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

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Solution

When 1 mole of ammonia is formed from its elements, the enthalpy change is known as the enthalpy of formation. It is equal to one half the value of the enthalpy change associated with formation of 2 moles of ammonia.
Hence, the enthalpy of formation of ammonia is $\frac{92.2}{2} = - 46.1$ kJ/mol.

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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?

Given

; Δ_rH^θ = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ gas?

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)}; △_{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}

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