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Question: Calculate the enthalpy of ammonia from the following bond energy data: \((N - H) = 389kJmo{l^{ - 1...

Calculate the enthalpy of ammonia from the following bond energy data:
(N−H)=389kJmol−1(N - H) = 389kJmo{l^{ - 1}} ; (H−H)bond=435kJmol−1(H - H)bond = 435kJmo{l^{ - 1}} and (N≡N)bond=945.36kJmol−1(N \equiv N)bond = 945.36kJmo{l^{ - 1}}

Explanation

Solution

Chemical formula of ammonia is NH3N{H_3}. First write the chemical equation for the formation of ammonia. You must know that enthalpy of formation of a molecule in a reaction is the difference of the sum of bond energies of the molecules on the reactant side and the sum of bond energies of the molecules on the product side.

Complete step by step solution:
We are asked to find the enthalpy of formation of ammonia and data of bond energies of some bonds are given. To calculate the enthalpy of formation of ammonia, we need to know the chemical equation where ammonia is formed.
Chemical equation for the formation of ammonia:
N2(g)+3H2(g)→2NH3(g){N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)
Thus, nitrogen gas is combined with three moles of hydrogen gas to form two moles of ammonia gas. Each nitrogen gas molecule contains N≡NN \equiv N bond, each hydrogen gas molecule contains one H−HH - Hbond and each ammonia molecule (NH3N{H_3}) contains 3 N−HN - H bonds.
Given data of bond enthalpies are: (N−H)=389kJmol−1(N - H) = 389kJmo{l^{ - 1}} ; (H−H)bond=435kJmol−1(H - H)bond = 435kJmo{l^{ - 1}} and (N≡N)bond=945.36kJmol−1(N \equiv N)bond = 945.36kJmo{l^{ - 1}}
Since, enthalpy of formation of a molecule in a reaction is the difference of sum of bond energies of the molecules on the reactant side and the sum of bond energies of the molecules on the product side.
Therefore, enthalpy of formation of ammonia (ΔH\Delta H) is:
ΔH\Delta H= [ΔH(N≡N)+3×ΔH(H−H)]−[6×ΔH(N−H)]\left[ {\Delta {H_{(N \equiv N)}} + 3 \times \Delta {H_{(H - H)}}} \right] - \left[ {6 \times \Delta {H_{(N - H)}}} \right]
Therefore, ΔH=[945.36+3×435]−[6×389]=−89.64kJ\Delta H = \left[ {945.36 + 3 \times 435} \right] - \left[ {6 \times 389} \right] = - 89.64kJ
Heat of formation of one ammonia molecule: ΔH2=−83.642=−41.82kJmol−1\dfrac{{\Delta H}}{2} = \dfrac{{ - 83.64}}{2} = - 41.82kJmo{l^{ - 1}}
Here, a negative sign indicates that the reaction is exothermic which means heat is released in the surroundings.

Therefore, the required answer is -41.82 kJmol-1.

Note: Enthalpy change is a very useful and important quantity. Knowledge of this quantity is required when one needs cooling or heating required to maintain or process an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.