8 th Week Chap(12-13) Thermodynamics and Spontaneous Processes • Standard State: The Thermodynamically stable state for pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at P= 1.0 atm and some specified T in each case • Reversible and Irreversible processes: Processes that occur through a series of equilibrium states are reversible. Adiabatic(q=0) paths are reversible • Thermodynamic Universe= System + Surrounding Isolated : No Energy and Matter can go in or out Adiabatic: No heat goes in or out • Entropy (S): measure of disorder Absolute Entropy S=k B lnNumber of Available Microstate • Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without work S ≥ q/T In-quality of Clausius S = q/T for reversible (Isothermal)
Transcript
8th Week Chap(12-13) Thermodynamics and Spontaneous Processes• Standard State: The Thermodynamically stable state for
pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at P= 1.0 atm and some specified T in each case
• Reversible and Irreversible processes:Processes that occur through a series of equilibrium states are reversible. Adiabatic(q=0) paths are reversible
• Thermodynamic Universe= System + Surrounding Isolated : No Energy and Matter can go in or outAdiabatic: No heat goes in or out
• Entropy (S): measure of disorderAbsolute Entropy S=kBln Number of Available Microstate
• Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without workS ≥ q/T In-quality of ClausiusS = q/T for reversible (Isothermal) processes S > q/T for irreversible processes
Midterm Friday: Ch 9, 10, 11.1-11.3, 11.5, 18, 12.1-12.6 One side of 1 page notes(must be hand written), closed book Review Session Today 6-7 pm, in FRANZ 1260
What is the heat of reaction when V is not constant:When the system can do work against and external pressure !
Use the Enthalpy H=U + PV
Since H = U + (PV) if P=const and not V
H = U + PV but w = -PV
Therefore H = U - w but U = q + w by the 1st Law
@ P=const. qP= H
Note that the Enthalpy is a state function and is thereforeIndependent of path; It only depends on other state functionsi.e.,
H=U + PV !
Energy transferred as heat @ constant pressure
A Flame: CH4 + 2O2 CO2 + 2H2O(l) combustion gives off energy that is transferred as heat(q) to the gas in the piston which can do work against the Pext but sinceV is held, no pressure volume
q>0
Energy transferred as heat @ constant pressureqP = H
For Chemical Reactions AB H=HB – HA= Hprod – Hreac
Heat transfer required to Change n mole of ice to steam at 1 atm
H2O P-T Phase Diagram
T1 T2
q= qice + nHfus + qwat + nHevap+ qvap
qice=ncp(s)T, qwat=ncp(l) T and qvap= ncp(g) T
For exampleIf a piece of hotmetal is placed ina container witha mole of water that was initially @Temperature T1 andThe water and metal came to equilibriumAt temperature T2
Physical Reaction/phase transitions
Fig. 12-14, p. 506
Hess’s Law applies to all State Function
A
B
D
C
AD (1)AB (2)BC
(3)CD
Example of Hess’s LawC(s,G) + O2(g) CO2(g) = -393.5 kJ CO2((g ) CO(g) + ½O2(g) = + 283 kJC(s,G) + O2(g) CO(g) + ½ O2(g) = ?
Example of Hess’s LawC(s,G) + O2(g) CO2(g) = -393.5 kJ CO(g) + ½O2(g) CO2((g) = -283 kJC(s,G) + O2(g) CO(g) + ½ O2(g) == -110.5 kJ
The Standard State:Elements are assigneda standard heat of Formation f= 0Solids/liquids in their stable form at p=1 atmIn species solution @ concentration of 1Molar
For compounds the f
Is defined by its formation From its elements in theirStandard states:
f = 0
In General for a reaction, with all reactants and products are at a partial pressure of one atm and/or concentration of 1 Molar
aA + bB fF + eE
The Standard Enthalpy Change at some specified Temperature
(rxn) f(prod) -
f(react)
(rxn) = f f(F) + e
f(E) – {af(A) + b
f(B)}
Elements in their standard statesElements in their standard states
