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CHAPTER 3 THERMODYNAMICSSome Important Terms.1. Thermo chemistry: A branch of chemistry deals with chemical reactions along with thermal changes. It is the practical approach off

1st law of thermodynamics.2. Thermo chemical Equation: A balanced chemical equation along with thermal change expression is called thermo chemical equation Eg.

C(s) +O2(g) CO2(g) ΔH = -393.50 kjEndothermic Reactions: Chemical reactions requiring heat energy to convert reactants into products are called endothermic reactions.

Eg. N2(g) +O2(g) +180.75 kj 2NO(g) Or N2(g) +O2(g) 2NO(g) -180.75 kjOr N 2(g) + O2(g) 2NO(g) ΔH= +180.75 kj

Since heat energy needs to be supplied it is placed with positive sign with reactants or with negative sign onproduct side.

4. Exothermic Reactions . Chemical reactions in which the heat is evolved along with the formation of products are called exothermic

reactions.Eg. C(s) +O2(g) CO2(g) +393.50 kj Or C(s) +O2{g) –393.50 kj CO2(g)

Or C(s) +O2(g) CO2(g) ΔH = –393.50 kj Since the reaction gives off heat energy, it is placed with negative sign with reactants side or positive sign with

product side.5. Explain graphical representation of Endothermic and exothermic reactions.Endothermic reaction If the heat is absorbed by the reactants to be converted into products, reaction is called endothermic reaction. Value of enthalpy change is positive. Thus enthalpy of products is higher than enthalpy of reactants. When water vapour (steam) is passed over burning coal at 600°C temperature, it produces water gas (i.e.

mixture of carbon monoxide and hydrogen gas). The above endothermic reaction is represented as under

Exothermic reaction If the heat is evolved by the reactants to be converted into products, reaction is called exothermic reaction. Value of enthalpy change is negative. Thus enthalpy of products is lower than enthalpy of reactants. When carbon monoxide undergoes combustion with oxygen gives carbon dioxide. The reaction is exothermic.

The above exothermic reaction is represented as under

CO + ½O2

CO2(g)

ΔH = –282.92KjEnthalpy

Reaction axis

Hr

Hp

C + H2O

ΔH =+131.38kjKcal

Enthalpy

CO + ½H2Hp

Hr

Reactio axis

2

6. What are the common features of chemical reactions? During the chemical reactions, reactants are converted into products. During the reaction, certain bonds break on the reactant-sides. New bonds are formed usually on product side.

Eg. H2 + Cl2 2HCl Breaking up of bond is an endothermic reaction while formation of a bond is exothermic reaction. Thus above

reaction can be understood as follows:i) H2 (g) + energy 2H•ii) Cl2(g) + energy 2Cl•iii) H• + Cl• HCl + energyiv) H2 + Cl2 2HCl + energy

Here energy consumed during step (i) + (ii) is less than energy released during step (iii) and hence overallreaction becomes exothermic

7. Define System. Part of universe with a definite boundary under experiment or observation (with which we are concerned) is

called system.8. Define: Open System The system with which energy as well as mass can be exchanged with the surrounding by its own is called open

system. A piece of ice placed in an open container is an open system9. Define: Closed System: The system with which energy can be exchanged but mass cannot with the surrounding by its own is called a

closed system Cooking in pressure cooker is an illustration of closed system.10. Define: Isolated System The system with which neither energy nor mass be exchanged with the surrounding by its own is called an

isolated system. A good quality of thermos flask provides an example of isolated system.11. Explain State Function Those properties of system which depend only on its state but not on how the state was reached are called state

function. If a body of mass (m) is raised to a height (h) has potential energy (mgh), irrespective of methods adopted to

take the body at a height (h). Thus potential energy of a substance is its state function. Similarly, the volume of 1 mole of ideal gas at STP is 22.4 litre, the volume of the gas under these conditions does

not depend on the condition of the gas before reaching at STP. Eg. Internal Energy, Enthalpy, Entropy and Free energy. Thus changes in values of the state function depend only on the final and initial state of a system and not on how

the change was carried out. Thus, pressure (P), temperature (t), volume (V), internal energy (E), enthalpy (H), entropy (S), free energy (G)

etc., are the state functions.

12. Explain Extensive properties Some properties of substances depend on the quantity of the matter. Eg. Mass, volume, enthalpy, entropy, free

energy etc. Thus, an extensive property depends upon the quantity of matter present in the system.13. Explain Intensive Properties Some properties of substances do not depend on the quantity of the matter. Temperature, pressure, density,

boiling point, freezing point, surface, tension etc are the properties of the substances of intensive type. These donot depend on the quantity of substance.

14. Define: Process Conversion of a system from one state to another is called a process.15. Define: Isothermal process If temperature of a system remains constant during a change of its state, process is called isothermal process. OR

Heat can flow in or out of the closed system but temperature of the system must remain constant during thechange. Process is called isothermal process. Δ T = 0

16. Define: Isobaric process If a gaseous system changes its state at constant pressure, the process is called isobaric process i.e. pressure

remains constant during the change then process is called isobaric process. ΔP = 0

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17. Define: Isochoric process Process in which volume of the system remains constant is called isochoric process. This process can be carried

out in a rigid container. ΔV = 018. Define: Adiabatic process If a system does not gain energy from surrounding or give up energy to the surrounding during change of its

state, the process is an adiabatic process.OR

No heat flows in or out of the closed system i.e. the system does not exchange heat with its surroundings.Process is called adiabatic process.

19. Explain Internal energy. Every substance is associated with certain amount of energy, known as internal energy (intrinsic energy).

Internal energy is represented by 'E' or 'U' Definition: A definite amount of energy stored in each substance depending upon its mass and characteristics of

substance is known as internal energy. The various forms of energies that contribute towards the internal energy of the substances are translation

energy, rotational energy, vibration energy and electronic energy, nuclear energies of constituent atoms and Substance possesses potential energy due to repulsion between nucleus-nucleus as well as attractive forces

between electron and nucleus. In addition the proton and neutron in the nucleus experience attractive forces. According to equation E= mc2 matter is also one form of energy. Therefore if matter of the substance is changed

in to energy than large amount of heat will be evolved. Internal energy is a state function. Thus internal energy depends upon the state of the substance but does not

depend on how the state is achieved. Internal energy is extensive property. Though every substance possess definite amount of internal energy its

absolute value cannot be determined, anyway change in internal energy (ΔU) is a measurable quantity. During any chemical reaction law of mass conservation is followed but as reactant and products have different

characteristics as a result their internal energies are different so energy change of reaction is given by ΔU = Up –Ur .

If there is any change in the volume of a mass during reaction, there is a change in its work energy. Hence, if heatevolved by the change in internal energy during a chemical reaction is to be measured, the volume of thesubstance must kept constant.

Thus in order to obtain exact energy change by using internal energy it is necessary to performed reaction atconstant volume and energy change of reaction at constant volume is given by ΔU=qv. Thus heat absorbed orevolved at constant volume is equal to internal energy change.

20. Explain Enthalpy The term enthalpy has been introduced to study the heat changes taking place during the reactions occurring

under constant pressure at constant temperature. Definition: Enthalpy is defined as sum of the internal energy and the pressure volume energy of the system

under a given set of conditions. Enthalpy is represented by 'H'. Enthalpy is extensive property and state function. As absolute value of internal energy cannot be determined, the same is true for enthalpy also Enthalpy (H) = E +

PV However, change in enthalpy is measurable quantity. Thus, ΔH = Hp –Hr

21. Explain Zeroth law of thermodynamics. When two thermally conducting substances having different temperatures come in contact with each other, the

heat flow from a substance having higher temperature to a substance having lower temperature till a thermalequilibrium is established.

Example When any person has fever, the temperature of the body is measured by clinical thermometer. When the thermometer comes in the contact with the body of a person, the heat from his body enters into the

mercury in the thermometer increasing volume of the mercury in thermo meter. When the temperature of the body and that of mercury in thermometer becomes equal, the increase in volume

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of mercury stops. This state is called state of thermal equilibrium thermometer.

22. Write the first law of thermodynamics. First law of thermodynamics is the law of conservation of energy. It can be written as the total quantity of the energy of the universe is constant. The energy can neither be

created nor can it be destroyed. It can only be converted from one form to another.

23. How energy of the system can be changed. The internal energy of the system can be changed by following two methods. By adding heat energy to the system or removing heat from the system. By doing work on the system or if work is done by the system. Here heat and work are on one side while energy is on other side. Heat and work indicate processes and hence

are not the state function while energy is illustration of state function. Suppose the internal energy in a definite state of a system is U1, and after a reaction it becomes U2. At final

state internal energy level, a new equilibrium is established. Internal energy level U2 is achieved either by gaining heat energy (q) or by losing the same amount of energy. If

work (w) is done on the system or work is done by the system, then first law of thermodynamics can be writtenas U2 =U1 + q + w

U2 –U1 = q + wΔU =q + w

Where ΔU is change in internal energy. This equation is symbolic representation of first law of thermodynamics.

24. Explain mathematical form of first law of thermodynamics. Symbolic representation of first law of thermodynamics is ΔU = q + w . When the symbolic form of law or is used for calculations, q and w are associated with conventional signs. If heat energy is added to the system q carry positive conventional sign i.e. for endothermic reaction, q is

positive. If heat energy is expelled by the system q carry negative conventional sign i.e. for exothermic reaction, q is

negative. If work is done on the system then value of w is positive. If work is done by the system then value of w is negative If internal energy does not change during the process i.e. ΔU = 0 then q= –w It indicates that all the heat absorbed by the system during the process is used to do work. OR For w = -q, it

indicates that whatever work is done on the system is available as heat energy. Since internal energy of a system is it state function the value of ΔU is independent of, path followed by the

process while q and w are not state functionExample: 1 A system by getting 112 calories heat from a surrounding does the work equivalent to 48 calories,calculate the change in internal energy.Calculation: The system obtains heat :. q = 112 calories

Work is done by the system :. W = -48According to the first, law of thermodynamics

ΔE =q+wΔE =q +(-w)ΔE = 112 -48:. ΔE = 64 calories

Exercise1. One system)ost 125.0 calorie heat and did 315.0 calorie work, calculate the change in its internal energy.

(Ans. : ΔE = -440 calorie)2. One system absorbed 170 calorie heat and did the work. The change in its internal energyis 110 calorie. Calculate how much will be the work done. (Ans. : w = -60 calorie)3. A system lost 27 calorie heat by 60 calorie work being done. Calculate its internal energy. (Ans. : ΔE = 33calorie)

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25. How first law of thermodynamics is related with work. Work is defined as the product of force and displacement. Two main type of work are found in chemistry : (i) Electrical work (ii) Mechanical work. The electrical work in reaction involves ions in a system while mechanical work is of importance in a reaction in

which external pressure changes the volume in gaseous system. At constant pressure if volume of gaseous system decreases then work is done on the system while if volume of

the system increases then work is done by the system. If the initial volume of the system is V1 and final volume is V2 under constant external pressure P then work done

by. the system is expressed as : w = P (V2 – V1)= PΔV According to first law of thermodynamics: ΔU = q + w :. ΔU = qv - PΔV (work done by the system) If volume of the system does not change during the reaction then Δ V = 0 :. ΔU =qv

Thus change in internal energy of a system in which reaction occurs at constant volume is equal to quantity ofheat absorbed or lost by the system.

26. Explain enthalpy and first law of thermodynamics. Usually chemical reaction takes place in open container i.e. under constant external pressure. A new state function called enthalpy (H) of a system can be derived. A relation between enthalpy (H) and internal energy (U) can be given as H = U + PV. Change in enthalpy of a system can be written as ::. ΔH = ΔU+Δ (PV)=ΔU+ΔPV +PΔV If reaction occurs under constant pressure then ΔP becomes zero.:. ΔH =ΔU+PΔV According to symbolic equation of first law of thermodynamics.ΔU = q+w . If reaction occur under constant pressure then q = q p and w = – PΔ V :. ΔU = q p –PΔV Substituting the value of ΔU in the equation Δ H= ΔU + P Δ V we get, ΔH =qp –PΔV +PΔV:. ΔH = qp

Thus in reactions occurring at constant pressure, the change in enthalpy of the system is equal to the value ofheat gained or lost by the system.

Thus, it is concluded that enthalpy of each substance is constant at constant temperature and at constantpressure.

Example: 2 : The volume of a gas at atmospheric pressure was 0.5 litres. If this gas obtains 29.0 calories heat, itsvolume becomes 2.0 litres at 1 atmospheric pressure. Calculate the change in internal energyCalculations : Atmospheric pressure = 1Initial volume of a gas V1 = 0.5 litreFinal volume of a gas V2 = 2.0 litre

Δ V = V2 – V1 =2.0-0.5 = 1.5 litreNow, w = P Δ V

w = 1 x 1 .5 atm litrew = 1.5 atm litre

Now, 1 atm pressure = 24.21 calories:. For 1.5 atm litre = 1.5 x 24.21 = 36.32 caloriesHere work is done by the system :. w = –36.32 caloriesHeat obtained by the system q = 29.0 calories.'. According to the first law of thermodynamics

Δ U=q+w:. Δ E = 29.0 - 36.32 = -7.32 calories

27. Derive Δ H = Δ U + Δ n(g)RT During the chemical reaction, the number of moles of gaseous reactant changes on product side and reaction

occur at constant temperature and under constant pressure, the volume of gaseous phase also changes. Q Suppose:

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n 1 = number of moles of gaseous reactants n2 = number of moles of gaseous products V1= Initial volume V2 = Final volume At constant temperature T and under constant pressure P According to ideal gas equation: PV = nRT

:. PV1= nl (g)RT while PV2 = n2(g)RT:. P(V2 – V2) = (n2(g) -nl(g) ) RT:. P Δ V = Δ n(g) RT

Substituting P Δ V = Δn(g)RT in the equation Δ H = Δ U + P ΔV we get,Δ H= Δ U + Δ n(g) RT

ORΔ U = Δ H – Δ n(g)RT where

Δ n(g) = (Number of moles of gaseous products) - (Number of moles of gaseous reactants) Conditions

If Δn(g) = 0 then ΔH = ΔUIf Δ n(g) > 0 then ΔH > ΔUIf Δ n(g) <0 then ΔH<ΔU

Example 2 : One mole of naphthalene was burnt in oxygen gas at constant volume to give carbon dioxide gas at250C. The heat evolved was found to be 1228.2 kcal. Calculate the heat of reaction at constant pressure. (R =0.002 kcal) ΔU = -1228.2 kcal.

Cl0H8 (s) + 12O2(g) 10 CO2(g) + 4H2O(1)

Calculation: ΔH = ΔU + Δ n(g)RTΔ n(g) = np -nr = 10-12 =.-2

T = 25 + 273 = 298 KΔH = -1228.2 +( -2)x 0.002 x 298

= -1228.2 -1.192ΔH = -1229.392 kcal.

Calculate the following:1. The heat associated with the combustion of liquid benzene at constant temperature is - 781.0 k.cal /mole.

Calculate the change in enthalpy taking place during the combustion of 1 mole of benzene at 270C. (R = 1.987calories) (Ans. : ΔH = -781.9 kcal / mole) C6H6(l) + 7.5O2(g) 6CO2(g) +3H2O(l)

2. Calculate the change in enthalpy of the following chemical reaction at 270C.C(s) +½ O2(g) CO(g) ΔU = -54.0 kcal (R = 1.987 calories) (Ans. : ΔU =-54.3 kcal / mole)

3. The enthalpy change of the following reaction at 1 atmosphere pressure and 270C temperature, ΔH = -20.0 kcal /mole. Calculate its change in internal energy . C(s) + 3H2(g) C2H6(g) (R = 1.987 calories)(Ans. : ΔU = -18.808 calorie)

4. The heat of formation of methane at constant pressure at 25°C is -17.89 kcal. What will be the heat of formationat constant volume? (R = 2 calorie) (Ans.: ΔU=-17.294 kcal)

5. On heating two moles of KClO3(s) in an open vessel, two moles of KCl(s) and three moles of O2(g) are obtained.What would be ΔE, ΔH and Δ n(g) if 21.4 kcal of heat are evolved during the reaction. (R = 2 x 10-3 kcal)

(Ans. : ΔU = -23.2 kcal)28. Discuss Heat Capacity. Heat capacity of the substance can be defined as the heat required to raise the temperature of any substance by

1°C. If the temperature of the substance is increased by 1°C at constant volume, heat capacity is mentioned as Cv If the temperature of the substance is increased by 1°C at constant pressure, heat capacity is mentioned as Cp . Difference between Cp and Cv for liquid and solid compound is very small. But if compound is in gaseous state,

difference between Cp and Cv is obtained to considerable extent. When the initial temperature t1 of a sample of gaseous substance is increased to final temperature t2, then the

relation between heat q, heat capacity C and temperature difference (t2 – t1) is shown by following equation.q=C× (t2-tl)

qC

t

Heat absorbedHeat Capacity

Temperature difference

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The unit of heat capacity is cal / C0

29. Define: Specific heat capacity. Specific heat capacity is defined as the quantity of heat required to increase the temperature of 1 gram

substance by 1°C.

. .

Heat absorbedspecific heat capacity

Temp difference Wt of sub in gram

Unit of Specific heat capacity is Calorie/C0 gram or cal °C-1gm-1

Define: Molar heat capacity Molar heat capacity is defined as the amount of heat required to increase the temperature of 1.0 mole of a

substance by 1°C.. .

. .

Heat absorbed mol wt of submolar heat capacity

Temp difference Wt of sub in gram

Unit of molar heat capacity is Calorie / °c mole or cal °C-1mole-1

Example 1 : The temperature of a sample of ethanol is increased by 2° when 23.43 calorie heat is supplied.Calculate the heat capacity of that sample.Calculation: Heat capacity

23.42

2

qC

t

= 11.71 Kcal / C0

Example 2 : Calculate the heat gained by water when 2 kilogram water kept at temperature 25°C is heated till itboils, the specific heat capacity of water is 1.0 calorie 1°C gram. Calculation

. .

Heat absorbedspecific heat capacity

Temp difference Wt of sub in gram

Heat absorbed = specific heat capacity × temp. difference × weight of substance in gram

= 1.0 × ( 100 – 25 ) × 2000= 1.0 × 75 × 2000= 150000 calorie= 150 Kcal = 1.50 × 102 Kcal

30 Define : enthalpy of reaction: Amount of heat absorbed or evolved during chemical reaction at constant temperature and pressures is called

enthalpy of reaction.( ΔH )31. Define: Standard State In scientific practice 1.0 bar pressure and 25°C temperature, are considered as standard state.32. Explain: Standard Form The stable form out of various forms of element or compound i.e. solid, liquid or gas is called standard form. Standard form of metal like sodium, copper, silver etc., is solid, which is expressed as Na (s) Cu(s), Ag(s) etc. The standard form of hydrogen, nitrogen, chlorine, oxygen etc., is gas, these elements exists as diatomic gas,

which are expressed as H2 (g), N2 (g), Cl2(g), O2 (g) etc. [ Br2(l) Hg(l) Ga(l) ] When elements exhibit allotropy, the most stable allotrope is considered as standard form. Eg. carbon

(graphite), rhombic sulphur, etc. Under standard conditions, the standard enthalpy of elemental state of element is considered as zero.

33. Define: Standard enthalpy of reaction. When products are obtained in standard form by reaction of reactant in standard form then net enthalpy change

of the reaction is called the standard enthalpy of reaction. The standard enthalpy change (ΔH0) associated with the reaction, when the reactants in their standard state are

converted into products in their standard state. It can be calculated by formula: ΔH0 = (Enthalpies of products) - (Enthalpies of reactants) Enthalpy change of reaction CO(g) +½O2(g) CO2(g) is called standard heat of reaction because here

reactants and products both are in standard form.

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Example 1 : In the following reaction, the standard enthalpies of Al2O3 and Fe3O4 are -399.9 and -267.0 k.cal /mole respectively. Calculate the heat 0f reaction of the given reaction

8Al+3Fe3O4 4Al2O3 +9FeCalculation : ΔH0 = (Enthalpies of products) - (Enthalpies of reactants)

=[ 4ΔH0Al2O3 +9 ΔH0

Fe -[ 8ΔH0Al +3ΔHo

Fe3O4 ]=[ 4( -399.0)+9(0)] -[8(0) + 3( -267) ]= –1596 +801

ΔH0 = –795 k.caVmole:. Heat of Reaction = –795 Kcal.

34. Explain Standard enthalpy of formation Standard enthalpy of formation (ΔH0) is defined as standard enthalpy associated with formation of one mole

compound in their standard state directly from the elements in their standard state. When 1 mole of substance is formed in standard state by direct reaction between its elements in their standard

form then net enthalpy change of the reaction is called standard heat of formation of that substance. Standard heat of formation of stable substance is normally negative. Eg: 1) C(s) + ½O2(g) CO(g) ΔHo = ΔH0

f (CO) = -26.4 kcal Here , the standard enthalpy change (ΔHo) in the formation of CO by reaction between carbon and oxygen is –

26.4 KcalΔHo = [std. Enthalpies of products ] - [std. Enthalpies of products]

= [ΔHo CO(g) ] – [ΔHo C(s) +½ ΔHoO2(g) ]–26.4 = [ΔHo CO(g) ] – [0 +0 ]ΔHo CO(g) = –26.4 kcal / mol heat of formation of CO ΔHo

f CO(g) = - 26.4 kcal / mol.2) H2(g) + + ½O2(g) H2O(l) ΔHo = ΔH0

f (H2O).= -68.3 kcal3) H2(g) +O2(g) + C HCOOH ΔHo = ΔH0

f (HCOOH) = -97.8 kcal

Exercise.1) The enthalpies of formation of NH3(g) H2O(l) and NO(g) are -11.0, -57.8 and +21.6 kcal / mole respectively.

Calculate enthalpy change in the reaction. 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) (Ans : ΔH = –216.4 kcal)2) Standard heat of formation for methanol, H2O(g) and CO2 (g) are -238.87,-241.84, -393.55 kJ/mole respectively.

Calculate the enthalpy change for the reaction i.e. heat of combustion of methanol giving carbon dioxide andwater vapour. (Ans : ΔH = –638.36 kcal )

3) Calculate heat of combustion of benzene by the help of given: ΔH0f C6H6 =11.72 kcal, ΔH0

f (CO2) = –94.05 kcal, ΔH0f

(H2O) = –68.3 kcal (Ans.: ΔH= –780.92 kcal)4) Calculate heat of formation of benzene from the data:ΔH0

f (C02) = –94.05 kcal, ΔH0f (H20) = –68.3 kcal, ΔH0 = –

781.2 kcal (Ans : ΔH0f (C6H6) = 11.72 Kcal )

35. Explain: enthalpy of combustion The quantity of heat evolved when one mole of an organic compound is burnt completely by oxygen under

constant pressure in standard condition is known as enthalpy of combustion. By comparing heat of combustions of different allotropes of some elements, their stabilities can be compared

i) Cgraphite(s) +O2(g) CO2 (g) ΔH(i) = -393.5 kJ mol-1

ii) Cdiamond(s) +O2(g) CO2 (g) ΔH(ii) = -395.4 kJ mol-1

Since diamond has higher internal energy than graphite thus, it is less stable. Value of heat of combustion is also useful to calculate energy change required to convert one allotrope in to

other for e.g. if in above equations, eqn 2 is subtracted from eqn1 then it give reactionCgraphite(s) Cdiamond(s) Where ΔH = ΔH(1) – ΔH(2) = 1.9 kJ mol-1

Heat of combustion is also useful to calculate standard heat of formation of organic compound if heat offormation of CO2(g) and H2O(l) are known.

For example heat of combustion of ethanol is ΔH0 = –348.76 Kcal / mole and heat of formation of CO2(g) andH2O(l) is –94.05 and – 68.32 kcal / mol respectively then heat of formation of ethanol can be calculated asfollows.

C2H5OH(s) +3O2(g) 2CO2(g) +3H2O(l)

ΔH0 = (Enthalpies of formation of products) - (Enthalpies of formation of reactants )]–348.76 = [ 2Δ H0

CO2 + 3ΔH0H2O ] - [ ΔH0

C2H5OH + ΔH0O2 ]

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–348.76 =[ 2(-94.05)+3( -68.32)]-[ ΔH0C2H5OH +3(0)]

ΔH0fC2H5OH = - 44.3

Example: 1 C2H5OH(s) +3O2(g) 2CO2(g) + 3H2O(l)In the above reaction the heat of formation of ethanol, CO2(g) and 3H2O(I) are -44.3, -94.0.5 and-=68.32 k.cal /mole respectively. Calculate the heat of combustion of ethanol.Calculation:

ΔH0 = (Enthalpies of formation of products) - (Enthalpies of formation of reactants )]= [ 2Δ H0

CO2 + 3ΔH0H2O ] - [ ΔH0

C2H5OH + ΔH0O2 ]

=[ 2(-94.05)+3( -68.32)]-[( -44.3)+3(0)]=-393.06+44.3Δ H=-348.76 kcal / mol

36. Define: Heat of fusion Amount of energy required to change solid physical state of substance into liquid state at its melting point under

constant pressure is called heat of fusion.H2O(s) + 1.436 kcal/mole H2O (l)

:. Heat of fusion of Ice is 1.436 kcal / mole

37. Define: Heat of vapourisation Amount of energy required to change liquid physical state of substance into gaseous state at its boiling point

under constant pressure is called heat of vapourisation.H2O(1) + 9.71 kcal/mole H2O(g)

Heat of evaporation of water is 9.71 kcal / mole.

38. Give Hess law of constant heat summation. Explain with illustration. Russian Scientist Hess has derived the law of heat summation in 1840. “The total heat change during a chemical reaction is the algebraic sum of the heat changes in different steps of

the reaction.” "The heat change associated with a chemical reaction is the same irrespective of the intermediate stage through

which system passes." The change in heat in chemical reaction at constant pressure and temperature is same as qp. Thus at constant T &

P reactant and products have constant value of enthalpies. As a result value of net enthalpy change of reactionremains constant whether reaction is taking place in single step or more than one step.i) C(s) + ½O2(g) CO (g) ΔH = –110.54 Kjmol-1

ii) CO(g) +½O2(g) CO2(g) ΔH = –282.96 Kjmol-1

iii) C(s) + ½O2(g) CO2(g) ΔH = – 393.5 Kjmol-1

Thus, method A and B have same amount of change in enthalpy Applications of Hess law

Thermo chemical equations can be operated mathematically. The heat liberated or adsorbed during different chemical reactions, such as heat of formation, heat of

combustion, heat of neutralization can be calculated on the basis of Hess law. Sometime, heat of reaction which cannot be determined experimentally, can be calculated on the basis of Hess

law by related reactionsExample: Calculate heat of formation of liquid CS2 on the basis of reactions given below:

ΔH = – 393.5 Kjmol-1

CO(g)

CO + ½O2

CO2(g) CO2(g)

ΔH = –282.96 Kjmol-1

ΔH = –110.54 Kjmol-1

C(s) + O2 (g)C(s) + ½O2 (g)

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i) C(s) + O2(g) CO2(g) ΔH1 = -94.05 k.calii) S(s) + O2(g) SO2(g) ΔH2 =-70.96kcaliii) CS2(1) + 3O2(g) CO2(g) + 2SO2(g) ΔH3= -265.97 kcalCalculation:Reaction showing formation of CS2 (1) is as under:C(s) +2S(s) CS2(1)

Above reaction can be derived by: Equation (1) + 2 x equation (2) - Equation (3):. ΔH =ΔH1+2(ΔH2)–ΔH3

i) - C(s) +O2(g) CO2 (g) ΔH1 = -94.05 k.calii) 2S(s) + 2O2(g) 2SO2(g) ΔH2 = -141.92 k.ca1(i) + 2(ii)

iii) C(s) + 2S (s) + 3O2.(g) CO2 (g) + 2SO2 (g) ΔH = -235.97 k.calNow (i) + 2(ii) - (iii)

C(s) + 2S (s) + 3O2.(g) CO2 (g) + 2SO2 (g) ΔH = -235.97 k.calCS2(l) +3O2 (g) CO2(g) +2SO2 (g) ΔH3 = -265.97 k.cal_ _ _ _ .C(s) + 2S (s) CS2 (l) ΔH = +30 k.cal

Exercise1) Calculate heat of formation of H2SO4 using-following results.

i) S(s) +O2(g) SO2(g) ΔH01 = -70.9 k.cal

ii) SO2(g) + ½O2(g) SO3(g) ΔH02 = -23.5 k.cal

iii) SO3(g) + H20(/) H2SO4 (I) ΔH°3 = -30.6 k.caliv) H2(g) + ½O2(g) H2O(l) ΔH4° = -68.3 k.cal

2) Calculate standard heat of formation of methane on the basis of following reaction.i) C(s) +O2(g) CO2(g) ΔH0

1 =-94.1 k.calii) H2(g) + ½O2(g) H2O(l) ΔH0

2 = -68.3 k.caliii) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH0

3 = -212.8 k.cal3) Compare the stability of two important allotropes of carbon on the basis of following data

i) C(graphite) + O2(g) CO2(g) ΔHlo = -393.514 k.cal

ii) C(diamond) +O2(g) CO2(g) ΔH02 = -395.408 K.cal

4) Calculate heat of sublimation of iodine on the basis of followingi) H2(g) + I2(S) 2HI(g) ΔHl

o = +11.95 k.cal ii) H2(g) + I2(g) 2HI(g) ΔH2° = -2.75 k.cal5) Calculate heat of formation of sucrose by following reactions.

i) C12H22O11(S) + 12O2(g) 12CO2(g) + 11H2O (l) ΔHlo = -1360.0 k.cal

ii) C(s)+O2(g) CO2(g) ΔH2° =-94.05 k.caliii) 2H2(g)+O2(g) 2H2O(/) ΔH3° =-136.64 k.cal

39. Explain Heat of Neutralization Amount of heat evolved when a dilute solution containing one gram equivalent of strong acid is neutralized by a

dilute solution containing one gram equivalent of strong base under standard condition is called heat ofneutralization. The value of heat of neutralization of strong acid – base reaction is 56 Kj/mole which is constant.

OR When a gram equivalent weight of strong acid dilute solution react with a gram equivalent weight of strong

base dilute solution, which results into one mole unionized water along with 56 Kj/mole heat evolved at 25°Ctemperature, the same is called heat of neutralization.

The value of heat of neutralization is constant for strong acid and strong base.For e.g HCl(aq) + NaOH(aq) Na+

(aq) + Cl־(aq) + H2O(l)

H+(aq) + Cl־(aq) +Na+

(aq) +OH-(aq) H2O(1) + Na+

(aq) + Cl־(aq)

HNO3(aq) + KOH(aq) K+(aq) + Cl־(aq) + H2O(l)

H+(aq) + NO3

־(aq) +K+

(aq) +OH-(aq) H2O(l) + K+

(aq) + NO3־(aq)

Net reaction H+(aq) +OH-

(aq) H2O(l) ΔH = -56 Kj/mole. Thus when any strong acid and strong base reacts with each other H+

(aq) of acid and OH–(aq) of base only reacts

with each other to give reaction H+(aq) +OH-

(aq) H2O(1)

As a result heat of neutralization of any strong acid and strong base is constant.

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During acid base reaction of strong acid and base anion of acid and cation of base do not take part in thereaction hence such ions are known as spectator ions.

When one mole H+(aq) ions combine with one mole OH–

(aq) ions, they give one mole unionized water, 13.36 kcalheat is evolved at 25°C temperature, the same is called heat of neutralization.H+

(aq) +OH-(aq) H20(l) . ΔH = -56 Kj/mole.

40. Define: Bond energy Bond energy is defined as the amount of energy required to break bonds in one mole of a gaseous covalent

substance to produce neutral atoms or free radicals from it.

41. How to calculate heat of reaction on the basis of bond energy? When experimental information of bond energy is available, the heats of reaction can be calculated theoretically

by applying the following equation.ΔH = [∑(Bond energies of Reactants)]-[ ∑ (Bond energies of Products)]

Example: The values of bond energies of C(s) → C(s) H - H, C - H and C– C are 170, 103, 98 and 80 kcal / mole.Calculate the enthalpy change of the following reaction utilizing these values. 3C(s) +4H2(g) C3H8(g)

Calculation:Δ.H = ∑ (Bond energies of reactants)] - [∑ (Bond energies of products)]= [3 (C(s) → C(g)+4(H-H)J-[ 2 (C -C)+8(C - H)]= [3(170) + 4(103)] -[2(80) + 8(98)]= 922.0 - 944.0Δ.H = -22.0 kcal.

Excercise1. With the help of enthalpy change in the following reaction, calculate the bond energy of N - H bond in NH3 (

N N = 226 kcal, H - H = 103 kcal) N2(g) +3H2(g) 2NH3(g) ΔH = -23.0 kcal.(Ans : N-H = 93.0 kcal / mole)

2. The bond energies of H-H, H-Cl and Cl- Cl are 103, 102 and 57 kcal / mole respectively in the reaction H2(g) + Cl2(g)

2HCl(g) . Calculate the heat of reaction of this reaction.(Ans : -44.0 kcal)

3. The values of bond energies of C(s) → C(g), H - H, C - C and C - H are 170, 103, 80 and 98 kcal/ mole respectively.Calculate the value of ΔH for the following reaction using these values of bond energies.2C(s) +3H2(g)

C2H6(g) (Ans. : -19.0 kcal / mole)4. Calculate the value of bond energy of C -C from the following information.

2C(graphite) +3H2(g) C2H6(g) ΔH = -20.23 kcal.i) C(graphite) → C(g) ΔH1 = 170 kcal ii) H2(g) → 2H(g) ΔH2 = 103 kcal iii) C – H bond energy = 98 kcal

5. The bond energies of O - H, H - H and O = O in the reaction H2 +½ O2(g) H2O(g) are 110, 103 and 118.Calculate the heat of formation of H2O(g) using these values of bond energies. (Ans. : -58.0 kcal)

Q 42: What is meant by thermodynamics and chemical thermodynamics? Explain it in brief. The science in which energy changes are associated with physical and chemical reactions are studied is called

thermodynamics. The science in which study of energy changes is done from chemistry point of view is called chemical

thermodynamics. The study of relation between energy and chemical reaction is mostly on the basis of four laws of

thermodynamics. These laws are 1st 2nd and 3rd law of thermodynamics The study of changes in energy during chemical reaction in chemistry and in industrial field is important.

Q 43: (a) State the 1st laws of thermodynamics and its limitations.(b) Discuss the limitation of 1st law with suitable examples.

First law of thermodynamics "It is not possible to create or destroy the energy but energy changes from one form to another form”, this

statement is known as 1st law of thermodynamics.OR

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Another statement is "the amount of total energy in universe is constant."Limitations

On the basis of 1st law of thermodynamics is not possible to know whether a reaction will occur by its own ornot.

It is also not possible to know the proportion of the conversion of reactant into product (i.e. up to what extent areactant will convert into product.)

Following examples will explain the limitations of 1st law of thermodynamics.1. If a piece of ice in a plate at room temperature it obtains heat from the atmosphere it and melts. The heat

received by the ice is equal to the heat lost by atmosphere. This reaction occurs by its own and obeys 1st law. Asagainst this the 1st law is obeyed when the water in the plate is converted to ice by loosing heat, but this reactiondoes not occur by its own.

2. Similarly the hot tea taken in a saucer kept at room temperature becomes cold of its own but, if it gets hot againthe 1st law is obeyed, but here also the reverse reaction does not occur by it self.

(a) (b)3. As shown in figure (a) and (b) a bulb filled with a gas is connected to an evacuated bulb through the valve and it

is closed. If the valve between the bulb is opened, the gas from one bulb is diffusing into other evacuated bulbuntil the pressure in both the bulb becomes equal. This reaction occurs by its own and also obeyed 1stlaw. If weassume that gas from one of the bulbs is diffused to the other bulb then the 1st law is obeyed but the reactiondoes not occur by its own.

4. If a drop of ink is added into beaker filled with water it diffuses by its own but its opposite reaction does notoccur.

5. When equal volumes of equal concentration of HCl and NaOH are mixed, NaCl and H2O will be formed byneutralization reaction by its own and 56 Kj/mole energy evolved. Here 1st law is obeyed. If NaCl dissolved inwater and 56 Kj/mole energy is supplied then reaction will not occur by its own and do not produce HCl andNaOH.

Thus whether the reaction will take place by its own or not cannot be predicted with the help of 1st law ofthermodynamics.

Explain limitation of 1st law of thermodynamics with the help of reaction involving chemical change. {Answer lasttwo points of above question.} [H.W.]

Q 44. Discuss characteristics of spontaneous reaction. Hot water or a tea taken in a vessel becomes cold by it's own. When temperature becomes equal to the

temperature of surrounding, the decrease the temperature of hot water or tea stops. Thus until there is nothermal equilibrium between the system and surrounding hot water continues losing heat.

Diffusion of a gas filled in a bulb into evacuated bulb until the pressure in both bulbs become equal is also a typeof equilibrium.

All chemical reactions occur spontaneously until the equilibrium is established. At equilibrium concentration ofreactant and products remain constant.

Thus all spontaneous reactions occur in the direction to establish equilibrium between reactant and productsand when equilibrium is established the reaction attains equilibrium.

In short the reaction which occurs in a particular one direction by itself till it gets equilibrium with thesurrounding is called spontaneous.

Q 45: Mention the second law of thermodynamics and state its importance.Second law of thermodynamics is as follows: "Entropy of universe increases in all spontaneous reactions."

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OR "The free energy of system decreases in all spontaneous reactions."

The examples of spontaneous processes When two substances are kept in contact with each other having with different temperature, the heat flows

spontaneously from higher temperature to the low temperature. Water flows from high level to lower level.

Importance of 2nd law With the help of 2nd law it is possible to determine spontaneity of reaction can be determined. It is also possible to calculate equilibrium constant (K) for spontaneous reaction. From the value of K it is determined up to what extent reactants will convert in to products.

Que46: Define entropy and explain its physical form. "Entropy is a measure of randomness (disorder) of substance. Randomness can be explained by following

illustration. (Entropy increases as the randomness increase (With the increase in temperature.).

Solid liquid gas As shown in the figure the entropy of a crystalline substance is minimum due to regular arrangement in a

crystalline substance. The gaseous states have maximum entropy because randomness is highest in them athigher temperature while liquid state has entropy between solid and gaseous.

Randomness and entropy increases from solid liquid gas. A gas filled in a bulb is diffuse into evacuated bulb attached to it so volume of gas will increase. Therefore the

molecules of gas have more space to move so entropy and randomness will increase. Such kind of literary description of entropy is not useful to understand spontaneity of reaction. For such purpose

mathematical form is more important.

Q 47: Explain mathematical of entropy and write a note on it. Entropy is measure of randomness of a substance. It is denoted by S and it is state function. Entropy is extensive property of substance. At definite temperature and pressure entropy of 1 mole of substance is constant. If state system changes entropy of system also changes. The change In entropy is denoted by ∆S.

It's mathematical form is Sfinal-Sinitial = revqΔS=T

where qrev = heat gained or lost reversibly by a system at temp T.

Unit of ∆S will be Cal/Kelvin or Kcal /Kelvin. The value of absolute entropies of substances can be calculated.

Q 48: Explain how spontaneity of reaction is determined by using entropy. To determine whether reaction will occur spontaneously or not the change in entropies of system and

surrounding is calculated. If the sum of these two values is positive then reaction is spontaneous.

∆Ssys + ∆Ssurr > 0 (spontaneous reaction) If the sum of these two values is negative then reaction is non spontaneous

∆Ssys + ∆Ssurr < 0 (non spontaneous reaction) If the sum of these two values is zero then reaction is in equilibrium.

∆Ssys + ∆Ssurr = 0 (reactionis in equilibrium)

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Q 49: Give formulas to calculate entropy change for the process like fusion vaporization and sublimation. When a solid melts into a liquid or a liquid into gas at definite temperature the change in entropy of the system

is calculated from the values of molar heat of fusion and molar heat of vaporization of the said substance. At constant pressure and temperature

ΔHfusionΔSfusion T

ΔHvapourisationΔSvapourisation T

When the solid substance directly convert into gaseous state without being converting into liquid then suchprocess is called sublimation

ΔHsublimationΔSsublimation T

Q 50:Write a short note on expansion of an ideal gas in vacuum and entropy changes." When 1 mole of an ideal gas filled in a vessel is connected to an evacuated flask, the gas expands by itself and no

work is done by the system. W = P∆V = 0 where P=0 In this process gas does not lose or gain heat q= 0 and therefore ∆U =0 (∆U = q + w) This process is not reversible hence qirrev = 0 If this process is reversible then increase in volume and change in entropy is expressed by following equation.

V1ΔS=2.303R logV2

Above equation is derived as below.For one mole of ideal gas,

V1q =RT lnVrev 2

q Vrev 1=RlnT V2

V1ΔS=R lnV2

V1ΔS=2.303R logV2

Q 51. Why free energy is more convenient than entropy to determine spontaneity of reaction? To determine spontaneity of chemical reaction on the basis of entropy it is necessary to considered entropy

change of system and surrounding both according to ∆Ssys + ∆Ssurr >0. While to determine the spontaneity of the reaction on the basis of free energy one has to consider only the

entropy and enthalpy changes of system only thus free energy is more convenient than entropy to determinespontaneity of reaction.

Q 52. What is meant by free energy? What is the relationship between free energy change, enthalpy change andentropy change during chemical reaction? Explain how spontaneity of reaction can determine on the basis of thisrelationship? To determine spontaneity of chemical reaction on the basis of entropy it is necessary to considered entropy

change of system and surrounding. While to determine the spontaneity of the reaction on the basis of free energy one has to consider only the

entropy and enthalpy of system only. Free energy is a state function and is an extensive property. Absolute value of free energy cannot be determined because absolute value of enthalpy can not be determined. Free energy is related with enthalpy and entropy as G = H –TS During chemical reaction at constant temperature change in free energy, change in enthalpy and change in

entropy is related as ∆G= ∆H-T ∆S Spontaneity of reaction can be determined on the basis of free energy.1. If free energy change of reaction is negative then reaction is spontaneous. (∆G =-ve)2. If free energy change of reaction is positive then reaction is non spontaneous. (∆G = +ve)3. If free energy change of reaction is zero then reaction is in equilibrium. (∆G = 0)

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Q 53 why absolute value of free energy can not be determined? Free energy depends on enthalpy and as absolute value of enthalpy can not be determined there fore absolute

value of free energy can not be determined.

Q 54 Define standard free energy of formation and explain it with suitable example. For elements in their standard state free energy is considered zero on the basis of this standard free energy of

formation of different substances is determined. "If one mole of a compound in standard state is directly produced from the elements in their standard states,

the value of change in free energy of reaction is called standard free energy of formation of that compound." For example if one mole of H2O is formed by reaction between 1moleH2 and ½mole O2 at 25°C and 1 atm

pressure the change in free energy is -56.69 kcal/molH2(g)+ ½ O2(g) H2O(l)

The free energy of liquid water 0 -1ΔG 56.69Kcalmolf H O2because free energy of formation of H2 and O2 is

zero.

Q 55 Explain relationship between standard free energy change and equilibrium constant of reaction. The relation between standard free energy change is related with equilibrium constant of reaction by following

formula :∆Go =−RT In K∆Go= -2.303 RT log K

If the system is consist of gaseous state then K = Kp and if reactants and products form homogenous liquidsolution then K = Kc.

If ∆G=0 then K= 1If ∆G = +ve then K<1If ∆G=-ve then K>1

Q 56 Write a note on Gibbs free energy and Useful work. The value of ∆G is the results of the tendency of chemical reaction to occur spontaneously. Thus ∆G is related with the spontaneity of reaction and it is also related with the useful work. The value of change in free energy associated with any process, is the maximum work can obtained from

process. ∆G = -Wmax

Thus any spontaneous process can be useful to perform work in the system. If the value of ∆G is negative thenmore work can be done by the process.

When any electrochemical cell is in operation, the electric work done by the cell is re1ated to free energy(∆G) ofthe reaction by ∆G = + Welect

But the electrical work done by a cell is related to the potential ( Ecell)of the cell and quantity of electricity ( nF)obtained from the cell as Welect = -nFEcell

Where F = Faraday constant 96500 coulomb, n= the no. of electrons passing from the external circuit. If the cell is in standard state, then the change in free energy associated with the reaction and the potential of

standard electrochemical cell as ∆G0= -nFEcell, Where Ecell= cell potential of standard cell.

Q 57 Give formulas to calculate free energy in a chemical reaction. The change in standard free energy in a chemical reaction can be calculated on the basis of the standard free

energy of formation of reactants and products,0 0 0ΔG = ΔG ΔGf (prod) f (react)

At constant temperature the volume of an ideal gas changes with change pressure. The change in free energyalong with this change can be determined by following equation.

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PΔG=nRT ln

P Where n= no.of moles of the gas, while P1 and P2 are pressure of gas for initial and final states respectively. For an ideal gas P1V1 = P2V2 so above equation can be written in the following form

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V1ΔG=nRT lnV2

Where n=no.of moles of the gas while V1 and V2 are volume of gas for initial and final states respectively.

Q58 Mention whether the value of ∆G positive or negative of the ice kept at 260K and 280K. The value of ∆G for ice at will be positive (260<273) while the value of ∆G for ice at 280K will be negative (280

>273).

Q 59 State the limitations of 2nd law of thermodynamics. If a chemical reaction is spontaneous then, values of equilibrium constant of such reaction can be determined on

the basis of 2nd law of thermodynamics but this law is not able to give information about the rates of reactions. Thus 2nd law of thermodynamics can predict about spontaneity of reaction and extent of reaction but it does not

give any information about the rate of reaction.

Q 60 State the 3rd law of thermodynamics and explain it. “The value of entropy of a perfectly crystalline substance at absolute zero temperature is zero." Generally values of entropy increase with the increase in temperature, because with the increase in temperature

the randomness increases in the molecules of substance. Randomness of molecules increases with the increase in temperature because with the increase in temperature,

rotational, vibrational, and linear velocity of the substance increases. While with the decrease in temperature the orderliness in the molecules of the substance increases and the

value of entropy decreases. On the basis of this in 1906 German scientist Nernst mentioned that with decrease in temperature the entropy

of perfectly pure crystalline substance decreases on the basis of which 3rdlaw of thermodynamics is expressedas under.

"The value of entropy of a perfectly crystalline substance at absolute zero temperature is zero. " Thus at absolute zero temperature arrangement of the constituent particles in perfectly pure crystalline

substance is completely in order. [i.e. all constituent particles are in their ordered position and there is no defectin the crystal at absolute zero temperature.]

At absolute temperature, the kinetic energy of molecules of a substance is almost negligible but their potentialenergy is not zero. As a result value of internal energy is not zero. In addition to this internal energy of substanceis its characteristic. There fore absolute value of entropy can be calculated but the absolute value of internalenergy can not be calculated

The entropy of 1 mole of substance is known as molar entropy Sm. At definite temperature and standard state, the entropy of one mole of a substance is called standard entropy

oSm . The standard entropy of substance is also called the absolute entropy. The unit molar entropy is calorie K-1mole1. It is normally observed that normally entropy of solids is less than that of liquid, which have even lower entropy

than gaseous substances.

∆H ∆S Temperature ∆G Reaction−Ve +ve At any temp −ve spontaneous+Ve −ve At any temp +ve Non spontaneous−Ve −Ve Low temp −Ve spontaneous−Ve −Ve High temp +Ve Non spontaneous+Ve +Ve Low temp +Ve Non spontaneous+Ve +Ve High temp −Ve spontaneous