Question

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

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

Answer 1

Answer: (D)

304 kJ

First write the balanced chemical; equations for he formation of dinitrogen oxide and nitric oxide.

$N_2+\frac{1}{2}{O}_2⟶N_{2}O;\Delta H=28kJ$...(i)

$\frac{1}{2}{N}_2+\frac{1}{2}{O}_2⟶NO;\Delta H=90kJ$...(ii)

Then multiply second equation with 4 and first equation with 2.

$2N_2+O_2⟶2N_{2}O;\Delta H=2\times 28kJ$...(iii)

$2N_2+2O_2⟶4NO;\Delta H=4\times 90kJ$...(iv)

Now substract third equation from fourth equation.

$2N_{2}O+O_2⟶4NO;\Delta H=304kJ$.

Thus,the enthalpy of the reaction $2N_{2}O\left(g\right)+O_2\left(g\right)⟶4NO\left(g\right)$ is 304 kJ.