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Heat of Formation

For the react...

Question

For the reaction: $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ , $Δ H^{\circ} = - 123.77 K J m o l^{- 1}$ at $727^{\circ} C$ . Given below is the data of $C_{p}$ . The heat of formation of ammonia in $K J m o l^{- 1}$ at $27^{\circ} C$ is
Substance $N_{2}$ $H_{2}$ $N H_{3}$
$C_{p}$ 3.5R 3.5R 4R

44.42

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-44.42

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88.85

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-88.85

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Solution

The correct option is B -44.42
Using Kirchhoff's equation,
$Δ H_{2} (1000 K) = Δ H_{1} (300 K) + Δ C_{p} (1000 - 300)$
Here, $Δ H_{2} (1000 K) = - 123.77 K J m o l^{- 1}$
$Δ H_{1} (300 K) = ?$
$Δ C p = 2 C_{p} (N H_{3}) - [C_{p} (N_{2}) + 3 C_{p} (H_{2})]$
$= - 6 R = - 6 \times 8.314 \times 10^{- 3} K J m o l^{- 1} K^{- 1}$
$∴ - 123.77 = Δ H_{1} (300 K) - 6 \times 8.314 \times 10^{- 3} \times 700$
$∴ Δ H_{1} (300 K) = - 88.85 K J$ for two moles of $N H_{3}$
$Δ H_{1} = - 44.42 K J m o l^{- 1}$

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Similar questions

1000 K

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

△ H = - 123.77 k J m o l^{- 1}

Substance	$N_{2}$	$H_{2}$	$N H_{3}$
$P / R$	$3.5$	$3.5$	$4$

N H_{3}

300 K

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); △_{f} H^{⊖} = - 92 k J

k J m o l^{- 1}

N H_{3} (g)

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ_{r} H = - 92 K J

K J m o l^{- 1}

N H_{3} (g)

N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3 (g)}; △_{r} H^{0}

- 91.8 k J m o l^{- 1}

N H_{3}

2 N H_{3 (g)} \to N_{2 (g)} + 3 H_{2 (g)}, △_{r} H^{0}

N H_{3}

N_{2}

H_{2}

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

[N_{2}] = 1.5 \times 10^{- 2} M

[H_{2}] = 3.0 \times 10^{- 2} M

[N H_{3}] = 1.2 \times 10^{- 2} M

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