Calculate the magnitude of proton affinity of NH 3 ( g ) from the following data (in kJ / mol ) Δ H ∘ dissociation: H 2 ( g ) = 436 Δ H ∘ formation NH 3 ( g ) = –46 Lattice energy of NH 4 Cl (

Question

Calculate the magnitude of proton affinity of $NH_{3} (g)$ from the following data (in $kJ / mol$ )

$Δ H^{\circ}$ dissociation: $H_{2} (g) = 436$

$Δ H^{\circ}$ formation $NH_{3} (g) = -46$

Lattice energy of $NH_{4} Cl (s) = -683$

Ionisation energy of $H = 130$

Electron affinity of $Cl = 380$

$Δ H^{\circ}$ dissociation: $Cl_{2} (g) = 240$

$Δ H^{\circ}$ formation $NH_{4} Cl (s) = -314$

TestHub · Accepted Answer

Formation of NH₄Cl(s):

NH₃(g) + HCl(g) → NH₄Cl(s)

For HCl formation:

$Δ H_{f} (H Cl) = \frac{436}{2} + \frac{240}{2} + 130 - 380 = 218 + 120 + 130 - 380 = 88 k J$

Using:

$Δ H_{f} (N H_{4} Cl) = Δ H_{f} (N H_{3}) + Δ H_{f} (H Cl) - P A + L . E .$

$- 314 = - 46 + 88 - P A - 683$

$P A = 327 k J / m o l$

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