Chemistry - Thermochemistry Question with Solution | TestHub

To calculate the proton affinity of NH₃(g), we need to determine the enthalpy change for the reaction:

NH₃(g) + H⁺(g) $\to$ NH₄⁺(g)

This can be achieved using a Born-Haber cycle or by manipulating given thermochemical equations.

Let's use the given data to construct a cycle:

1. Formation of NH₃(g):

$\frac{1}{2}$N₂(g) + $\frac{3}{2}$H₂(g) $\to$ NH₃(g); $\Delta$H₁ = -46 kJ mol⁻¹

2. Formation of NH₄Cl(s):

$\frac{1}{2}$N₂(g) + 2H₂(g) + $\frac{1}{2}$Cl₂(g) $\to$ NH₄Cl(s); $\Delta$H₂ = -314 kJ mol⁻¹

3. Subtracting (1) from (2) gives:

NH₃(g) + $\frac{1}{2}$H₂(g) + $\frac{1}{2}$Cl₂(g) $\to$ NH₄Cl(s); $\Delta$H₃ = $\Delta$H₂ - $\Delta$H₁ = -314 - (-46) = -268 kJ mol⁻¹

Now, let's consider the Born-Haber cycle for the formation of NH₄Cl(s) from gaseous ions:

NH₄⁺(g) + Cl⁻(g) $\to$ NH₄Cl(s); $\Delta$H = Lattice Energy = -683 kJ mol⁻¹

We need to form NH₄⁺(g) and Cl⁻(g) from NH₃(g), H₂(g), and Cl₂(g).

Consider the following steps:

a) Dissociation of H₂(g):

$\frac{1}{2}$H₂(g) $\to$ H(g); $\Delta$H = $\frac{1}{2}\times 218=109$ kJ mol⁻¹

b) Dissociation of Cl₂(g):

$\frac{1}{2}$Cl₂(g) $\to$ Cl(g); $\Delta$H = $\frac{1}{2}\times 124=62$ kJ mol⁻¹

c) Ionization of H(g):

H(g) $\to$ H⁺(g) + e⁻; $\Delta$H = Ionization Energy = 1310 kJ mol⁻¹

d) Electron gain by Cl(g):

Cl(g) + e⁻ $\to$ Cl⁻(g); $\Delta$H = Electron Gain Enthalpy = -348 kJ mol⁻¹

e) Protonation of NH₃(g) (the target reaction):

NH₃(g) + H⁺(g) $\to$ NH₄⁺(g); $\Delta$H = Proton Affinity (let's call it x)

Now, we can set up an equation based on Hess's Law, relating $\Delta$H₃ to the sum of the enthalpies of the steps that lead from NH₃(g), $\frac{1}{2}$H₂(g), and $\frac{1}{2}$Cl₂(g) to NH₄Cl(s) via gaseous ions.

$\Delta$H₃ = (enthalpy to form H⁺(g) from $\frac{1}{2}$H₂(g)) + (enthalpy to form Cl⁻(g) from $\frac{1}{2}$Cl₂(g)) + (proton affinity of NH₃) + (lattice energy of NH₄Cl)

$\Delta$H₃ = ($\Delta$H from a + $\Delta$H from c) + ($\Delta$H from b + $\Delta$H from d) + x + (Lattice Energy)

-268 = (109 + 1310) + (62 + (-348)) + x + (-683)

-268 = 1419 + (-286) + x - 683

-268 = 1419 - 286 - 683 + x

-268 = 450 + x

x = -268 - 450

x = -718 kJ mol⁻¹

The proton affinity of NH₃(g) is -718 kJ mol⁻¹.