1.

Standard Gibb's energy of reaction (Delta_(r)G^(@)) at a certain temperature can be computed as Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows Delta_(r)H_(T_(2))^(@)-Delta_(r)H_(T_(1))^(@)=Delta_(r)C_(p)^(@)(T_(2)-T_(1))Delta_(r)S_(T_(2))^(@)-Delta_(r)S_(T_(1))^(@)=Delta_(r)C_(p)^(@)ln((T_(2))/(T_(1)))Delta_(r)G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) "andby "Delta_(r)G^(@)=-RT lnK_(eq)Consider the following reaction : CO(g)+2H_(2)(g)subCH_(3)OH(g)Given :Delta_(f)H^(@)(CH_(3)OH),g=-201 kJ//mol Delta_(f)H^(@)(CO, g)=-114 kJ//mol S^(@)(CH_(3)OH, g)=240 J//K-mol, S^(@)(H_(2),g)=29 JK^(-1)mol^(-1)S^(@)(CO, g)=198 J//mol-K, C_(p,m)^(@)(H_(2))=28.8J//mol-KC_(p,m)^(@)(CO)=29.4 J//mol-K, C_(p,m)^(@)(CH_(3)OH)=44 J//mol-Kand ln((320)/(300))0.06,all data at 300 KDelta_(r)H^(@) at 320 K is

Answer»

`-288.86 kJ//mol`
`-289.1 kJ//mol`
`-87.86 kJ//mol`
NONE of these

Solution :`Delta_(r)H^(@)" at 320 K = "Delta_(r)H_(320)^(@)+Delta_(r)C_(P)^(@)(T_(2)-T_(1))`
`=-87+((-43)xx20)/(1000)=-87.86KJ//mol`.


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