A chemistry student needs to calculate the equilibrium constant for the following reaction:2 H2O(g) ⇌ 2 H2(g) + O2(g)The student has the initial concentration of H2 and O2 as 0.25 M and the initial concentration of H2O as 1.00 M. After the system reaches equilibrium, the student measures the concentration of H2 as 0.10 M. Using this information, what is the value of the equilibrium constant, Kc, for the above reaction?

Question

A chemistry student needs to calculate the equilibrium constant for the following reaction:2 H2O(g) ⇌ 2 H2(g) + O2(g)The student has the initial concentration of H2 and O2 as 0.25 M and the initial concentration of H2O as 1.00 M. After the system reaches equilibrium, the student measures the concentration of H2 as 0.10 M. Using this information, what is the v

Answer 1

To calculate the equilibrium constant, Kc, we first need to determine the changes in concentration for each species at equilibrium. We are given the initial concentrations and the final concentration of H2.Initial concentrations:[H2O] = 1.00 M[H2] = 0.25 M[O2] = 0.25 MFinal concentration:[H2] = 0.10 MSince the stoichiometry of the reaction is 2 H2O  2 H2 + O2, the changes in concentration can be determined as follows:[H2] = -[H2O] = 0.10 M - 0.25 M = -0.15 M[H2O] = -[H2] / 2 = 0.15 M / 2 = 0.075 M[O2] = -[H2] / 2 = 0.15 M / 2 = 0.075 MNow we can find the equilibrium concentrations:[H2O] = 1.00 M - 0.075 M = 0.925 M[H2] = 0.10 M[O2] = 0.25 M + 0.075 M = 0.325 MFinally, we can calculate the equilibrium constant, Kc, using the expression:Kc = [H2]^2 * [O2] / [H2O]^2Kc =  0.10 ^2 *  0.325  /  0.925 ^2Kc  0.0038The value of the equilibrium constant, Kc, for the given reaction is approximately 0.0038.