Question

Given: 2SO₂ + O₂ <------> 2 SO₃   K_eq=2.9

A Certain amount of SO₃ is introduced into 2.5 L container. if [O₂] at equilibrium is 0.95 mol/L.

a) What was the initial concentration of [SO₃] in the container?

b) What percent of SO₃ reacted?

Answer 1

Ans. #a. # Step 1: Create an ICE table as shown in figure.

Given – Equilibrium [O₂] = 0.5X = 0.95 mol/ L = 0.95 M

            So, X = 0.95 M / 0.5 = 1.90 M

# Also, Equilibrium, [SO₂] = X = 1.90 M

            And, Equilibrium [SO₃] = Y – X = (Y – 1.90) M

# Step 2: When a reaction is reversed, its keq value is inverse of the original value.

Now,

            Keq’ = 1 / 2.9 = [O₂] [SO₂]² / [SO₃]²

            Or, 0.3448 = 0.95 x (1.90)² / (Y – 1.90)² = 3.4295 / (Y – 1.90)²

            Or, (Y – 1.90)² = 3.4295 / 0.3448 = 9.9463

            Or, Y – 1.90 = square root (9.9463) = 3.154

            Or, Y = 3.154 + 1.90 = 5.054

Therefore, Initial [SO₃] = 5.054 M

#b. Equilibrium [SO₃] = 5.054 M – 1.90 M = 3.154 M

[SO₃] reacted = (Initial – Equilibrium) = 5.054 M – 3.154 M = 1.90 M

Now,

            % [SO₃] reacted = ([SO₃] reacted / Initial [SO₃]) x 100

                                                = (1.90 M / 5.054 M) x 100 = 37.59 %

+ 2 SO2 Initial Change Eq 0 +0.5X +0.5X 0 +0.5X +0.5X -X uilibrium Y X