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: Δ$$rHT2$$∘−Δ$$rHT1$$∘$$=$$ΔrCp∘$$T2$$−$$T1$$Δ$$rST2$$∘−Δ$$TST1$$∘$$=$$ΔrCp∘ln⁡$$T2T1$$ΔrG∘$$=$$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 $$300K$$Δ,S∘ at $$300K$$ for the reaction is ΔrH∘ at $$300K$$ for the reaction is : ΔrS∘ at $$320K$$ is :

Question

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: Δ$$rHT2$$∘−Δ$$rHT1$$∘$$=$$ΔrCp∘$$T2$$−$$T1$$Δ$$rST2$$∘−Δ$$TST1$$∘$$=$$ΔrCp∘ln⁡$$T2T1$$ΔrG∘$$=$$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 $$300K$$Δ,S∘ at $$300K$$ for the reaction is  ΔrH∘ at $$300K$$ for the reaction is : ΔrS∘ at $$320K$$ is :

Infinity Learn · Accepted Answer

Δ$$rS=240$$−$$198$$−$$29$$×$$2=$$−$$16$$Δ$$rH=$$−$$201+114=$$−$$87$$Δ$$rS=$$−$$16+$$∫$$300320$$ ΔrCpdTTΔ$$rS=$$−$$16+(44$$−$$29.4$$−$$2$$×$$28.8)ln$$⁡$$320300$$Δ$$rS=$$−$$16+($$−$$43)ln$$⁡$$320300=$$−$$18.78$$

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