The standard enthalpy of formation of $\mathrm{NH}_{3}$ is $-46.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If the enthalpy of formation of $\mathrm{H}_{2}$ from its atoms is $-436 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and that of $\mathrm{N}_{2}$ is $-712 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the average bond enthalpy of $\mathrm{N}-\mathrm{H}$ bond in $\mathrm{NH}_{3}$ is

(B) $+352 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(A) $-964 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(C) $+1056 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(D) $-1102 \mathrm{~kJ} \mathrm{~mol}^{-1}$

JEE Main · 2010 · 11 Apr 2010 · Offline · Chemical Thermodynamics

The standard enthalpy of formation of

NH_{3}

- 46.0 kJ mol^{- 1}

. If the enthalpy of formation of

H_{2}

from its atoms is

- 436 kJ mol^{- 1}

and that of

N_{2}

- 712 kJ mol^{- 1}

, the average bond enthalpy of

N - H

bond in

NH_{3}

Answer: (B)

Write the making of two moles of ammonia from gaseous atoms so that only

N - H

bonds are formed. First,

\frac{1}{2} N_{2} + \frac{3}{2} H_{2} \to N H_{3}

has

Δ H_{f} = - 46

kJ, so for two moles

Δ H = - 92

kJ. Breaking one

N_{2}

and three

H_{2}

into atoms costs

712 + 3 (436) = 2020

kJ, and forming the six

N - H

bonds releases

6 x

, where

x

is the bond enthalpy. The balance

2020 - 6 x = - 92

gives

6 x = 2112

, so

x = 352

kJ mol

^{- 1}

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option