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

2SO2(g) + O2(g) ⇌ 2SO3(g)

What is Kc at this temperature?

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

KP = 2.0 x 1010 bar-1

R = 0.083 L bar K-1 mol-1

T = 450 K

Δng = 2 - 3 = -1

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

Kc = (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.

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