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Question: The enthalpy of formation of ammonia is -46.0 \[KJ\,mo{l^{ - 1}}\]. The enthalpy change for the reac...

The enthalpy of formation of ammonia is -46.0 KJ mol−1KJ\,mo{l^{ - 1}}. The enthalpy change for the reaction 2NH3(g)→N2(g)+3H2(g)2N{H_3}_{(g)} \to {N_2}_{(g)} + 3{H_2}_{(g)} is:
A. 46.0 KJ mol−1KJ\,mo{l^{ - 1}}
B. 92.0 KJ mol−1KJ\,mo{l^{ - 1}}
C. -23.0 KJ mol−1KJ\,mo{l^{ - 1}}
D. -92.0 KJ mol−1KJ\,mo{l^{ - 1}}

Explanation

Solution

To solve this question, we must first work with the given data. Since, enthalpy of formation has been provided, construct a chemical equation for the same. Then establish a relationship between the given reaction, and the reaction of formation.

Complete Step-by-Step answer:
Let us consider a reaction in which we produce ammonia from nitrogen and hydrogen. The chemical equation for this reaction can be given as:
N2(g)+3H2(g)→2NH3(g){N_2}_{(g)} + 3{H_2}_{(g)} \to 2N{H_3}_{(g)}
The data that has been provided to us state that enthalpy of formation of ammonia is -46.0 KJ mol−1KJ\,mo{l^{ - 1}}. Since nitrogen and hydrogen both are present in their standard states, the only contributor to the enthalpy of the reaction would be the enthalpy of formation of ammonia. Let us consider ΔH1\Delta {H_1}to be enthalpy of the reaction and ΔHf(NH3)\Delta {H_f}(N{H_3})to be the enthalpy of formation of ammonia. Then,
ΔH1\Delta {H_1}=[ΔHf(NH3)\Delta {H_f}(N{H_3})] (number of moles of ammonia formed) = (-46.0) (2) = - 92.0 KJ mol−1KJ\,mo{l^{ - 1}}
Now let us consider the equation given to us:
2NH3(g)→N2(g)+3H2(g)2N{H_3}_{(g)} \to {N_2}_{(g)} + 3{H_2}_{(g)}
This reaction can be considered to be the exact reverse of the reaction for formation of ammonia. If ΔH2\Delta {H_2}can be considered to be the enthalpy for this reaction, then we can establish the following relationship:
ΔH2\Delta {H_2} = - ΔH1\Delta {H_1}… (due to the reversing of the reaction)
Hence, ΔH2\Delta {H_2} = - (-92.0) = 92.0 KJ mol−1KJ\,mo{l^{ - 1}}
Hence, the enthalpy change for the given reaction is calculated to be 92.0 KJ mol−1KJ\,mo{l^{ - 1}}.

Hence, Option B is the correct option.

Note: The enthalpy of formation is the standard reaction enthalpy for the formation of the compound from its elements (atoms or molecules) in their most stable reference states at the chosen temperature (298.15K) and at 1bar pressure.