Question

For the reaction 2H2O(g)↽−−⇀ 2H2(g)+O2(g)2⁢H2O(g)⁢↽−−⇀ 2⁢H2⁡(g)+O2⁡(g) the equilibrium concentrations were found to be [H2O]=0.250 M[H2O]=0.250 M, [H2]=0.330 M[H2]=0.330 M, and [O2]=0.800 M[O2]=0.800 M. What is the equilibrium constant for this reaction?

Answer 1

The equilibrium constant (Kc) for the given reaction 2H2O(g) ⇌ 2H2(g) + O2(g) with the provided concentrations is calculated as 1.394.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the reaction 2H2O(g) ⇌ 2H2(g) + O2(g), use the equilibrium concentrations provided: [H2O] = 0.250 M, [H2] = 0.330 M, and [O2] = 0.800 M. The expression for the equilibrium constant is Kc = ([H2]2[O2]) / [H2O]2. Substituting the given concentrations into this expression gives Kc = ((0.330)2(0.800)) / (0.250)2.

After performing the calculations, Kc = (0.1089 × 0.800) / 0.0625 = 0.08712 / 0.0625 = 1.394. Therefore, the equilibrium constant for this reaction is 1.394.