Question

Suppose a 500. mL flask is filled with 0.70 mol of H2 and 0.60 mol of HI. The following reaction becomes possible: -2H1(g) The equilibrium constant K for this reaction is 2.38 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places.

Answer 1

Ans. Initial [H₂] = Moles of H₂ / Volume of reaction vessel in L

                        = 0.70 mol / 0.500 L

                        = 1.40 M

Initial [HI] = 0.60 mol / 0.500 L = 1.20 M

# Since there is no I₂ gas on the reactant side, the reaction can’t go in forward direction because there is no I₂as reactant at all. So, the reaction must go to the backward direction to establish equilibrium.

So, we have to proceed with the reversed reaction        2 HI(g) -----> H₂(g) + I₂(g)

Reversing a reaction inverses the original value of equilibrium constant.

So, equilibrium constant for 2 HI(g) -----> H₂(g) + I₂(g)         , K’ =1 / 2.38 = 0.420

# Create an ICE table for the reversed reaction as shown below-

2 HI (g)H2(g) + 12(g) Initial (M) Change (M) Equilibrium (M) 1.2 (-2x) 1.2- 2x 1.4 0 +X

Now,

            K’ = [X (1.4 + X) ] / (1.2 – 2X)²

            Or, 0.420 = (1.4X + X²) / (1.44 + 4X² – 4.8X)

            Or, 0.420 (1.44 + 4X² – 4.8X) = 1.4X + X²

            Or, 0.6048 + 1.68X² – 2.016X = 1.4X + X²

            Or, 0.6048 + 1.68X² – 2.016X - 1.4X - X² = 0

            Or, 0.68X² – 3.416X + 0.6048 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 4.839                 ; X2 = 0.184

Since X can’t be greater than 1.2, reject X1.

Hence, X = 0.184

# Equilibrium [H₂] = 1.4 + X = 1.4 + 0.184 = 1.584

            Hence, equilibrium [H₂] = 1.58 M