The molar heat of formation of ${N H}_{4} {N O}_{3} (s)$ is $- 367.5 k J$ and those of $N_{2} O (g)$ and $H_{2} O (l)$ are $+ 81.46 k J$ and $- 285.78 k J$ respectively. at $25^{o} C$ and $1$ atmospheric pressure. Calculate the $Δ H$ and $Δ U$ for the reaction,
${N H}_{4} {N O}_{3} (s) ⟶ N_{2} O (g) + 2 H_{2} O (l)$

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Solution

$Δ H^{o} = Δ H_{f (p r o d u c t s)}^{o} - Δ H_{f (r e a c t a n t s)}$
$[Δ H_{(N_{2} O)}^{o} + 2 \times Δ H_{(H_{2} O)}^{o}] - [Δ H_{({N H}_{4} {N O}_{3})}^{o}]$
$= 81.46 + 2 \times (- 285.78) - (- 367.5) = - 122.56 k J$
We know that $Δ H = Δ U + Δ n R T$
$Δ U = Δ H - Δ n R T$
$Δ n = 2, R = 8.314 \times 10^{- 3} k J {m o l}^{- 1} K^{- 1}, T = 298 K$
$Δ U = - 122.56 - (1) (8.314 \times 10^{- 3}) (298) = - 125.037 k J$

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The molar heat of formation of $N H_{4} N O_{3} (s)$ is $- 367.5 K J$ and those of $N_{2} O (g)$ and $H_{2} O (l)$ are $+ 81.46 K J$ and $- 285.78 K J$ respectively at $25^{o} C$ and 1 atmospheric pressure. The $△$ U for the reaction
$N H_{4} N O_{3} (s) \to N_{2} O (g) + 2 H_{2} O (l)$