Answer
look at the table below.
Work Step by Step
First, write the equilibrium expression for the reaction $N_{2} + 3H_{2} \leftrightharpoons 2NH_{3}$
equilibrium constant $K_{eq} = \frac{[NH_{3}]^2}{[N_{2}][H_{2}]^{3}}$
Now, look at the 1st row. Here,$ [N_{2}], [H_{2}], [NH_{3}]$ are given. we put those value on the equilibrium expression and find $K_{eq}$.
$\therefore K_{eq} = \frac{0.439^{2}}{0.115\times0.105^{3}} \approx 1447.65$
For the second row, we have to find $[H_{2}]$.
$[H_{2}] =\sqrt[3]{\frac{[NH_{3}]^{2}}{[N_{2}][K_{eq}]}} \approx 0.249$
For the third row, we have to find $[NH_{3}]$.
$ [NH_{3}] = \sqrt{[K_{eq}][N_{2}][H_{2}]^{3}} = \sqrt{0.0584\times0.120\times0.140^{3}} \approx 0.0139$
