Determine the amount of heat (in kcal) given off at constant volume when $0.5 m o l$ of $N_{2}$ & $1.5$ mol $H_{2}$ reacted ording to equation at $300 K$ .
$N_{2} (g) + 3 H_{2} (g) \to {2 N H}_{3} (g); Δ_{r} H_{300} = - 380 k c a l / m o l$
(Given: R=2 cal/ mol-K)

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Solution

since only $\frac{1}{2}$ mole of reaction takes place,
$∴$ heat released = $\frac{Δ_{r} H_{300}}{2}$
= $190 k c a l$

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