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Heat of Reaction

Calculate the...

Question

Calculate the heat change in the reaction,
$4 {N H}_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (l)$
at $298 K$ given that heats of formation at $298 K$ for ${N H}_{3} (g)$ and $H_{2} O (l)$ are $- 46.0$ and $- 286 k J {m o l}^{- 1}$ respectively.

Δ H^{o} = - 1532 k J

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Δ H^{o} = 1532 k J

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Δ H^{o} = - 132 k J

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Δ H^{o} = - 152 k J

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Solution

The correct option is A $Δ H^{o} = - 1532 k J$
$4 N H_{3} (g) + 3 O_{2} (g) ⟶ 2 N_{2} (g) + 6 H_{2} O (l) Δ H_{r} = \sum Δ H_{f} (p r o d u c t) - \sum Δ H_{f} (r e a c t a n t) Δ H_{r} = (2 Δ H_{f} (N_{2}) + 6 Δ H_{f} (H_{2} O (l))) - (4 Δ H_{f} (N H_{3} (g)) + 3 Δ H_{f} (O_{2} (g))) Δ H_{r e x} = [2 \times 0 + 6 \times (- 286)] - [4 \times (- 46) + 3 \times 0] = - 1716 + 186 Δ H_{r e x} = - 1532 k J / m o l$

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