Question

Determine $\Delta G^o$ for the following reaction.

$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)$; $\Delta H^o=-282.84kJ$

[Given: $S_{C{O}_2}^o=213.8$ JK${}^{-1}$mol${}^{-1}$, $S^o{}_{CO\left(g\right)}=197.9$ J K${}^{-1}$mol${}^{-1}$ $S^o{}_{O_2}=205.8$ J K${}^{-1}$mol${}^{-1}$]

• A. $-157.33$ kJ
• B. $+201.033$ kJ
• C. $-256.91$ kJ
• D. $+257.033$ kJ

Answer 1

The correct option is C $-256.91$ kJ

The reaction is:
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{O}_2\left(g\right)$
The standard entropy change for the reaction is:

$\Delta S^o=S_{C{O}_2}^0-[S_{CO}^o+0.5S_{O_2}^o]$

$\Delta S^0=213.8-[197.9+0.5\times 205.8]=-87J/K/mol$
The change in standard Gibbs free energy for the reaction is:

$\Delta G^o=\Delta H^0-T\Delta S^o$

$\Delta G^o=-282840-298\times (-87)=-256914J/mol=-256.914kJ/mol$