Question

Calculate the equilibrium concentration of each species (in mol/L) in a system containing 0.200 M KI and 0.200 M Cl2 that forms KCl and I2 at equilibrium, given that the equilibrium constant (Kc) is 9.0 x 10^-2 at a certain temperature.

Answer 1

To calculate the equilibrium concentrations, we can use an ICE  Initial, Change, Equilibrium  table. The balanced chemical equation for the reaction is:2 KI + Cl2  2 KCl + I2Initial concentrations:[KI] = 0.200 M[Cl2] = 0.200 M[KCl] = 0 M  assuming no KCl is present initially [I2] = 0 M  assuming no I2 is present initially Change in concentrations:Let x be the change in concentration for KI and Cl2.Then, the change for KCl and I2 will be 2x and x, respectively.Equilibrium concentrations:[KI] = 0.200 - x[Cl2] = 0.200 - x[KCl] = 2x[I2] = xNow, we can use the equilibrium constant  Kc  expression to solve for x:Kc = [KCl]^2 * [I2] /  [KI]^2 * [Cl2] 9.0 x 10^-2 =  2x ^2 * x /   0.200 - x ^2 *  0.200 - x  Now, we need to solve for x:9.0 x 10^-2 = 4x^3 /  0.200 - x ^3Let's assume that x is much smaller than 0.200, so we can simplify the equation:9.0 x 10^-2  4x^3 /  0.200 ^3Now, solve for x:x   9.0 x 10^-2 *  0.200 ^3  / 4x  0.0016 MNow, we can find the equilibrium concentrations:[KI] = 0.200 - x  0.200 - 0.0016  0.1984 M[Cl2] = 0.200 - x  0.200 - 0.0016  0.1984 M[KCl] = 2x  2 * 0.0016  0.0032 M[I2] = x  0.0016 MSo, the equilibrium concentrations are approximately:[KI]  0.1984 M[Cl2]  0.1984 M[KCl]  0.0032 M[I2]  0.0016 M