The plot shows the variation of - ln K_p vers | vedclass

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(D) The relationship between Gibbs free energy change and equilibrium constant is $\Delta G^\circ = -RT \ln K_p$,which can be rewritten as $-\ln K_p =

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At $T > 1200 \ K$,carbon will reduce $MO_{(s)}$ to $M_{(s)}$.

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(D) The relationship between Gibbs free energy change and equilibrium constant is $\Delta G^\circ = -RT \ln K_p$,which can be rewritten as $-\ln K_p = \frac{\Delta G^\circ}{RT}$.For a reaction to be spontaneous,$\Delta G^\circ From the Ellingham diagram,at $T > 1200 \ K$,the value of $-\ln K_p$ for the reaction $C_{(s)} + \frac{1}{2} O_{2(g)} 	o CO_{(g)}$ is lower than that for $M_{(s)} + \frac{1}{2} O_{2(g)} 	o MO_{(s)}$.This indicates that the formation of $CO$ is more thermodynamically stable than $MO$ at $T > 1200 \ K$.Therefore,carbon can reduce $MO_{(s)}$ to $M_{(s)}$ at temperatures above $1200 \ K$ according to the reaction $MO_{(s)} + C_{(s)} 	o M_{(s)} + CO_{(g)}$.

The plot shows the variation of $- \ln K_p$ versus temperature for the two reactions:
$M_{(s)} + \frac{1}{2} O_{2(g)} \to MO_{(s)}$ and $C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$
Identify the correct statement.

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The plot shows the variation of $- \ln K_p$ versus temperature for the two reactions:$M_{(s)} + \frac{1}{2} O_{2(g)} \to MO_{(s)}$ and $C_{(s)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}$Identify the correct statement.