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Standad Gibbs energy of the reaction, ΔrGo at a certain temperature can be computed as ΔrGo=ΔrHoTΔrSo and the change in the value of ΔrHo and ΔrSo for a reaction with temperature can be copmuted as follows :
ΔrHoT2ΔrHoT1=ΔrCop(T2T1)
ΔrSoT2ΔrSoT1=ΔrCoplnT2T1
ΔrGo=ΔrHoTΔrSo
and by ΔrGo=RTlnKeq
Consider the following reaction,
CO(g)+2H2(g)CH3OH(g)
Given : ΔfHo(CH3OH(g))=201 kJ mol1
ΔfHo(CO(g))=114 kJ mol1
So(CH3OH(g))=240 kJ mol1
So(H2(g))=29 J K1 mol1
So(CO(g))=198 J mol1 K1
Cop, m(H2)=28.8 J/mol-K
Cop, m(CO)=29.4 J/mol-K
Cop, m(CH3OH)=44 J/mol-K
and ln(320300)=0.06
All data were taken at 300 K.
The ΔrGo (in kJ/mol) at 320 K is x. Find the value of x. (Report the answer to the closest integer)

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Solution

ΔrSo=SoCH3OHSoCOSoH2=16 J K1 mol1
ΔrHo=ΔHoCH3OHΔoCO2ΔHofH2=87 kJ mol1
ΔrSo320ΔrSo300=ΔrCP[T2T1]
Where,
ΔrCop=4429.42×28.8
=43 J/K-mol
ΔrSo320=16+(43)ln(320320)
=18.58

ΔrHo320=ΔrHo300T.ΔrCoP[T2T1]
=87+(43)×201000
=87.86 kJ/mol

ΔrGo320=ΔrHo320T.ΔrSo320
=87.86320×(18.58)1000
=81.91 kJ/mol
Hence, x=81.91


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