A chemistry student wants to investigate the effect of changing the volume on the equilibrium position for the reaction between nitrogen and hydrogen gas to form ammonia. If the student initially has 1 mole of nitrogen gas and 3 moles of hydrogen gas in a 2-liter container at equilibrium, and then decreases the volume to 1 liter while maintaining the same temperature, what will happen to the equilibrium concentration of ammonia in moles per liter?

Question

A chemistry student wants to investigate the effect of changing the volume on the equilibrium position for the reaction between nitrogen and hydrogen gas to form ammonia. If the student initially has 1 mole of nitrogen gas and 3 moles of hydrogen gas in a 2-liter container at equilibrium, and then decreases the volume to 1 liter while maintaining

Answer 1

To answer this question, we need to consider the balanced chemical equation for the formation of ammonia:N2 g  + 3H2 g   2NH3 g Initially, the student has 1 mole of N2 and 3 moles of H2 in a 2-liter container. Let's assume that at equilibrium, x moles of N2 and 3x moles of H2 have reacted to form 2x moles of NH3.Now, when the volume is decreased to 1 liter, the concentrations of all species will double, since concentration is inversely proportional to volume. So, the new concentrations at equilibrium will be:[N2] =  1 - x  / 1 L[H2] =  3 - 3x  / 1 L[NH3] =  2x  / 1 LAccording to Le Chatelier's principle, when the volume of the container is decreased, the equilibrium will shift in the direction that reduces the number of moles of gas. In this case, the equilibrium will shift to the right, favoring the formation of ammonia.Since the equilibrium will shift to the right, the concentration of ammonia will increase. To determine the new equilibrium concentration of ammonia, we would need the equilibrium constant  K  for this reaction at the given temperature. However, we can conclude that the equilibrium concentration of ammonia in moles per liter will increase due to the decrease in volume.