Find [C] at equilibrium for A(g) + 2B(g) ⇌ 3C(g)

Practice question

Question

For the reaction A(g) + 2B(g) <=> 3C(g) , Kc = 27 at 400 K. If 1 mole of A and 3 moles of B are placed in a 1 L vessel, what is [C] at equilibrium?

Options

Choose one · Correct answer highlighted

2.25 M Correct 1.5 M 3.0 M 0.75 M

Explanation

Initial: [A] = 1 M , [B] = 3 M , [C] = 0 . Let 3x be moles of C formed, so A decreases by x , B by 2x . At equilibrium: [A] = 1 - x , [B] = 3 - 2x , [C] = 3x . Kc = ([C]³/[A][B]²) = ((3x)³/(1 - x)(3 - 2x)²) = 27 , (27x³/(1 - x)(3 - 2x)²) = 27 , (x³/(1 - x)(3 - 2x)²) = 1 . Solving, x³ = (1 - x)(3 - 2x)² , test x = 0.5 : (0.5)³ = 0.125 , (1 - 0.5)(3 - 1)² = 0.5 × 4 = 2 (not equal). Solving numerically, x ≈ 0.75 , [C] = 3 × 0.75 = 2.25 M .

Share this question