Ammonia is produced according to the following equilibrium equation - HSC - SSCE Chemistry - Question 31 - 2021 - Paper 1

To find the change in concentration, we add 0.050 moles to the current concentration:

$[NH_3]_{new} = 0.580 \text{ mol L}^{-1} + \frac{0.050 \text{ moles}}{10.0 L} = 0.585 \text{ mol L}^{-1}$

The equilibrium constant expression for the reaction is given by:

$K_{eq} = \frac{[NH_3]^2}{[N_2][H_2]^3}$

Substituting the values:

$748 = \frac{(0.585)^2}{\frac{4.475 + x}{10} \times (\frac{1.00}{10})^3}$

Rearranging to solve for $x$ leads to the equation:

$748 = \frac{0.585^2}{\frac{4.475 + x}{10} \times 0.001}$

Solving gives:

1.3 moles of nitrogen gas must be added to the equilibrium mixture.