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Question: Given the bond energies \(N \equiv N,H - H\) and N–H bonds are 945, 436 and 391 kJ mole<sup>–1</sup>...

Given the bond energies N≡N,H−HN \equiv N,H - H and N–H bonds are 945, 436 and 391 kJ mole–1 respectively the enthalpy of the following reaction N2(g)+3H2(g)→2NH3(g)N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g) is

A

– 93 kJ

B

102 kJ

C

90 kJ

D

105 kJ

Answer

– 93 kJ

Explanation

Solution

N \equiv N + 3H - H \\ 945 + 3 \times 436 \end{matrix}}{︸}}\overset{\quad\quad}{\rightarrow}\underset{\text{energy released}}{\overset{\underset{2 \times (3 \times 391) = 2346}{\overset{\underset{H}{H}}{\overset{\underset{|}{|}}{2N—H}}}}{︸}}$$ Net energy (enthalpy) = BE<sub>reactants</sub> – BE<sub>products</sub> = 2253 – 2346 =– 93 $$\Delta H = - 93kJ$$