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Heat of Formation

Given N 2 g +...

Question

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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Solution

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of $N H_{3 (g)}$ ,
$\frac{1}{2} N_{2 (g)} + \frac{3}{2} H_{2 (g)} \to N H_{3} (g)$
$∴$ Standard entalpy of formation of $N H_{3 (g)}$
$= \frac{1}{2} Δ_{r} H^{θ} = \frac{1}{2} (- 92.4 k J m o l^{- 1}) = - 46.2 k J m o l^{- 1}$

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; Δ_rH^θ = –92.4 kJ mol^–1

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Heat of Formation

Standard XII Chemistry

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