Consider the reaction:$$\mathrm{N_2(g) + 3H_2(g) <=> 2NH_3(g)}$$At a certain temperature, the equilibrium constant $K_c$ for this reaction is $5.6 \times 10^{-5}$. If a reaction vessel initially contains only $\mathrm{0.10\,mol}$ of $\mathrm{N_2}$ and $\mathrm{0.20\,mol}$ of $\mathrm{H_2}$, what will be the equilibrium concentrations of $\mathrm{N_2}$, $\mathrm{H_2}$, and $\mathrm{NH_3}$ once the reaction reaches equilibrium?

Question

Consider the reaction:$$\mathrm{N_2(g) + 3H_2(g) <=> 2NH_3(g)}$$At a certain temperature, the equilibrium constant $K_c$ for this reaction is $5.6 \times 10^{-5}$. If a reaction vessel initially contains only $\mathrm{0.10\,mol}$ of $\mathrm{N_2}$ and $\mathrm{0.20\,mol}$ of $\mathrm{H_2}$, what will be the equilibrium conce

Answer 1

Let's denote the change in the concentration of N, H, and NH as x, 3x, and 2x, respectively, as the reaction proceeds. At equilibrium, the concentrations will be as follows:[N] = 0.10 - x[H] = 0.20 - 3x[NH] = 2xNow, we can use the equilibrium constant expression to find the value of x:$$K_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3}$$Substitute the equilibrium concentrations:$$5.6 \times 10^{-5} = \frac{ 2x ^2}{ 0.10 - x  0.20 - 3x ^3}$$Solving for x is quite challenging algebraically. However, we can make an assumption that x is small compared to the initial concentrations of N and H, which allows us to simplify the equation:$$5.6 \times 10^{-5} = \frac{4x^2}{ 0.10  0.20 ^3}$$Now, solve for x:$$x^2 = \frac{5.6 \times 10^{-5} \times 0.10 \times 0.20^3}{4}4}$$$$x^2 = 1.792 \times 10^{-7}$$$$x = \sqrt{1.792 \times 10^{-7}}$$$$x = 1.34 \times 10^{-4}$$Now we can find the equilibrium concentrations:[N] = 0.10 - x = 0.10 - 1.34  10  0.0999 M[H] = 0.20 - 3x = 0.20 - 3 1.34  10   0.1996 M[NH] = 2x = 2 1.34  10   2.68  10 MSo, the equilibrium concentrations are approximately 0.0999 M for N, 0.1996 M for H, and 2.68  10 M for NH.