Question

At 450 K, K_p = 2.0 x 10¹¹/bar for the given reaction at equilibrium

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

What is K_c at this temperature?

Answer 1

K_p = K_c(RT)^Δng or K_c = \(\frac{K_p}{(RT)^{\Delta ng}}\)

K_P = 2.0 x 10¹⁰ bar^-1

R = 0.083 L bar K^-1 mol^-1

T = 450 K

Δ^ng = 2 - 3 = -1

K_c = \(\frac{(2.0 \times 10^{10}\,bar^{-1})}{[(0.083\, L\, bar\, k^{-1}\, mol^{-1})\times (450\, K)]^{-1}}\)

K_c = (2.0 x 10¹⁰ bar^-1) x (0.083 L bar K^-1 mol^-1) x (450 K)

= 7.47 x 10¹¹ mol L^-1 = 7.47 x 10¹¹ M^-1.