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)

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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

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