JEE Main 2014 · 9 Apr · Shift 1 · Chemistry Q40

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ChemistrySingle correct+4 / -1Question 40 of 90

The standard enthalpy of formation of $\ce{NH3}$ is $-46.0$ kJ/mol. If the enthalpy of formation of $\ce{H2}$ from its atoms is $-436$ kJ/mol and that of $\ce{N2}$ is $-712$ kJ/mol, the average bond enthalpy of N-H bond in $\ce{NH3}$ is :

-1102

kJ/mol

-964

kJ/mol

+352 kJ/mol

+1056 kJ/mol

Answer and worked solution

Official answer: C

For $\ce{1/2N2 + 3/2H2 -> NH3}$ (−46.0 kJ), go through the atoms. The bond enthalpies are N≡N 712 and H-H 436 kJ/mol:
$-46=\tfrac12(712)+\tfrac32(436)-3E_{\ce{N-H}}\implies3E_{\ce{N-H}}=1056\implies E_{\ce{N-H}}=+352\text{ kJ/mol}.$

Topics:Bond enthalpy Hess's law and enthalpies of reaction

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JEE Main 2014 · 9 Apr · Shift 1, Chemistry question 40