Question

The enthalpies of formation of $N_{2}O$ and NO are 28 and 90 kJ $mo{l}^{-1}$ respectively. The enthalpy of the reaction $2N_{2}O(g)+O_2(g)\to 4NO(g)$ is equal to

• A. 8 kJ
• B. 88 kJ
• C. -16 kJ
• D. 304 kJ

Answer 1

Answer: (D)

${\text{N}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\to {\text{N}}_{\text{2}}\text{O;}\text{ΔH}\text{=}\text{28 kJ}$ …(i)

$\frac{\text{1}}{\text{2}}{\text{N}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\to \text{NO;}\text{ΔH}\text{=}\text{90}\text{kJ}$ …(ii)

By equation [4 $\times$ (ii)] - [2 $\times$ (i)],

$2N_{2}O+O_2\to 4NO$

$\Delta H=360-56$

$\Delta H=304kJ$ .