Question

Show that the reaction
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶C{O}_2\left(g\right)$ at $300K$
is spontaneous and exothermic, when the standard entropy change is $-0.094kJmo{l}^{-1}{K}^{-1}$. The values of standard Gibbs free energy of formation of $C{O}_2$ and $CO$ are $-394.4$ and $-137.2kJmo{l}^{-1}$, respectively.

• A. $\Delta G=-257.2kJ$, spontaneous
• B. $\Delta G=+257.2kJ$,non spontaneous
• C. $\Delta G=-257.2kJ$, non spontaneous
• D. None of these

Answer 1

The correct option is A $\Delta G=-257.2kJ$, spontaneous

The balanced chemical reaction for the combustion of CO to $C{O}_2$ is as shown.
$CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶C{O}_2\left(g\right)$
The expression for the standard Gibbs free energy change during the reaction is
$\Delta G^{\ominus }=\left\{G^{\ominus }\right(C{O}_2)-\left[G^{\ominus }\right(CO)+\frac{1}{2}{G}^{\ominus }(O_2\left)\right]\}$
$=\{-394.4-[-137.2+0]\}$
$=-257.2kJ$
The standard entropy change for the reaction is
$\Delta S^{\ominus }=-0.094kJmo{l}^{-1}{K}^{-1}$
Gibbs-Helmholtz equation gives the relationship between the standard Gibbs free energy change, the standard enthalpy change and the standard entropy change for the reaction.
$\Delta G^{\ominus }=\Delta H^{\ominus }-T\Delta S^{\ominus }$
or $-257.2=\Delta H^{\ominus }-300\times (-0.094)$
or $\Delta H^{\ominus }=-285.4kJ$
The reaction is spontaneous because $\Delta G^{\ominus }$ is negative.
The reaction is exothermic because $\Delta H^{\ominus }$ is negative.