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Q.

Standard Gibb's energy of reaction (∆r ,G°) at a certain temperature car computed and as ΔTG∘=ΔΓH∘−T⋅ΔrS∘and the change in the value of ΔrH∘and ΔrS∘ for a reaction with temperature can be computed as follows: ΔrHT2∘−ΔrHT1∘=ΔrCp∘T2−T1ΔrST2∘−ΔTST1∘=ΔrCp∘ln⁡T2T1ΔrG∘=ΔrH∘−T⋅ΔrS∘and  by Δ−G∘=−RTln⁡Keq.Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Given :ΔfH∘CH3OH,g=−201kJ/mol′;ΔfH∘(CO,g)=−114kJ/molS∘CH3OH,g=240J/K−mol; S∘H2,g=29JK−1mol−1S∘(CO,g)=198J/mol−K; Cp∘mH2=28.8J/mol−KC∘n−m(CO)=29.4J/mol−K; C0,m∘CH3OH=44J/mol⋅K and ln⁡320300=0.06, all data at 300KΔ,S∘ at 300K for the reaction is  ΔrH∘ at 300K for the reaction is : ΔrS∘ at 320K is :

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a

152.6 J/K-mol

b

181.6 J/K-mol

c

-16 J/K-mol

d

None of these

e

-87 kJ/mol

f

87 kJ/mol

g

-315 kJ/mol

h

-288 kJ/mol

i

155.18 J/mol-K

j

150.02 J/mol-K

k

172 J/mol-K

l

None of these

answer is , , .

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Detailed Solution

ΔrS=240−198−29×2=−16ΔrH=−201+114=−87ΔrS=−16+∫300320 ΔrCpdTTΔrS=−16+(44−29.4−2×28.8)ln⁡320300ΔrS=−16+(−43)ln⁡320300=−18.78
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