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Q. At constant $T$ and $p$ which one of the following statements is correct for the reaction?
$CO ( g )+\frac{1}{2} O _{2}( g ) \longrightarrow CO _{2}( g )$

AMUAMU 2001

Solution:

Key Idea The relation between $\Delta H$ and $\Delta E$ is given by
$\Delta H=\Delta E+\Delta n_{g} R T$
where $\Delta n_{g}=\Sigma$ product of gaseous mole
$- \Sigma$ reactant of gaseous mole
For the reaction
$CO (g)+\frac{1}{2} O _{2}(g) \longrightarrow CO _{2}(g)$
$\Delta n_{g}=\left(1 mol\right.$ of $\left.CO _{2}\right)$
- $\left(1 mol\right.$ of $CO +0.5\, mol$ of $O _{2}$ ) $\Delta n_{g}=-0.5$, ie, it is negative.
Hence, $\Delta H < \Delta E$
If $\Delta n_{g}=0, \Delta H =\Delta E$
$\Delta n_{g} =+\text { ve } \Delta H > \Delta E$