A chemistry student wants to understand the effect of changing the volume on the equilibrium position of a chemical reaction. Accordingly, the student designs an experiment where they mix 0.2 mol of nitrogen gas with 0.1 mol of hydrogen gas in a 1 L container and allow them to react to form ammonia gas. If the reaction reaches equilibrium at a certain temperature and the volume of the container is suddenly decreased to 0.5 L, what will be the effect on the concentration of ammonia gas?

Question

A chemistry student wants to understand the effect of changing the volume on the equilibrium position of a chemical reaction. Accordingly, the student designs an experiment where they mix 0.2 mol of nitrogen gas with 0.1 mol of hydrogen gas in a 1 L container and allow them to react to form ammonia gas. If the reaction reaches equilibri

Answer 1

To understand the effect of changing the volume on the equilibrium position of the reaction, we need to consider the balanced chemical equation for the formation of ammonia gas:N g  + 3H g   2NH g Now, let's calculate the initial concentrations of the reactants and product:[N] = 0.2 mol / 1 L = 0.2 M[H] = 0.1 mol / 1 L = 0.1 M[NH] = 0 mol / 1 L = 0 M  since no ammonia is formed initially When the volume of the container is decreased to 0.5 L, the concentrations of the reactants and product will change:[N] = 0.2 mol / 0.5 L = 0.4 M[H] = 0.1 mol / 0.5 L = 0.2 M[NH] = 0 mol / 0.5 L = 0 M  initially Now, let's apply Le Chatelier's principle. When the volume is decreased, the pressure increases, and the system will try to counteract this change by shifting the equilibrium position to the side with fewer moles of gas. In this case, the forward reaction  formation of ammonia  has fewer moles of gas  2 moles of NH  compared to the reverse reaction  1 mole of N and 3 moles of H . Therefore, the equilibrium will shift to the right, favoring the formation of ammonia.As a result, the concentration of ammonia gas will increase at the new equilibrium position.