Question

Consider the reaction: 2H2O (g) ⇄ 2H2 (g) + O2 (g) At 1000 K, the equilibrium constant Kc for this reaction is 4.17 x 10^-5. Suppose that initially, 0.10 moles of H2O, 0.20 moles of H2, and 0.20 moles of O2 are placed in a 1.0 L container at 1000 K. What will be the equilibrium concentrations of H2O, H2, and O2?

Answer 1

To find the equilibrium concentrations, we can use an ICE  Initial, Change, Equilibrium  table:``          H2O   H2    O2Initial   0.10  0.20  0.20Change   -2x    +2x   +xEquilibrium 0.10-2x 0.20+2x 0.20+x``Now, we can write the expression for the equilibrium constant Kc:Kc =  [H2]^2 * [O2]  / [H2O]^2Plug in the equilibrium concentrations and the given value of Kc:4.17 x 10^-5 =   0.20+2x ^2 *  0.20+x   /  0.10-2x ^2This is a quadratic equation that needs to be solved for x. Solving the equation, we get:x  0.004Now, we can find the equilibrium concentrations:[H2O] = 0.10 - 2x  0.10 - 2 0.004   0.092 M[H2] = 0.20 + 2x  0.20 + 2 0.004   0.208 M[O2] = 0.20 + x  0.20 + 0.004  0.204 MSo, the equilibrium concentrations are approximately 0.092 M for H2O, 0.208 M for H2, and 0.204 M for O2.