Question

The standard enthalpy of formation of $N_{2}O$ and $NO$ are $28{\text{kJ mol}}^{-1}$ and $90{\text{kJ mol}}^{-1}$ respectively. The enthalpy change of the reaction, $2N_{2}O\left(g\right)+O_2\left(g\right)\to 4NO\left(g\right)$ is equal to:

• A. $62{\text{kJ mol}}^{-1}$
• B. $88{\text{kJ mol}}^{-1}$
• C. $-252{\text{kJ mol}}^{-1}$
• D. $304{\text{kJ mol}}^{-1}$

Answer 1

The correct option is D $304{\text{kJ mol}}^{-1}$

First write the balanced chemical equation for the formation $N_{2}O$ and $NO$:

$N_2+\frac{1}{2}{O}_2\to N_{2}O;\Delta H^o=28{\text{kJ mol}}^{-1}...\left(i\right)$

$\frac{1}{2}{N}_2+\frac{1}{2}{O}_2\to NO;\Delta H^o=90{\text{kJ mol}}^{-1}...\left(ii\right)$

Then multiply equation $\left(i\right)$ by $2$ and $\left(ii\right)$ equation by $4$ we get,
$2N_2+O_2\to 2N_{2}O;\Delta H^o=2\times 28{\text{kJ mol}}^{-1}....\left(iii\right)$

$2N_2+2O_2\to 4NO;\Delta H^o=4\times 90{\text{kJ mol}}^{-1}....\left(iv\right)$

Now substract $\left(iii\right)$ equation from $\left(iv\right)$ equation, we get:
$2N_{2}O\left(g\right)+O_2\left(g\right)\to 4NO\left(g\right);\Delta H^o=304{\text{kJ mol}}^{-1}$
Thus, the enthalpy change of the reaction is $304{\text{kJ mol}}^{-1}$.