Question

Determine the equilibrium concentrations of reactants and products for the reaction 2H2(g) + O2(g) ⇌ 2H2O(g) at 500K when initial concentrations of H2 and O2 are 0.2M and 0.1M respectively, and the equilibrium constant (Kc) is 4.67 x 10^3.

Answer 1

To determine the equilibrium concentrations of the reactants and products, we can use an ICE  Initial, Change, Equilibrium  table. Initial concentrations:[H2] = 0.2 M[O2] = 0.1 M[H2O] = 0 MChange in concentrations:Let x be the change in concentration of H2 and O2. Since the stoichiometry of the reaction is 2:1:2, the change in concentration of H2O will be 2x.Equilibrium concentrations:[H2] = 0.2 - 2x[O2] = 0.1 - x[H2O] = 2xNow, we can write the expression for the equilibrium constant  Kc :Kc = [H2O]^2 /  [H2]^2 * [O2] Plug in the equilibrium concentrations and the given Kc value:4.67 x 10^3 =  2x ^2 /   0.2 - 2x ^2 *  0.1 - x  Now, we need to solve for x. Since this is a quadratic equation, it might be difficult to solve it algebraically. Instead, we can use a numerical method or a calculator to find the value of x.Using a calculator, we find that x  0.0485.Now, we can find the equilibrium concentrations:[H2] = 0.2 - 2x  0.2 - 2 0.0485   0.103 M[O2] = 0.1 - x  0.1 - 0.0485  0.0515 M[H2O] = 2x  2 0.0485   0.097 MSo, the equilibrium concentrations are approximately [H2] = 0.103 M, [O2] = 0.0515 M, and [H2O] = 0.097 M.