Ammonia, NH₃, is manufactured from nitrogen and hydrogen by the Haber process - CIE - A-Level Chemistry - Question 1 - 2017 - Paper 1

To calculate the N–H bond energy, we can use the provided bond energies:

1. Given:

   • N≡N bond energy = 944 kJ mol⁻¹
   • H–H bond energy = 436 kJ mol⁻¹
   • ΔH for the reaction = -92 kJ mol⁻¹

2. The reaction can be represented as:

   N₂ + 3H₂ → 2NH₃

   according to bond breaking and forming:

   Bond energy required to break bonds = Energy released from forming bonds + ΔH

   E(N–H) = (944 + 3*436) - (-92)

   Rearranging gives:

   E(N–H) = 944 + 1308 + 92 = 2344 kJ

   Therefore,

   2 E(N–H) = 2344 kJ

   Hence,

   E(N–H) = 1172 kJ mol⁻¹.

   So, the N–H bond energy is 1172 kJ mol⁻¹.

To indicate the Boltzmann distribution at a higher temperature, the second curve should be drawn to the right of the initial curve. It will be broader and flatter, showing that at a higher temperature, more molecules have energies exceeding the activation energy, Ea. Label the x-axis as 'molecular energy' and the y-axis as 'proportion of molecules with a given energy'.

As the temperature increases, the rate of production of ammonia increases. This is because a higher temperature results in a greater proportion of molecules having energy equal to or greater than the activation energy (Eₐ). This leads to more frequent and successful collisions between the reacting molecules, thus increasing the reaction rate.

Increasing temperature typically reduces the yield of ammonia. According to Le Chatelier's principle, increasing the temperature shifts the equilibrium position to favor the endothermic reaction, which in this case is the reverse reaction (breaking down NH₃ into N₂ and H₂). Thus, there is a reduction in the amount of ammonia produced.

1. Starting amounts: N₂ = 1.00 mol, H₂ = 3.00 mol
2. Reaction: N₂ + 3H₂ ⇌ 2NH₃
3. Since 0.300 mol of NH₃ is formed, using the stoichiometry of the reaction:
   • From 2 mol NH₃, 1 mol N₂ is consumed → from 0.300 mol NH₃, 0.150 mol N₂ is consumed
   • From 2 mol NH₃, 3 mol H₂ is consumed → from 0.300 mol NH₃, 0.225 mol H₂ is consumed
4. Therefore,
   • Amount of N₂ remaining = 1.00 - 0.150 = 0.850 mol
   • Amount of H₂ remaining = 3.00 - 0.225 = 2.775 mol

To calculate the partial pressure of NH₃:

1. Total pressure (P) = 2.00 × 10² Pa
2. The fraction of ammonia, x(NH₃) = 0.300 / (0.850 + 2.775 + 0.300) = 0.300 / 3.925 = 0.0764
3. Therefore, the partial pressure of NH₃ = x(NH₃) * P = 0.0764 * 2.00 × 10² = 15.28 Pa.
4. Rounding this, we find pNH₃ = 15.3 Pa.

The expression for the equilibrium constant Kc for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is given by:

$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Using the partial pressures given:

1. Partial pressures:
   • pN₂ = 2.20 × 10⁶ Pa
   • pH₂ = 9.62 × 10⁵ Pa
   • pNH₃ = 1.40 × 10⁴ Pa
2. Calculate Kc: $K_c = \frac{(1.40 \times 10^4)^2}{(2.20 \times 10^6)(9.62 \times 10^5)^3}$
3. This calculation yields Kc approx = 2.88 × 10⁻¹⁴ Pa⁻¹.
4. The units of Kc are Pa⁻¹.

The effect on the yield of ammonia is that it will increase due to the smaller volume, which shifts the equilibrium position towards producing more NH₃. For the value of Kc, it remains constant as Kc is only affected by temperature change, not by changes in pressure or concentration.