Question

A chemistry student wants to find out the effect on the equilibrium position when the volume of a reaction mixture is changed. They have an equilibrium mixture of hydrogen and iodine gases forming hydrogen iodide gas according to the equation:$$H_2 (g) + I_2 (g) \leftrightharpoons 2HI (g)$$At 298 K, the equilibrium constant for this reaction is 54.3. Initially, the reaction mixture contains 0.5 moles of hydrogen gas, 0.5 moles of iodine gas, and 0 moles of hydrogen iodide gas in a container of volume 2L. If the volume is reduced to 1L while maintaining the temperature at 298 K, what will be the new concentration of hydrogen, iodine, and hydrogen iodide gases, and in which direction will the equilibrium shift?

Answer 1

To solve this problem, we need to first find the initial concentrations of the reactants and products, and then use the reaction quotient  Q  to determine the direction of the equilibrium shift. Finally, we will use the equilibrium constant  K  to find the new concentrations of the reactants and products.1. Calculate the initial concentrations of the reactants and products:Initial concentration of H = moles/volume = 0.5 moles / 2 L = 0.25 MInitial concentration of I = moles/volume = 0.5 moles / 2 L = 0.25 MInitial concentration of HI = moles/volume = 0 moles / 2 L = 0 M2. Calculate the reaction quotient  Q  using the initial concentrations:$$Q = \frac{[HI]^2}{[H_2][I_2]} = \frac{0^2}{ 0.25  0.25 } = 0$$Since Q < K  0 < 54.3 , the reaction will shift to the right  toward the products  to reach equilibrium.3. Calculate the new concentrations after reducing the volume to 1 L:When the volume is reduced to 1 L, the concentrations of the reactants and products will double:New concentration of H = 0.25 M * 2 = 0.5 MNew concentration of I = 0.25 M * 2 = 0.5 MNew concentration of HI = 0 M * 2 = 0 M4. Use the equilibrium constant  K  to find the new equilibrium concentrations:Let x be the change in concentration of H and I, and 2x be the change in concentration of HI. At equilibrium:[H] = 0.5 - x[I] = 0.5 - x[HI] = 0 + 2xNow, we can use the equilibrium constant expression:$$K = \frac{[HI]^2}{[H_2][I_2]}$$Substitute the equilibrium concentrations and the value of K:$$54.3 = \frac{ 2x ^2}{ 0.5 - x  0.5 - x }$$Solve for x:$$54.3 = \frac{4x^2}{ 0.5 - x ^2}$$$$54.3 0.5 - x ^2 = 4x^2$$This is a quadratic equation that can be solved either by factoring, using the quadratic formula, or by numerical methods. The positive solution for x is approximately 0.436.5. Calculate the new equilibrium concentrations:[H] = 0.5 - x = 0.5 - 0.436  0.064 M[I] = 0.5 - x = 0.5 - 0.436  0.064 M[HI] = 2x = 2 * 0.436  0.872 MSo, the new equilibrium concentrations are approximately 0.064 M for H, 0.064 M for I, and 0.872 M for HI. The equilibrium shifted to the right, favoring the formation of hydrogen iodide gas.