Question

In which of the following reactions, standard reaction entropy change $\left(\Delta S^0\right)$ is positive and standard Gibbs's energy change $\left(\Delta G^0\right)$ decreases sharply with increasing temperature?

• A. $Mg\left(s\right)+\frac{1}{2}{O}_2\left(g\right)⟶MgO\left(s\right)$
• B. $\frac{1}{2}C\left(\text{graphite}\right)+\frac{1}{2}{O}_2\left(g\right)⟶\frac{1}{2}C{O}_2\left(g\right)$
• C. $C\left(\text{graphite}\right)+\frac{1}{2}{O}_2\left(g\right)⟶CO\left(g\right)$
• D. $CO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶C{O}_2\left(g\right)$

Answer 1

The correct option is C $C\left(\text{graphite}\right)+\frac{1}{2}{O}_2\left(g\right)⟶CO\left(g\right)$

$C_{\left(s\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to {CO}_{\left(g\right)};\Delta \eta =+\frac{1}{2};$
Also the moles of gases increase and therefore entropy change $\left(\Delta S\right)$ is positive.
An increase in temperature will cause more change in $T\Delta S$.
Also it is a combustion reaction and thus $\Delta H=-ve$

Since $\Delta G=\Delta H-T\Delta S$

$=-ve-(+ve)=-ve$