Question

A mixture of $H_2,N_2$ and $N{H}_3$ with molar concentration $5\times {10}^{-3}mol{L}^{-1},4\times {10}^{-3}mol{L}^{-1}and2\times {10}^{-3}mol{L}^{-1}$ respectively was prepared and heated to 500K.The value of $K_c$ for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ at this given temperature is 60.Predict whether ammonia tends to form or decompose at this stage of concentration

Answer 1

$N_2\left(g\right)+3H_2\left(g\right)=2N{H}_3\left(g\right)$; Kc = 60 at 500K

Given, $\left[N_2\right]=5\times {10}^{-3}$ mol/L

$\left[H_2\right]=4\times {10}^{-3}$ mol/L

$\left[N{H}_3\right]=2\times {10}^{-3}$ mol/L

$Q_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$

$Q_c=\frac{(2\times {10}^{-3}{)}^2}{5\times {10}^{-3}\times (4\times {10}^{-3}{)}^2}=100$

but $K_c=60$

so, $K_c$ is not equal to $Q_c$ , equilibrium has not been attained.

here, $K_c<Q_c$ , the reaction will proceed from right to left. e.g., ammonia decomposes at this stage.