If bond enthalpies of N equiv N,H - H and N - | vedclass

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(B) The formation reaction of $NH_3$ is: $\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) \rightarrow NH_3(g)$.Bond enthalpy is defined as the energy required

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$\frac{1}{2}x_1 + \frac{3}{2}x_2 - 3x_3$

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(B) The formation reaction of $NH_3$ is: $\frac{1}{2} N_2(g) + \frac{3}{2} H_2(g) ightarrow NH_3(g)$.Bond enthalpy is defined as the energy required to break one mole of bonds.For the reaction,$\Delta H_f^o = \sum 	ext{Bond Enthalpies of Reactants} - \sum 	ext{Bond Enthalpies of Products}$.Reactants: $\frac{1}{2}$ mole of $N \equiv N$ and $\frac{3}{2}$ moles of $H - H$.Products: $3$ moles of $N - H$ bonds (since $NH_3$ has $3$ $N - H$ bonds).Therefore,$\Delta H_f^o = [\frac{1}{2}x_1 + \frac{3}{2}x_2] - [3x_3]$.

If bond enthalpies of $N \equiv N$,$H - H$ and $N - H$ bonds are $x_1, x_2$ and $x_3$ respectively,$\Delta H_f^o$ for $NH_3$ will be

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