Given

; Δ_rH^θ = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ 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 NH₃₍_g_),

Standard enthalpy of formation of NH₃₍_g₎

= ½ Δ_rH^θ

= ½ (–92.4 kJ mol^–1)

= –46.2 kJ mol^–1

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Similar questions

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?

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g) : △_{r} H^{⊖} = - 92.4 k J m o l^{- 1}

N H_{3}

N_{2} (g) + 3 H_{2} (g) \to 2 {N H}_{3} (g) : △ H^{θ} = 92.4 k J {m o l}^{- 1}

{N H}_{3}

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)

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