Question

$\Delta G^{0}vsT$ plot in the Ellinghm's diagram slopes downwards for the reaction

• A. $Mg+\frac{1}{2}{O}_2\to MgO$
• B. $2Ag+\frac{1}{2}{O}_2\to A{g}_{2}O$
• C. $C+\frac{1}{2}{O}_2\to CO$
• D. $CO+\frac{1}{2}{O}_2\to C{O}_2$

Answer 1

The correct option is C $C+\frac{1}{2}{O}_2\to CO$

During the reaction the gas is utilized so, the entropy becomes negative which leads to the positive value of the term, TΔS. So, slope ΔS comes positive for most of the reduction reaction.

For $C+\frac{1}{2}{O}_2\to CO$, the oxygen gas is consuming but the formed product is also a gas so, the number of gaseous species in increasing hence entropy increasing so, for the oxidation reaction of carbon to carbon monoxide the slope is downward. So, option (A) is correct.