Question

A mixture of ${NH}_3\left(g\right)$ and $N_2{H}_4\left(g\right)$ is placed in a sealed container at 300 K .The total pressure of the gaseous mixture is 0.5atm . The container is heated to 1200 K at which both ${NH}_3$ and $N_2{H}_4$ decomposed completely according to the following equation.
$2{NH}_3\left(g\right)⟶N_2\left(g\right)+3H_2\left(g\right)$
$N_2{H}_4\left(g\right)⟶N_2\left(g\right)+2H_2\left(g\right)$
After decomposition is complete, the total pressure at 1200 K is found to be 4.5 atm, The mole % of $N_2{H}_4$ in the initial mixture at 300 K was:

• A. 20%
• B. 35%
• C. 50%
• D. 75%
• E. 25%

Answer 1

The correct option is E 25%

Let no. of moles of ${NH}_3=n_1$ and $N_2{H}_4=n_2$ in the initial mixture

Total no. of moles initially =$n_1+n_2$

Total no. of moles after decomposition =${2n}_1+{3n}_2$

$\therefore 0.5\times V=({2n}_1+{3n}_2)R\times 300$

$4.5\times V=({2n}_1+{3n}_2)R\times 1200$

$\frac{{2n}_1+{3n}_2}{n_1+n_2}=\frac{9}{4},n_1=3n_2$

$\frac{9n_2}{n_1+n_2}=\frac{9}{4}\{\frac{n_2}{n_1+n_2}=\frac{1}{4}\}$

$\therefore$ % mole of $N_2{H}_4=25$%