Chemistry - Chemical Equilibrium Question with Solution | TestHub

$N_2+3H_2\rightleftharpoons 2N{H}_3$

$t=0$: $1$ $3$ $0$

At equilibrium: $(1-x)$ $(3-3x)$ $2x$

According to the question,

$meq.ofammonia=meq.of{H}_{2}S{O}_4$

$=1\times 2\times 100=200$

Millimoles of ammonia = $200$

Moles of ammonia = $\frac{200}{1000}=0.20$

At equilibrium, $2x=0.2$, so $x=0.1$.

$K_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}=\frac{(0.2{)}^2}{(0.9)(2.7{)}^3}=2.26\times 10^{-3}$