Standard Gibb's energy of reaction `(Delta_(r )G^(@))` at a certain temperature can be computed `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^(@)`
and `" by "Delta_(r )G^(@)=-"RT " ln K_(eq)`.
Consider the following reaction : `CO(g)+2H_(2)(g)iffCH_(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))=28.8 " J"//"mol-K"`
`C_(p,m)^(@)(CO)=29.4 " J"//"mol-K, "C_(p,m)^(@)(CH_(3)OH)=44 " J"//"mol-K"`
and `" "ln ((320)/(300))=0.06`, all data at 300 K
`Delta_(r )H^(@)` at 320 K is :

Options

A. `-288.86 kJ//mol`
B. `-289.1 kJ//mol`
C. `-87.86 kJ//mol`
D. None of these

Answer

C

Explanation

`Delta H_(320)^(0) - Delta H_(300)^(0) = Delta C_(p) (T_(2) - T_(1))`