Question

The enthalpy change for a reaction $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ is -92.2 kJ/mol. Calculate the enthalpy of formation of ammonia.

• A. -92.2 kJ/mol
• B. -46.1 kJ/mol
• C. 46.1 kJ/mol
• D. -92.2 kJ/mol

Answer 1

The correct option is B -46.1 kJ/mol

Given, $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$ $\Delta H_r=-92.2kJ$

Here 2 moles of ammonia are forming but according to definition of enthalpy of formation we know it is for formation of 1 mole of substance. i.e for formation of 1 mole of ammonia

dividing whole equation by 2 we get
$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\to N{H}_3\left(g\right)$
$\Delta H_f\left(N{H}_3\right)=\frac{-92.2}{2}=-46.1kJ$
hence, heat of formation of ammonia is $-46.1kJ/mol$