Question

A chemistry student needs to determine the equilibrium concentrations of reactants and products for the following reaction:2 NOCl (g) ⇌ 2 NO (g) + Cl2 (g)The initial concentration of NOCl is 0.50 M and the equilibrium constant, Kc = 4.68 x 10^-3. What are the equilibrium concentrations of NO and Cl2?

Answer 1

To solve this problem, we will use an ICE  Initial, Change, Equilibrium  table to keep track of the concentrations of the reactants and products.Initial concentrations:[NOCl] = 0.50 M[NO] = 0 M[Cl2] = 0 MChange in concentrations:Since the stoichiometry of the reaction is 2:2:1, for every 2 moles of NOCl that react, 2 moles of NO and 1 mole of Cl2 are produced. Let x be the change in concentration of NOCl.-2x  moles of NOCl are consumed +2x  moles of NO are produced +x  moles of Cl2 are produced Equilibrium concentrations:[NOCl] = 0.50 - 2x[NO] = 2x[Cl2] = xNow we can use the equilibrium constant expression to solve for x:Kc = [NO]^2 * [Cl2] / [NOCl]^24.68 x 10^-3 =  2x ^2 *  x  /  0.50 - 2x ^2Solve for x:4.68 x 10^-3 *  0.50 - 2x ^2 = 4x^30.00234 - 0.01872x + 0.03744x^2 = 4x^3Rearrange the equation:4x^3 - 0.03744x^2 + 0.01872x - 0.00234 = 0Solving this cubic equation, we find that x  0.002  ignoring the negative and non-physical solutions .Now we can find the equilibrium concentrations:[NOCl] = 0.50 - 2x  0.50 - 2 0.002   0.496 M[NO] = 2x  2 0.002   0.004 M[Cl2] = x  0.002 MSo, the equilibrium concentrations are approximately [NOCl] = 0.496 M, [NO] = 0.004 M, and [Cl2] = 0.002 M.