Chemistry · 2016
JEE Main · 9 April 2016, Shift 1 · Q42
The plot shows the variation of - ln K_p versus temperature for the two reactions. M ( s )+1/2 O_2( g ) → MO ( s ) and C ( s )+1/2 O_2( g ) → CO ( s…
The plot shows the variation of $\displaystyle -\ln \mathrm{K}_{\mathrm{p}}$ versus temperature for the two reactions.
$$\begin{aligned}
& \mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{MO}(\mathrm{~s}) \text { and } \\
& \mathrm{C}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{~s})
\end{aligned}
$$
Identify the correct statement :
Identify the correct statement :Official answer
(2)
At T<$\displaystyle 1200$ K, the reaction $\displaystyle \mathrm{MO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{M}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$ is spontaneous.
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JEE Main 2016 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.