Question:
At 450K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

2SO₂(g) + O₂(g) ↔ 2SO₃ (g)

What is K_c at this temperature ?

Answer:

For the given reaction,

Δn = 2 - 3 =-1

T = 450 K

R = 0.0831 bar L bar K^-1 mol^-1

K_p = 2.0 × 10¹⁰ bar^-1

We know that,

K_p = K_c (RT)Δn

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Question:
At 450K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

2SO₂(g) + O₂(g) ↔ 2SO₃ (g)

What is K_c at this temperature ?

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Question: At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) ↔ 2SO3 (g) What is Kc at this temperature ?

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Question:
At 450K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

2SO₂(g) + O₂(g) ↔ 2SO₃ (g)

What is K_c at this temperature ?