Limiting reagent is $A$ so if $5$ mol of $A$ reacts then energy liberated $=10^{2} \,\mathrm{kJ}$ then for $3$ mole is $60\, \mathrm{kJ}$.
$\boxed{\Delta H = \Delta E + \Delta {n_g}RT}$
$\Delta E=-60\,kJ$
$\Delta n_{e}=-5$
$\Delta \mathrm{H}=-60+(-5) \,\mathrm{R} \times 300 \times 10^{-3}=-60-1500$
$\times 10^{-3} \,\mathrm{R}=-(60+1.5\, \mathrm{R})$
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$l$ $m$
$\Delta_{f} \mathrm{H}^{\ominus}$ for $\mathrm{KCl}=-436.7 \,\mathrm{~kJ}\, \mathrm{~mol}^{-1}$
$\Delta_{\text {sub }} \mathrm{H}^{\ominus}$ for $\mathrm{K}=89.2 \,\mathrm{~kJ}\, \mathrm{~mol}^{-1}$
$\Delta_{\text {ionization }} \,\mathrm{H}^{-}$ for $\mathrm{K}=419.0\, \mathrm{~kJ}\, \mathrm{~mol}^{-1}$
$\Delta_{\text {electron gain }} \mathrm{H}^{\ominus}$ for $\mathrm{Cl}_{(\text {e) }}=-348.6 \,\mathrm{~kJ} \,\mathrm{~mol}^{-1}$
$\Delta_{\mathrm{bond}} \mathrm{H}^{-}$ for $\mathrm{Cl}_{2}=243.0 \,\mathrm{~kJ} \,\mathrm{~mol}^{-1}$
The magnitude of lattice enthalpy of $\mathrm{KCl}$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is ..... . (Nearest integer)