Question

The plot shows the variation of $-\ln K_p$ versus temperature for the two reactions.
$M\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to MO\left(s\right)$ and
$C\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to CO\left(s\right)$
Identify the correct statement:

• A. At $T>1200$ $K,$ carbon will reduce $MO\left(s\right)$ to $M\left(s\right)$
• B. At $T<1200$ $K,$ the reaction
  $MO\left(s\right)+C\left(s\right)\to M\left(s\right)+CO\left(g\right)$ is
  spontaneous.
• C. At $T<1200K,$
  oxidation of carbon is unfavourable.
• D. Oxidation of carbon is favourable at
  all temperatures.

Answer 1

Answer: (A)

The correct option is A At $T>1200$ $K,$ carbon will reduce $MO\left(s\right)$ to $M\left(s\right)$

solution:

$(ln{K}_p=\frac{\Delta G^{\circ }}{RT}$)

At $(T>1200$) K, carbon will reduce $\left(MO\right(s)$) to $\left(M\right(s)$)$(\Delta G^{\circ }=-RTln{K}_p$)

$(\Delta G^{\circ }\propto Temperature$)

At high temperature $(\Delta G^{\circ }$) value is very high so non-feasible or non-spontaneous process takes place and carbon will reduce $\left(MO\right(s)$) to $M\left(s\right)$.

Hence, option A is correct.