Consider the reaction 2 Cl2 (g) + 2 H2O (g) 4 HCl (g) + O2(g) At 480.0 C, a reac

Question

Consider the reaction 2 Cl2 (g) + 2 H2O (g) 4 HCl (g) + O2(g) At 480.0 C, a reaction flask containing 1.00 M Cl2, 0.80 M H2O, 0.70 M HCl, and 0.20 M O2,was at equilibrium. Some H2O (g) was then added to the flask. When the equilibrium was re-established, the concentration of HCl became 1.10 M. What was the new equilibrium concentration of H2O (g) and how much H2O (g) was added to the flask? (Hint: First calculate the equilibrium constant of the reaction using the old equilibrium concentrations.)

Explanation / Answer

As the hint states, please calculate the equilibrium in the inital condition

Keq = [products]/[reactants]

Keq = [HCl]^4[O2]/[H2O]^2[CL2]^2

Keq = [1]^4[0.2]/[0.8^]2[0.7]^2= 0.637

Keq = 0.637

Now, for a change in Water, there will be some changes as well that change call it "x"

Initial Concentrations [HCl, O2, H2O, Cl2] = 1, 0.2, 0.8, 0.7

Change of Concentrations [HCl, O2, H2O, Cl2] = 4x, x, -2x, -2x (be sure to assign the correct stoichiometric factor)

Final Concentration [HCl, O2, H2O, Cl2] = 1+4x, 0.2+x, 0.8-2x, 0.7-2x

From here, our value of x can be calculated:

1.10 = 1+4x -> x = 0.10/4 = 0.025

Now substitute these final concentrations in the Keq NEW

Keq NEW = [1+x]^4[O2]/[H2O]^2[CL2]^2

Keq NEW = [1.1]^4[0.2+0.025]/[0.8+0.025]^2[0.7+0.025]^2

Keq New = 0.92

x will be the value added: 0.025 mol/L... you will need the vlaue of the total volume to know exactly how much was that

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