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Heat of Formation

Given that ...

Question

Given that $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); △_{f} H^{⊖} = - 92 k J$ , the standard molar enthalpy of formation in $k J m o l^{- 1}$ of $N H_{3} (g)$ is:

- 92

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+ 46

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+ 92

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- 46

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Solution

The correct option is D $- 46$
$N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} (g); Δ_{f} H^{\circ} = - 92 k J$

$Δ_{f} H^{\circ} = \sum Δ_{f} H^{\circ} (p r o d u c t s) - \sum Δ_{f} H^{\circ} (r e a c t a n t s)$

$= 2 Δ_{f} H^{\circ} (N H_{3} (g)) - [Δ_{f} H^{\circ} (N_{2} (g)) + 3 Δ_{f} H^{\circ} (H_{2} (g))]$

$- 92 = 2 Δ_{f} H^{\circ} (N H_{3} (g))$

$Δ_{f} H^{\circ} (N H_{3} (g)) = - \frac{92}{2} = - 46 k J / m o l$

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Given
$N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3 (g)}; Δ_{r} H^{θ} = - 92.4 k J m o l^{- 1}$
What is the standard enthalpy of formation of $N H_{3}$ gas?

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