N2(g) H2(g) NH3(g)CP JK-1 mol129.1 28.8 35.1If CP is independent of temperature, then reaction at 100C as compared to that of 25C will be :(A) More endothermic (B) Less endothermic (C) More exothermic (D) Less exothermic4.3* Which of the reaction defines molar "Hf?(A) CaO(s) + CO2(g) 01 0 CaCO3 (s) (B) 21 Br2 (!) + 21H2 (g) 01 0 HBr(g)(C) N2 (g) + 2H2 (g) + 23 O2 (g) 01 0 NH4 NO3 (s) (D) 21!2 (s) + 21H2 (g) 01 0 H! (g)4.4 For the allotropic change represented by the equation C (graphite) 01 C (diamond), "H = 1.9 kJ. If 6 g ofdiamond and 6 g of graphite are separately burnt to yield CO2, the enthalpy liberated in first case is(A) less than in the second case by 1.9 kJ (B) more than in the second case by 11.4 kJ(C) more than in the second case by 0.95 kJ (D) less than in the second case by 11.4 kJETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 344.5 In the reaction, CO2(g) + H2(g) 1 CO(g) + H2O(g); "H = 2.8 kJ, "H represents(A) heat of reaction (B) heat of combustion (C) heat of formation (D) heat of solution4.6 NH3(g) + 3Cl2(g) ! NCl3(g) + 3HCl (g); "H1N2(g) + 3H2(g) ! 2NH3(g); "H2H2(g) + Cl2(g) ! 2HCl (g) ; "H3The enthalpy of formation of NCl3 (g) in the terms of "H1, "H2 and "H3 is(A) "Hf = "H1 + 232H2)""H3(B) "Hf ="H1 + 232H2)""H3(C) "Hf = "H1 232H2)""H3(D) None4.7 Given, H2(g) +Br2(g)01 02HBr(g),"H01 andstandard enthalpy ofcondensation of bromineis "H02,standard enthalpy of formation of HBr at 250C is(A) "H01 / 2 (B) "H01 / 2 + "H02(C) "H01 / 2 ) "H02(D) ("H01)"H02) / 24.8 For the following reaction, C (diamond)+ O2 01 0 CO2(g); "H = 94.3 kcalC (graphite) + O2 01 0 CO2(g); "H = 97.6 kcalThe heat required to change 1 g of C (diamond) 01 0 C (graphite) is(A) 1.59 kcal (B) 0.1375 kcal (C) 0.55 kcal (D) 0.275 kcal4.9 Hydrazine, a component of rocket fuel, undergoes combustion to yield N2 and H2O.N2H4 (l) + O2 (g) 01 N2 (g) + 2H2O (l)What is the enthalpy combustion of N2H4 (kJ/mole)Given Reaction !H/kJ2NH3 (g) + 3N2O (g) 01 4N2 (g) + 3H2O (l) 1011 kJN2O (g) + 3H2 (g) 01 N2H4 (l) + H2O (l) 317 kJ4NH3 (g) + O2 (g) 01 2N2H4 (l) + 2H2O (l) 286 kJH2 (g) + 21O2 (g) 01 H2O (l) 285 kJ(A) 620.5 (B) 622.75 (C) 1167.5 (D) + 622.754.10 C (s) + O2 (g)01 0 CO2, (g); "H = 94.3 kcal/molCO (g) + 21O2(g)01 0 CO2 (g); "H = 67.4 kcal/mo!O2(g) 01 0 2O (g); "H = 117.4 kcal/molCO (g) 01 0 C (g) + O(g) ; "H = 230.6 kcal/molCalculate "H for C (s) 01 0 C (g) in kcal/mol.(A) 171 (B)154 (C)117 (D)1454.11 Find "rU for the reaction 4HCl (g) + O2 (g) ! 2Cl2(g) + 2H2O (g) at 300 K. Assume all gases are ideal.Given: H2(g) + Cl2(g) 01 2HCl (g)o300 rH " = 184.5 kJ/mole2H2(g) + O2(g) 01 2H2O (g)o300 rH " = 483 kJ/mole (Use R = 8.3 J/mole)(A) 111.5 kJ/mole (B) 109.01 kJ/mole (C) 111.5 kJ/mole (D) None4.12 The difference between "H and "E (on a molar basis) for the combustion of n-octane (!) at 25C would be :(A) 13.6 kJ (B) 1.14 kJ (C) 11.15 kJ (D) + 11.15 kJ4.13 The standard heat of combustion of solid boron is equal to :(A) "Hf (B2O3) (B) 1/2 "Hf (B2O3) (C) 2"Hf (B2O3) (D) 1/2 "Hf (B2O3)ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 354.14 The molar heat capacities at constant pressure (assume constant with respect to temperature) of A, B andCareinratioof1.5:3.0:2.0.Ifenthalpychangefortheexothermicreaction A+2B013Cat300 K is 10 kJ/mol & Cp,m (B) is 300 J/mol then enthalpy change at 310 K is :(A) 8.5 kJ/mol (B) 8.5 kJ/mol (C) 11.5 kJ/mol (D) none of these4.15 From the following data of "H, of the following reactions,C(s) + 21O2 (g) 01 0 CO (g) "H = 110 kJC(s) + H2O (g) 01 0 CO (g) + H2(g) "H = 132 kJWhat is the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, tomaintain constant temperature :(A) 0.5 : 1 (B) 0.6 : 1 (C) 0.8 : 1 (D) 1 : 14.16 2 mole of zinc is dissolved in HCl at 25C. The work done in open vessel is :(A) 2.477 kJ (B) 4.955 kJ (C) 0.0489 kJ (D) None4.17 The lattice enthalpy of solid NaCl is 772 kJmol1 and enthalpy of solution is 2 kJmol1. If the hydrationenthalpy of Na+ & Cl ions are in the ratio of 3:2.5, what is the enthalpy of hydration of chloride ion?(A) 140 kJmol1(B) 350 kJmol1(C) 351.81 kJmol1(D) none4.18 If heat of dissociation of CHCl2COOH is 0.7 kcal/mole then "H for the reaction :CHCl2COOH + KOH 01 0 CHCl2COOK + H2O(A) 13kcal (B) + 13 kcal (C) 14.4 kcal (D) 13.7 kcal4.190fH " ofwateris285.8kJmol1.Ifenthalpyofneutralisationofmonoacidstrongbaseis57.3 kJ mol1, 0fH "of OH ion will be(A) 228.5 kJ mol1(B) 228.5 kJ mol1(C) 114.25 kJ mol1(D) 114.25 kJ mol14.20 A solution is 500 ml of 2 M KOH is added to 500 ml of 2 M HCl and the mixture is well shaken. The rise intemperature T1 is noted. The experiment is then repeated using 250 ml of each solution and rise in temperatureT2 is again noted. Assume all heat is taken by the solution(A) T1 = T2(B) T1 is 2 times as large as T2(C) T2 is twice of T1(D) T1 is 4 times as large as T24.210fH " ofwateris285.8kJmol1.Ifenthalpyofneutralisationofmonoacidstrongbaseis57.3 kJ mol1, 0fH "of OH ion will be(A) 228.5 kJ mol1(B) 228.5 kJ mol1(C) 114.25 kJ mol1(D) 114.25 kJ mol14.22 One mole of anhydrous MgCl2 dissolves in water and librates 25 cal/mol of heat. "Hhydration of MgCl2 = 30 cal/mol. Heat of dissolution of MgCl2.H2O is(A) +5 cal/mol (B) 5 cal/mol (C) 55 cal/mol (D) 55 cal/mol4.23 In the reaction CS2 (!)+ 3O2 (g) 01 0 CO2 (g) + 2SO2 (g)"H = 265 kcalThe enthalpies of formation of CO2 and SO2 are both negative and are in the ratio 4 : 3. The enthalpy offormation of CS2 is +26kcal/mol. Calculate the enthalpy of formation of SO2.(A) 90 kcal/mol (B) 52 kcal/mol (C) 78 kcal/mol (D) 71.7 kcal/mol4.24 Consider the "G and "H (kJ/mol) for the following oxides. Which oxide can be most easily decomposedto form the metal and oxygen gas ?(A) ZnO ("G = 318.4, "H = 348.3) (B) Cu2O ("G = 146.0, "H = 168.8)(C) HgO ("G = 58.5, "H = 90.8) (D) PbO ("G = 187.9, "H = 217.3)4.25 If "G = 177 K cal for (1) 2 Fe(s) + 23O2 (g) 01 0 Fe2O3 (s)and "G = 19 K cal for (2) 4 Fe2O3 (s) + Fe(s) 01 0 3 Fe3O4 (s)What is the Gibbs free energy of formation of Fe3O4(s) ?(A) + 229.6 kcal/mol (B) 242.3 kcal/mol (C) 727 kcal/mol (D) 229.6 kcal/molETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 365.Bond enthalpies5.1 The reaction CH4(g) + Cl2(g) 01 CH3Cl(g) + HCl(g) has "H = 25 kCal.BondBond EnergykCal4CCl 844HCl 1034CH x4ClCl yx : y = 9 : 5From the given data, what is the bond enthalpy of ClCl bond(A) 70 kCal (B) 80 kCal (C) 67.75 kCal (D) 57.75 kCal5.2 ThebonddissociationenergyofgaseousH2,Cl2andHClare104,58and103kcalmol1respectively. The enthalpy of formation for HCl gas will be(A) 44.0 kcal (B) 22.0 kcal (C) 22.0 kcal (D) 44.0 kcal5.3 If x1, x2 and x3 are enthalpies of HH, O=O and OH bonds respectively, and x4 is the enthalpy of vaporisationof water, estimate the standard enthalpy of combustion of hydrogen(A) x1+2x22x3+x4(B) x1+2x22x3x4(C) x1+2x2x3+x4(D) 2x3x12x2x45.4 The average OH bond energy in H2O with the help of following data(1) H2O(!) 01 0 H2O(g) ; "H = + 40.6 KJ mol1(2) 2H(g) 01 0 H2 (g) ; "H = 435.0 KJ mol1(3) O2(g) 01 0 2O(g) ; "H = + 489.6 KJ mol1(4) 2H2 (g) + O2 (g)01 0 2H2O(!) ; "H = 571.6 KJ mol1(A) 584.9 KJ mol1(B) 279.8 KJ mol1(C) 462.5 KJ mol1(D) 925 KJ mol15.5 What is the ratio of the enthalpy yield on combustion of hydrogen atoms to steam to the yield on combustionof an equal mass of hydrogen molecules to steam?Given : H2(g) + 21O2(g) 01H2O(g) "H = 242 kJB.E. (H H) = 436 kJ(A) 0.80 : 1 (B) 1 : 0.80 (C) 1.80 : 1 (D) 2.80 : 15.6 Heat of hydrogenation of ethene is x1 and that of benzene is x2. Hence, resonance energy is :(A) x1 x2(B) x1 + x2(C) 3x1 x2(D) x1 3x26. IInd Law of thermodynamics:Entropy calculation for different types ofphysical process on an ideal gas, solid and liquid6.1 Predict which of the following reaction (s) has a positive entropy change ?I. Ag+ (aq) + Cl (aq)01 0AgCl (s)II. NH4Cl (s)01 0NH3 (g) + HCl (g)III. 2NH3 (g)01 0 N2 (g) + 3H2 (g)(A) I and II (B) III (C) II and III (D) II6.2 When two mole of an ideal gas #$%&'(. R25Cm , pheated from 300 K to 600 K at constant pressure. The changein entropy of gas ("S) is :(A) 23R ln 2 (B) 23R ln 2 (C) 5R ln 2 (D) 25R ln 2ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 376.3 The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300C reversibly andisochorically(A) 23R ln#$%&'(200300(B) 25R ln#$%&'(273573(C) 3R ln#$%&'(473573(D) #$%&'(473573ln R236.4 When one mole of an ideal gas is comressed to half of its initial volume and simultaneously heated to twiceits initial temperature, the change in entropy of gas ("S) is :(A) Cp, m ln 2 (B) Cv, m ln 2 (C) R ln 2 (D) (Cv, m R) ln 26.5 The entropy change when two moles of ideal monoatomic gas is heated from 200 to 300C reversibly andisochorically ?(A) 23 R ln#$%&'(200300(B) 25R ln#$%&'(273573(C) 3R ln#$%&'(473573(D) 23 R ln#$%&'(4735736.6 If one mole of an ideal gas#$%&'(. R25Cm , p is expanded isothermally at 300 K until its volume is tripled, thenchange in entropy of gas is :(A) zero (B) infinity (C) 25R ln 3 (D) R ln 36.7 Two mole of an ideal gas is expanded irreversibly and isothermally at 37C until its volume is doubled and3.41 kJ heat is absorbed from surrounding. "Stotal (system + surrounding) is :(A) 0.52 J/K (B) 0.52 J/K (C) 22.52 J/K (D) 06.8 1 mole of an ideal gas at 25C is subjected to expand reversibly and adiabatically to ten times of its initialvolume. Calculate the change in entropy during expansion (in J k1 mol1)(A) 19.15 (B) 19.15 (C) 4.7 (D) zero6.9 One mole of an ideal diatomic gas (Cv = 5 cal) was transformed from initial 25C and 1 L to the state whentemperature is 100C and volume 10 L. The entropy change of the process can be expressed as (R = 2calories/mol/K)(A) 3 ln 373298 + 2 ln 10 (B) 5 ln 298373 + 2 ln 10(C) 7 ln 298373 + 2 ln 101(D) 5 ln 298373 + 2 ln 1016.10 What is the change in entropy when 2.5 mole of water is heated from 27C to 87C ? Assume that the heatcapacity is constant. (Cp,m (H2O) = 4.2 J/g-K ln (1.2) = 0.18)(A) 16.6 J/K (B) 9 J/K (C) 34.02 J/K (D) 1.89 J/K6.11 Calculate the total entropy change for the transition at 368 K of 1 mol of sulphur from the monoclinic to therhombic solid state and "H = 401.7 J mol1 for the transition. Assume the surroundings to be an ice-water.Both at 0C :(A) 1.09 JK1(B) 1.47 JK1(C) 0.38 JK1(D) None of these6.12 Calculate standard entropy change in the reactionFe2O3 (s) + 3H2 (g)01 02Fe (s) + 3H2O (!)Given : Sm (Fe2O3, S) = 87.4 , Sm (Fe, S) = 27.3, Sm (H2, g) = 130.7, Sm (H2O, !) = 69.9 JK1 mol1.(A) 212.5 JK1 mol1(B) 215.2 JK1 mol1(C) 120.9 JK1 mol1(D) None of these6.13 Given "rS = 266 and the listed [Sm values]calculate S for Fe3O4 (s) :4Fe3O4 (s) [..............] + O2 (g) [205]01 06Fe2O3 (s) [87](A) +111.1 (B) +122.4 (C) 145.75 (D) 248.25ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 386.14 C2H6 (g) + 3.5 O2 (g) 1 2CO2 (g) + 3H2O (g)"Svap (H2O, !) = x1 cal K-1 (boiling point = T1)"Hf (H2O, !) = x2"Hf (CO2) = x3"Hf (C2H6) = x4Hence, "H for the reaction is -(A) 2x3 + 3x2 x4(B) 2x3 + 3x2 x4 + 3x1T1(C) 2x3 + 3x2 x4 3x1T1(D) x1T1 + X2 + X3 x47. !G calculation, spontaneity of chemical reaction, significance of!G and IIIrd Law of thermodynamics7.1 For the reaction at 300 KA(g) + B(g) 01 C (g)"E = 3.0 kcal ; "S = 10.0 cal/Kvalue of "G is(A) 600 cal (B) 6600 cal (C) 6000 cal (D) None7.2* For isothermal expansion in case of an ideal gas :(A) "H = 0 (B) "E = 0 (C) "G = T."S (D) Tfinal = Tinitial7.3 For the gas - phase decomposition, PCl5 (g) PCl3 (g) + Cl2 (g) :(A) "H < 0, "S < 0 (B) "H > 0, "S > 0 (C) "H > 0, "S < 0 (D) "H < 0, "S > 07.4 If "Hvaporisation of substance X (l) (molar mass : 30 g/mol) is 300 J/g at it's boiling point 300 K, then molarentropy change for reversible condensation process is(A) 30 J/mol.K (B) 300 J/mol.K (C) 30 J/mol.K (D) None of these7.5 What is the free energy change ("G) when 1.0 mole of water at 100C and 1 atm pressure is converted intosteam at 100C and 1 atm pressure ?(A) 80 cal (B) 540 cal (C) 620 cal (D) Zero7.6 What can be concluded about the values of "H and "S from this graph?(A) "H > 0, "S > 0 (B) "H > 0, "S < 0(C) "H < 0, "S > 0 (D) "H < 0, "S < 07.7 The enthalpy change for a given reaction at 298 K is x J mol1 (x being positive). If the reaction occursspontaneously at 298 K, the entropy change at that temperature(A) can be negative but numerically larger than x/298(B) can be negative but numerically smaller than x/298(C) cannot be negative(D) cannot be positive7.8 The change in entropy of 2 moles of an ideal gas upon isothermal expansion at 243.6 K from 20 litre until thepressure becomes 1 atm, is :(A) 1.385 cal / K (B) 1.2 cal / K (C) 1.2 cal / K (D) 2.77 cal / K7.9 A reaction has "H = 33 kJ and "S = 58 KJ. This reaction would be :(A) spontaneous at all temperatures (B) non-spontaneous at all temperatures(C) spontaneous above a certain temperature (D) spontaneous below a certain temperatureETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 39Single Correct Answer Type:1. A piece of zinc at a temperature of 20C weighing 65.38 g is dropped into 180 g of boiling water (T = 100C).Thespecificheatofzincis0.4Jg1 C1 andthatofwateris4.2Jg1C1.Whatisthefinalcommontemperature reached by both the zinc and water ?(A) 97.3 C (B) 33.4 C (C) 80.1 C (D) 60.0C2. The heat capacity of liquid water is 75.6 J / mol.K, while the enthalpy of fusion of ice is 6.0 kJ/mol. What isthe smallest number of ice cubes at 0C, each containing 9.0 g of water, needed to cool 500 g of liquid waterfrom 20C to 0C?(A) 1 (B) 7 (C) 14 (D) None of these3. A certain mass of gas is expanded from (1L, 10 atm) to (4L, 5 atm) against a constant external pressure of1 atm. If initial temperature of gas is 300 K and the heat capacity of process is 50 J/C. Then the enthalpychange during the process is (1L atm ~ 100 J)(A) "H = 15 kJ (B) "H = 15.7 kJ (C) "H = 14.4 kJ (D) "H = 14.7 kJ4. A heating coil is immersed in a 100 g sample of H2O (l) at 1 atm and 100C in a closed vessel. In this heatingprocess, 60% of the liquid is converted to the gaseous form at constant pressure of 1 atm. The densities ofliquid and gaseous water under these conditions are 1000 kg/m3 and 0.60 kg/ m3 respectively. Magnitude ofthe work done for the process is :(A) 4997 J (B) 4970 J (C) 9994 J (D) None of these5. 10 litres of a monoatomic ideal gas at 0C and 10 atm pressure is suddenly released to 1 atm pressure and thegas expands adiabatically against this constant pressure. The final temperature and volume of the gas respectivelyare.(A) T = 174.9 K, V = 64 L (B) T = 153 K, V = 57 L(C) T = 165.4 K, V = 78.8 L (D) T = 161.2 K, V = 68.3 L6. Consider a classroom that is roughly 5 m 10m 3m. Initially t = 27C and P = 1 atm. There are 50 peoplein an insulated class loosing energy to the room at the average rate of 150 watt per person. How long canthey remain in class if the body temperature is 42C and person feels uncomfortable above this temperature.Heat capacity of air = (7/2) R.(A) 4.34 minutes (B) 5.91 minutes (C) 6.86 minutes (D) 7.79 minutes7. At 5 105 bar pressure, density of diamond and graphite are3 g/cc and 2 g/cc respectively, at certaintemperature T. Find the value of "U "H for the conversion of 1 mole of graphite to 1 mole of diamond attemperature T :(A) 100 kJ/mol (B) 50 kJ/mol (C) 100 kJ/mol (D) None of these8. A new flurocarbon of molar mass 102 g mol)1 was placed in an electrically heated vessel. When the pressurewas 650 torr, the liquid boiled at 770C. After the boiling point had been reached, it was found that a current of0.25 A from a 12.0 volt supply passed for 600 sec vaporises 1.8g of the sample. The molar enthalpy & internalenergy of vaporisation of new flourocarbon will be :(A) "H = 102 kJ/mol, "E = 99.1 kJ/mol (B) "H = 95 kJ/mol, "E = 100.3 kJ/mol(C) "H = 107 kJ/mol, "E = 105.1 kJ/mol (D) "H = 92.7 kJ/mol, "E = 97.4 kJ/molETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 409. Two rigid adiabatic vessels A and B which initially, contain two gases at different temperatures are connectedby pipe line with valve ofnegligible volume. The vessel 'A' contain 2 moles Ne gas #$%&'(. R25Cm , pat 300 K,vessel 'B' contain 3 moles of SO2 gas (Cp,m = 4 R) at 400 K.The volume of A & B vessel is 4 and 6 litre respectively.The final total pressure (in atm) when valve is opened and 12 Kcal heat supplied through it to vessels.[Use : R = 2 cal / mol. K and R = 0.08 L . atm / mol K as per desire]Ne SO2AB(A) 3.5 atm (B) 7 atm (C) 35 atm (D) 70 atm10. The maximum efficiency of a heat engine operating between 100C and 25C is(A) 20.11% (B) 22.2% (C) 25.17% (D) None11. A heat engine operating between 227C and 27C absorbs 2 Kcal of heat from the 227Creservoir reversiblyper cycle. The amount of work done in one cycle is(A) 0.4 Kcal (B) 0.8 Kcal (C) 4 Kcal (D) 8 Kcal12. A reversible heat engine A (based on carnot cycle) absorbs heat from a reservoir at 1000K and rejects heatto a reservoir at T2. A second reversible engine B absorbs, the same amount of heat as rejected by the engineA, from the reservoir at T2 and rejects energy to a reservoir at 360K.If the efficiencies of engines A and B are the same then the temperature T2 is(A) 680 K (B) 640 K (C) 600 K (D) 550 K13. Pressure of 10 moles of an ideal gas is changed from 2 atm to 1 atm against constant external pressurewithout change in temperature. If surrounding temperature (300 K) and pressure (1 atm) always remainsconstant then calculate total entropy change ("Ssystem + "Ssurrounding) for given process.[Given : !n2 = 0.70 and R = 8.0 J/mol/K](A) 56 J/K (B) 14 J/K (C) 16 J/K (D) None of these14. For a perfectly crystalline solid Cp,m = aT3 + bT, where a and b constant. If Cp,m is 0.40 J/K mol at 10 K and0.92 J/K mol at 20 K, then molar entropy at 20 K is :(A) 0.92 J/K mol (B) 8.66 J/K mol (C) 0.813 J/K mol (D) None of these15. When two equal sized pieces of the same metal at different temperatures Th (hot piece) and Tc(cold piece)are brought into contact into thermal contact and isolated from it's surrounding. The total change in entropyof system is given by(A) C lnch cT 2T T/(B) C ln12TT(C) C lnc h2h cT . T 2) T T ( /(D) C lnc h2h cT . T 4) T T ( /16. Reactions involving gold have been of particular interest to a chemist . Consider the following reactions,Au(OH)3 + 4 HCl 01 HAuCl4 + 3 H2O , "H = ) 28 kCalAu(OH)3 + 4 HBr 01HAuBr4 + 3 H2O , "H = ) 36.8 kCalInanexperimenttherewasanabsorptionof0.44kCalwhenonemoleofHAuBr4 wasmixedwith4 moles of HCl . What is the percentage conversion ofHAuBr4 into HAuCl4 ?(A) 0.5 % (B) 0.6 % (C) 5 % (D) 50 %ETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 4117. Ethanol can undergoes decomposition to form two sets of productsC2H5OH (g) 01 if the molar ratio of C2H4 to CH3CHO is 8 : 1 in a set of product gases, then the enthalpy involved in thedecomposition of 1 mole of ethanol is(A) 65.98 kJ (B) 48.137 kJ (C) 48.46 kJ (D) 57.22 kJ18. (i) Cis)2 ) butene 1trans ) 2 ) butene, "H1(ii) Cis ) 2) butene 11 ) butene, "H2(iii) Trans ) 2 ) butene is more stable than cis ) 2 ) butene.(iv) Enthalpy of combustion of 1)butene, "H = )649.8 kCal/mol(v) 9"H1+5 " H2 = 0(vi) Enthalpy of combustion of trans 2 ) butene, "H = )647.0 kCal/mol.The value of "H1 & "H2 in KCal/mole are(A) )1.0 , 1.8 (B) 1.8, -1.0 (C) 5, 9 (D) 2, 3.619. The enthalpy changes of the following reactions at 27C areNa(s) + 21Cl2 (g) 01 NaCl (s) "rH = 411 kJ/molH2(g) + S (s) + 2O2 (g) 01 H2SO4 (l) "rH = 811 kJ/mol2Na(s) + S(s) + 2O2 (g) 01 Na2SO4 (s)"rH = 1382 kJ/mol21H2(g) + 21Cl2(g) 01 HCl (g) "rH = 92 kJ/mol; R = 8.3 J/K-molfrom these data, the heat change of reaction at constant volume ( in kJ/mol) at 27C for the process2NaCl (s) + H2SO4 (l) 01 Na2SO4 (s) + 2HCl (g) is(A) 67 (B) 62.02 (C) 71.98 (D) None20. The standard enthalpy of formation of FeO & Fe2O3 is ) 65 kcal mol)1 and )197kcalmol)1 respectively.A mixture of two oxides contains FeO & Fe2O3 in the mole ratio 2 : 1. If by oxidation, it is changed into a1 : 2 mole ratio mixture, how much of thermal energy will be released per mole of the initial mixture ?(A) 13.4 kcal/mole (B) 14.6 kcal/mole (C) 15.7 kcal/mole (D) 16.8 kcal/mole21. An athelete is given 100 g of glucose (C6H12O6) of energy equivalent to 1560 kJ. He utilises 50 percent of thisgained energy in the event. In order to avoid storage of energy in the body, the weight of water he would needto perspire is- (The enthalpy of evaporation of water is 44 kJ/mole.)(A) 319 gm (B) 422 gm (C) 293 gm (D) 378 gm22. TheheatofformationofC2H5OH(!)is)66kcal/mole.TheheatofcombustionofCH3OCH3(g)is 348 kcal/mole. "Hf for H2O and CO2 are )68 kcal/mole and )94 kcal/mole respectively. Then, the "H forthe isomerisation reaction C2H5OH (!)01 0CH3OCH3(g), and "E for the same are(A) "H = 18 kcal/mole, "E = 17.301 kcal/mole (B) "H = 22 kcal/mole, "E = 21.408 kcal/mole(C) "H = 26 kcal/mole, "E = 25.709 kcal/mole (D) "H = 30 kcal/mole, "E = 28.522 kcal/mole23. AB,A2andB2arediatomicmolecules.IfthebondenthalpiesofA2,AB&B2areintheratio1 : 1 : 0.5 and enthalpy of formation of AB from A2 and B2 is 100 kJ/mol1. What is the bond enthalpy of A2.(A) 400 kJ/mol (B) 200 kJ/mol (C) 100 kJ/mol (D) 300 kJ/mol24. Enthalpy of polymerisation of ethylene, as represented by the reaction, nCH2=CH2 01 5)CH2)CH2))n is)100 kJ per mole of ethylene. Given bond enthalpy of C=C bond is 600 kJ mol)1, enthalpy of C)C bond (in kJmol) will be :(A) 116.7 (B) 350 (C) 700 (D) indeterminateETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 4225. Substance A2B(g) can undergoes decomposition to form two set of products :If the molar ratio of A2(g) to A(g) is 5 : 3 in a set of product gases, then the energy involved in the decompositionof 1 mole of A2B2 , is :(A) 48.75 kJ/mol (B) 43.73 kJ/mol (C) 46.25 kJ/mol (D) None of these26. For the hypothetical reactionA2(g) + B2(g) 2AB(g)If "rG and "rS are 20 kJ/mol and 20 JK1 mol1 respectively at 200 K."rCp is 20 JK1 mol1 then "rH at 400 K is :(A) 20 kJ/mol (B) 7.98 kJ/mol (C) 28 kJ/mol (D) None of these27. Enthalpy of neutralization of H3PO3 acid is 106.68 kJ/mol using NaOH. If enthalpy of neutralization of HClby NaOH is 55.84 kJ/mol. Calculate "Hionization of H3PO3 into its ions :(A) 50.84 kJ/mol (B) 5 kJ/mol (C) 2.5 kJ/mol (D) None of these28. The enthalpy of neutralisation of a weak acid in 1 M solution with a strong base is 56.1 kJmol1.If theenthalpy of ionization of the acid is 1.5 kJ mol1 and enthalpy of neutralization of the strong acid with astrong base is 57.3 kJ equiv1, what is the % ionization of the weak acid in molar solution (assume theacid to be monobasic)?(A) 10 (B) 15 (C) 20 (D) 2529. The heat of formation of HCl at 348 K from the following data, will be0.5 H2(g) + 0.5 Cl2 (g)01 0HCl "H298 = 22060 calThe mean heat capacities over this temperature range are,H2(g), CP = 6.82 cal mol1 K1 ; Cl2 (g), Cp = 7.71 cal mol1 K1 ; HCl (g), CP = 6.81 cal mol1 K1(A) 20095 cal (B) 32758 cal (C) 37725 cal (D) 22083 cal30. TheaverageXe)Fbondenergyis34kcal/mol,firstI.E.ofXeis279kcal/mol,electronaffinityofFis85 kcal/mol & bond dissociation energy of F2 is 38 kcal/mol. Then, the enthalpy change for the reactionXeF4 01 0Xe+ + F) + F2 + F will be(A) 367 kcal/mole (B) 425 kcal/mole (C) 292 kcal/mole (D) 392 kcal/mole31. Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining processes.Ifall thecapturingofenergy fromthereaction isdonethrough electricalprocess(nonPV work)thencalculate maximum available energy which can be captured by combustion of 34.2 g of sucroseGiven : "Hcombustion (sucrose) = 6000 kJ mol1"Scombustion = 180 J/K - mol and body temperature is 300 K(A) 600 kJ (B) 594.6 kJ (C) 5.4 kJ (D) 605.4 kJ32. TheenthalpyoftetramerizationofXingasphase(4X(g)1X4(g))is100kJ/molat300K.The enthalpy of vaporisation for liquid X and X4 are respectively 30 kJ/mol and 72 kJ/mol respectively."S for tetramerization of X in liquid phase is 125 J / K mol at 300 K.What is the "G at 300 K for tetramerization of X in liquid phase ?(A) 52 kJ/mol (B) 89.5 kJ/mol (C) 14.5 kJ/mol (D) None of theseOne or More Than One Correct Answer Type:33. The normal boiling point of a liquid 'A' is 350 K. "Hvap at normal boiling point is 35 kJ/mole. Pick out thecorrect statement(s). (Assume DHvap to be independent of pressure).(A) "Svaporisation > 100 KJ/mole at 350 K and 0.5 atm(B) "Svaporisation < 100 KJ/mole at 350 K and 0.5 atm(C) "Svaporisation < 100 KJ/mole at 350 K and 2 atm(D) "Svaporisation = 100 KJ/mole at 350 K and 2 atmETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 4334. Which of the following statement(s) is/are false :(A) "rS for 21N2(g) 1 N(g) is positive(B) "Gsystem is always zero for a reversible process in a closed system(C) "G for an ideal gas is a function of temperature and pressure(D) entropy of a closed system is always maximized at equilibrium35. Which statement is / are correct:(A) Final temperature in reversible adiabatic expansion is lesser than in irreversible adiabatic expansion.(B) When heat is supplied to an ideal gas in an isothermal process, kinetic energy of gas will increase(C) When an ideal gas is subjected to adiabatic expansion it gets cooled(D) Entropy increases in atomisation of dihydrogen36. Which statement is / are correct:(A) wadiabatic > wisothermal in an ideal gas compression from same initial state to same final state(B) The value of , ##$%&&'(. ,vpCC remains constant for diatomic gas at all temperature(C) Entropy increases when an ideal gas expanded isothermally.(D) "rH & "rS both are + ve for the decomposition of MgCO3(s) .37. In isothermal ideal gas compression:(A) w is +ve (B) "H is zero (C) "Sgas is +ve (D) "G is +ve38. Which of the following is true for reversible adiabatic process involving an ideal gas?(A) Gas with higher , has high magnitude of slope in a P (y-axis) v/s T (x-axis)curve(B) Gas with higher , has high magnitude of slope in a V (y-axis) v/s T (x-axis)curve(C) Gas with higher , has high magnitude of slope in a P (y-axis) v/s V (x-axis) curve(D) Gas with higher , has low magnitude of slope in a P (y-axis) v/s T (x-axis)curve39. One mole of an ideal diatomic gas (Cv = 5 cal) was transformed from initial 25C and 1 L to the state whentemperature is 100C and volume 10 L. Then for this process(R = 2 calories/mol/K) (take calories as unit ofenergy and kelvin for temp)(A) "H = 525 (B) "S = 5 ln 298373 + 2 ln 10 (C) "E = 525(D) "G of the process can not be calculated using given information.40. Consider the reactions(i) S (rhombic) + 3/2 O2(g)01 0SO3(g), "H1(ii) S (monoclinic) + 3/2O2(g)01 0SO3(g), "H2(iii) S (rhombic) + O3(g)01 0SO3(g), "H3(iv) S (monoclinic) + O3(g)01 0SO3(g), "H4(A) "H1 < "H2 < "H4 (magnitude only) (B) "H1 < "H3 < "H4 (magnitude only)(C) "H1 < "H2 = "67 8 "H4 (magnitude only) (D) "H1 + "H4 = "69 / "H341. The normal boiling point of a liquid `X` is 400 K. Which of the following statement is true about the processX (l) 01 0 X(g)?(A) at 400 K and 1 atm pressure "G = 0 (B) at 400 K and 2 atm pressure "G = + ve(C) at 400 K and 0.1 atm presure "G = ve (D) at 410 K and 1 atm pressure "G = + ve42. 100 ml 0.5 N H2SO4 (strong acid) is neutralised with 200 ml 0.2M NH4OH in a constant pressure Calorimeterwhichresultsintemperatureriseof 1.4C.If heatcapaci tyofCalori metercontenti s1.5 kJ/C. Which statement is/are correctGiven : HCl + NaOH01NaCl + H2O + 57 kJCH3COOH + NH4OH01CH3COONH4 + H2O + 48.1 kJ(A) Enthalpy of neutralisation of HCl v/s NH4OH is 52.5 kJ/mol(B) Enthalpy of dissociation (ionization) of NH4OH is 4.5 kJ/mol(C) Enthalpy of dissociation of CH3COOH is 4.6 kJ/mol(D) "H for 2H2O(l) 01 2H+ (aq.) + 2OH(aq.) is 114 kJETOOS ACADEMY Pvt. LtdF-106,RoadNo.2IndraprasthaIndustrialArea,EndofEvergreenMotor,BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303THERMODYNAMICS_JEE MAINS # 4443. From the following data at 25CReaction !rH kJ/mol21H2(g) + 21O2(g) 01 OH (g)42H2(g) + 21O2(g) 01 H2O(g) 242H2(g) 01 2H(g)436O2(g) 01 2O(g)495Which of the following statement(s) is/are correct:(A) "rH for the reaction H2O (g) 01 2H(g) + O(g) is 925.5 kJ/mol(B) "rH for the reactionOH(g) 01 H(g) + O(g) is 502 kJ/mol(C) Enthalpy of formation of H(g) is 218 kJ/mol(D) Enthalpy of formation of OH(g) is 42 kJ/mol44. Which of the following is true ?(A) For the reaction CaCO3(calcite) 01 CaCO3 (aragonite)given : o298 fG " (calcite) = 1128.8 kJ/mol, o298 fG " (aragonite) = 1127.75 kJ/mol,then calcite form is thermodynamically more stable at standard conditions.(B) For the reaction,(a) C(diamond) + 2H2 (g) 01 CH4(g) "H1(b) C(g) + 4H(g) 01 CH4(g) "H2then more heat is evolved in reaction (b).(C) "fH (I2, g) = "subH [I2, s] at 25C.(D) For the exothermic reaction 2Ag(s) + 1/2 O2(g) 01 2Ag2O(s) at 298 K. "H < "U45. Which of the following do(es) not represent "H formation of the product.(A)21H2(g) + (aq) 01 H+(aq) (B) 32O3(g) 01 O2(g) + e(C)/4NH (g) + Cl(g) 01 NH4Cl(s) (D) P4(black) + 5O2 (g) 01 P4O10(s)46. Which of the following statement(s) is/are true?(A) When ("Gsystem)T,P < 0; the reaction must be exothermic(B) "f H (S, monoclinic) 2 0(C) If dissociation enthalpy of CH4(g) is 1656 kJ/mole and C2H6 (g) is 2812 kJ/mole, then value of CCbond enthalpy will be 328 kJ/mole(D) If H+(aq) + OH(aq) 01 H2O(l)"rH = 56 kJ/mol"f H(H2O, g) = 242 kJ/mole; Enthalpy of vaporization of liquid water= 44 kJ/molthen, "f H(OH, aq) will be 142 kJ/mole47. Select the option in which heat evolved is maximum.Given : "fH (CO2,g) = 75 kCal/mol ; "fH (CO,g) = 25 kCal/molThe product will be CO if excess amount of carbon is present and CO2 if excess O2 is present(A) 10 moles o