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 )G^(@)` at 320 K is :

Options

A. `-48295.2` kJ/mol
B. `240.85` kJ/mol
C. 240.85 kJ/mol
D. `-81.91` kJ/mol

Answer

D

Explanation

`Delta_(r )S^(@)=S_(CH_(3)OH)^(@)-S_(CO)^(@)-2S_(H_(2))^(@)=-J//"K-mol"`
`Delta_(r )H^(@)=Delta_(f)H^(@)(CH_(3)OH)-Delta_(f)H^(@)(CO)-2.Delta_(f)H^(@)(H_(2))`
=-87 kJ/mol
`Delta_(r )S_(320)^(@)-Delta_(r )S_(300)^(@)=Delta_(r )C_(p)[T_(2)-T_(1)]`
where `Delta_(r )C_(p)^(@)=44-29.4 -2 xx28.8`
`=-43 J//"K-mol"`
`Delta_(r )S_(320)^(@)=-16+(-43)ln.(320)/(320)`
`=-18.58`
`Delta_(r)H_(320)^(@)=Delta_(r)H_(300)^(@)+Delta_(r)C_(p)^(@)[T_(2)-T_(1)]`
`=-87+((-43)xx20)/(1000)`
=-87.86 kJ/mol
`Delta_(r )G_(320)^(@)=Delta_(r )H_(320)^(@)-T.Delta_(r)S_(320)^(@)`
`=-87.86-(320xx(-18.58))/(1000)`
`=-81.91 " kJ"//"mol"`