JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
The standard enthalpy of formation of \(NH_3\) is \(-46.0\, kJ/mol\). If the enthalpy of formation of \(H_2\) from its atoms is \(-436\, kJ/mol\) and that of \(N_2\) is \(-712\, kJ/mol\), the average bond enthalpy of \(N - H\) bond in \(NH_3\) is......\(kJ/mol\)
- A \(- 1102\)
- B \(- 964\)
- C \(+352\)
- D \(+1056\)
Answer & Solution
Correct Answer
(B) \(- 964\)
Step-by-step Solution
Detailed explanation
Given \(\frac{1}{2}{N_2} + \frac{3}{2}{H_2} \leftrightarrow N{H_3};\) \(\Delta {H_f} = - 46.0\,kJ/mol\) \(H + H \leftrightarrow {H_2};\,\Delta {H_f} = - 436\,kJ/mol\) \(N + N \leftrightarrow {N_2};\,\Delta {H_f} = - 712\,kJ/mol\)…
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