Question

In a mixture of $N_2$ and $H_2$ in the ratio of $1:3$ at $64$ atmospheric pressure and ${300}^{o}C$, the percentage of ammonia under equilibrium is $33.33$ by volume. Calculate the equilibrium constant of the reaction using the equation, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$

Answer 1

The reaction takes place as:

$N_2+3H_2\leftrightharpoons 2{NH}_3$

The 33% of ammonia have mole fraction of 0.333

${\rho }_{{NH}_3}=0.334\times 34=21.3$

The other 66.7% of gases splits in the ratio of 1:3 between $H_2$ and

$N_2$

The mole fractions can be calculated for $N_2$

$\frac{1}{4}\times 66.7$

$=0.166$

The mole fractions of $H_2$ $=0.677-0.167=0.500$

$P_{N_2}=-0.16\times 64=10.624$

$P_{H_2}=-0.50\times 64=32$

$K_P=\frac{{\left[{NH}_3\right]}^2}{P_{N_2}\times P_{H_2}^3}$

$K_P=\frac{{\left[21.312\right]}^2}{10.624\times {32}^3}$

$K_P=0.0013$