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Question
Standad Gibbs energy of the reaction, ΔrGo at a certain temperature can be computed as ΔrGo=ΔrHo−TΔrSo and the change in the value of ΔrHo and ΔrSo for a reaction with temperature can be copmuted as follows :
ΔrHoT2−ΔrHoT1=ΔrCop(T2−T1)
ΔrSoT2−ΔrSoT1=ΔrCoplnT2T1
ΔrGo=ΔrHo−TΔrSo
and by ΔrGo=−RTlnKeq
Consider the following reaction,
CO(g)+2H2(g)⇌CH3OH(g)
Given : ΔfHo(CH3OH(g))=−201 kJ mol−1
ΔfHo(CO(g))=−114 kJ mol−1
So(CH3OH(g))=240 kJ mol−1
So(H2(g))=29 J K−1 mol−1
So(CO(g))=198 J mol−1 K−1
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)
ΔrHoT2−ΔrHoT1=ΔrCop(T2−T1)
ΔrSoT2−ΔrSoT1=ΔrCoplnT2T1
ΔrGo=ΔrHo−TΔrSo
and by ΔrGo=−RTlnKeq
Consider the following reaction,
CO(g)+2H2(g)⇌CH3OH(g)
Given : ΔfHo(CH3OH(g))=−201 kJ mol−1
ΔfHo(CO(g))=−114 kJ mol−1
So(CH3OH(g))=240 kJ mol−1
So(H2(g))=29 J K−1 mol−1
So(CO(g))=198 J mol−1 K−1
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=SoCH3OH−SoCO−SoH2=−16 J K−1 mol−1
ΔrHo=ΔHoCH3OH−ΔoCO−2ΔHofH2=−87 kJ mol−1
ΔrSo320−ΔrSo300=ΔrCP[T2−T1]
Where,
ΔrCop=44−29.4−2×28.8
=−43 J/K-mol
ΔrSo320=−16+(−43)ln(320320)
=−18.58
ΔrHo320=ΔrHo300−T.ΔrCoP[T2−T1]
=−87+(−43)×201000
=−87.86 kJ/mol
ΔrGo320=ΔrHo320−T.ΔrSo320
=−87.86−320×(−18.58)1000
=−81.91 kJ/mol
Hence, x=81.91
ΔrHo=ΔHoCH3OH−ΔoCO−2ΔHofH2=−87 kJ mol−1
ΔrSo320−ΔrSo300=ΔrCP[T2−T1]
Where,
ΔrCop=44−29.4−2×28.8
=−43 J/K-mol
ΔrSo320=−16+(−43)ln(320320)
=−18.58
ΔrHo320=ΔrHo300−T.ΔrCoP[T2−T1]
=−87+(−43)×201000
=−87.86 kJ/mol
ΔrGo320=ΔrHo320−T.ΔrSo320
=−87.86−320×(−18.58)1000
=−81.91 kJ/mol
Hence, x=81.91
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