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At constant $T$ and $P$, which one of the following statements is correct for the reaction, $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)$
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The correct answer is:
$\Delta H \lt \Delta E$
$\Delta n_g=1-\frac{3}{2}=\frac{-1}{2}$, As $\Delta n_g$ is negative, thus $\Delta H \lt \Delta E$
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