InterviewSolution
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InterviewSolution
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\begin{array}{l}{\text { Calculate the bond enthalpy of } \mathrm{N}-\mathrm{H} \text { in } \mathrm{NH}_{3}(\mathrm{g})} \\ {\text { Given } \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{NH}_{3} ;(\mathrm{g}) \mathrm{D}_{f} \mathrm{H}^{\circ}=-46 \mathrm{kJmol}^{-1}} \\ {\frac{1}{2} \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g}) ; \Delta r H^{\ominus}=218 \mathrm{kJmCl}^{-1}} \\ {\frac{1}{2} \mathrm{N}_{2}(\mathrm{g}) \rightarrow \mathrm{N}(\mathrm{g}) ; \Delta r H^{\ominus}=973 \mathrm{kJmol}^{-1}}\end{array} |
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Answer» To find the bond enthalpy, you subtract the energy of the bonds formed (since they release energy) and add the energy of the bonds broken (since you need to supply energy). 1/2 an N2 bond is 941/2 = 470.5 kJ/mol3/2 H2 bonds = 3×436/2 = 654 kJ/molTotal energy = bonds broken-bonds formed = (470.5 + 654) - 3(N-H) Total energy = -46.3 kJ/mol Therefore, -3(N-H) + 1124.5 = -46.3 3(N-H) = 1124.5 + 46.3 = 1170.8 (N-H) = 1170.8/3 = 390.2667 ≈ 390 kJ/mol Remember: total energy of reaction = (total energy of bonds broken) - (total energy of bonds formed) |
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