Question

A mixture of $N{H}_3\left(g\right)$ and $N_2{H}_4\left(g\right)$ is placed in a sealed container at $300K$. The total pressure is $0.5atm$. The container is heated to $1200K$ at which both gases decompose completely according to the equation:
$2N{H}_3\left(g\right)\to N_2\left(g\right)+3H_2\left(g\right)$ and $N_2{H}_4\left(g\right)\to N_2\left(g\right)+2H_2\left(g\right)$
After complete decomposition, the total pressure at $1200K$ is found to be $4.5atm$. Find the moles % of $N_2{H}_4$ in the original mixture :

• A. 20%
• B. 25%
• C. 50%
• D. 75%

Answer 1

Answer: (B)

25%

Let the initial mixture contains $n_1$ & $n_2$ moles of $N{H}_3$ & $N_2{H}_4$ respectively.

Total moles of gases originally present $=n_1+n_2$.

Total moles of gases after decomposition of gases $=2n_1+3n_2$.

Now, using ideal gas law: $PV=nRT$

$0.5\times V=(n_1+n_2)R\times 300$........(i)

$4.5\times V=(2n_1+3n_2)R\times 1200$........(ii)

Solving these two equations, we get-

$\frac{2n_1+3n_2}{n_1+n_2}=\frac{9}{4}$

$=>\frac{n_1}{n_2}=\frac{3}{1}$