The enthalpy of formation of ammonia is $-46.0\mathrm{KJ}{\mathrm{mol}}^{-1}$. The enthalpy change for the reaction $2{\mathrm{NH}}_3\to {\mathrm{N}}_2+3{\mathrm{H}}_2$ is

The standard enthalpy of formation of ammonia is $-46.0\mathrm{ }\mathrm{k}\mathrm{J}{\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$. The enthalpy change for the reaction:

Because the formation of 1 mole of ammonia releases $-46.0\mathrm{ }\mathrm{k}\mathrm{J}{\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$ of energy, the breakdown of 1 mole of ammonia consumes the same amount of energy, i.e. $+46.0\mathrm{ }\mathrm{k}\mathrm{J}{\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$. The energy spent will be $+92.0\mathrm{ }\mathrm{k}\mathrm{J}{\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$ since the described reaction implies the breakdown of two moles of ammonia.

Hence the correct answer is (B) $+92\mathrm{KJ}$.