The standard enthalpy of formation of NH_3 is - 46.0 ,kJ , mol^ -… - Vidyadip

MCQ

The standard enthalpy of formation of $NH_3$ is $- 46.0 \,kJ\, mol^{-1}.$ If the enthalpy of formation of $H_2$ from its atoms is $- 436\, kJ\, mol^{-1}$ and that of $N_2$ is $- 712 \,kJ\, mol^{-1},$ the average bond enthalpy of $N - H $ bond in $NH_3$ is ................ $\mathrm{kJ\,mol}^{-1}$

A
$- 964$
✓
$+ 352$
C
$+1056$
D
$-1102$

✓

Answer

Correct option: B.

$+ 352$

b
$N_{2}+3 H_{2} \rightarrow 2 N H_{3}$

$\Delta H=2 \times-46.0 \mathrm{\,kJ}\, \mathrm{mol}^{-1}$

Let $x$ be the bond enthalpy of $N-H$ bond then

$\Delta H=\sum$ Bond energies of products

$-\sum$ Bond energies of reactants

$2 \times-46=712+3 \times(436)-6 x$

$-92=2020-6 x$

$6 x=2020+92$

$\Rightarrow 6 x=2112$

$\Rightarrow x=+352\; k J / m o l$

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