Question

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

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

Answer 1

Answer: (D)

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

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

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

$2N_{2}O+O_{2} \rightarrow 4NO$

$\Delta H=360-56$

$\Delta H=304 \,kJ$ .