Question

A student wants to investigate the effect of volume change on the equilibrium position of the reaction 2SO2(g) + O2(g) ⇄ 2SO3(g). The initial equilibrium concentration of SO3 is 0.1 M at a constant temperature. The volume is reduced by half at equilibrium. Calculate the new equilibrium concentrations of SO2, O2 and SO3 assuming the equilibrium constant (Kc) of the reaction remains constant.

Answer 1

To solve this problem, we need to use the reaction quotient  Q  and the equilibrium constant  Kc . The reaction is:2SO2 g  + O2 g   2SO3 g Let's denote the initial equilibrium concentrations as follows:[SO2] = x[O2] = y[SO3] = 0.1 MWhen the volume is reduced by half, the concentrations will double:[SO2] = 2x[O2] = 2y[SO3] = 0.2 MNow, let's denote the changes in concentration at the new equilibrium as follows:[SO2] = 2x - 2a[O2] = 2y - b[SO3] = 0.2 + 2aSince the equilibrium constant  Kc  remains constant, we can write the expression for Kc as follows:Kc = [SO3]^2 /  [SO2]^2 * [O2] Now, we need to find the expression for Q, the reaction quotient, which has the same form as Kc but uses the initial concentrations after the volume change:Q = [ 0.2 + 2a ^2] /  [ 2x - 2a ^2] *  2y - b  At equilibrium, Q = Kc. Therefore, we can set up the following equation:[ 0.2 + 2a ^2] /  [ 2x - 2a ^2] *  2y - b   = KcUnfortunately, we cannot solve this equation without knowing the values of Kc, x, and y. However, if the student can determine the initial equilibrium concentrations of SO2 and O2, as well as the equilibrium constant, they can use this equation to calculate the new equilibrium concentrations of SO2, O2, and SO3 after the volume change.