If bond dissociation energies of N≡N, H-H, and N-H are x1, x2 and x3 respectively, hence enthalpy of formation of NH3(g) is

# If bond dissociation energies of N$\equiv$N, H-H, and N-H are x1, x2 and x3 respectively, hence enthalpy of formation of NH3(g) is

1. A

${\mathrm{x}}_{1}+3{\mathrm{x}}_{2}-6{\mathrm{x}}_{3}$

2. B

$3{x}_{3}-\frac{1}{2}{x}_{1}-\frac{3}{2}{x}_{2}$

3. C

$\frac{{x}_{1}}{2}+\frac{3}{2}{x}_{2}-3x$3

4. D

$6{x}_{3}-{x}_{1}-3{x}_{2}$

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### Solution:

$\begin{array}{l}\frac{1}{2}{\mathrm{N}}_{2}\left(\mathrm{g}\right)+\frac{3}{2}{\mathrm{H}}_{2}\left(\mathrm{g}\right)\right)\to {\mathrm{NH}}_{3}\left(\mathrm{g}\right)\\ {\mathrm{\Delta }}_{r}H={\mathrm{\Delta }}_{\mathrm{form}}H\left({\mathrm{NH}}_{3},\mathrm{g}\right)\\ {\mathrm{\Delta }}_{\mathrm{form}}H\left({\mathrm{NH}}_{3},\mathrm{g}\right)=\frac{1}{2}{D}_{\mathrm{N}\equiv \mathrm{N}}+\frac{3}{2}{D}_{\mathrm{H}-\mathrm{H}}-3{D}_{\mathrm{N}-\mathrm{H}}\\ =\frac{1}{2}×{x}_{1}+\frac{3}{2}{x}_{2}-3{x}_{3}\end{array}$

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