Question

At \(450\ K\), \(K_p= 2.0 × 10^{10}/bar\) for the given reaction at equilibrium.
\(2SO_2(g) + O_2(g) ⇋ 2SO_3 (g)\)
What is \(K_c\) at this temperature ?

Answer 1

Given:

Temperature, T = 450 K
K_p = 2.0 × 10¹⁰ bar⁻¹
Reaction:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

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Step 1: Determine change in number of moles of gaseous species (Δn)

Δn = (moles of gaseous products) − (moles of gaseous reactants)

Δn = 2 − (2 + 1) = −1

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Step 2: Relation between K_p and K_c

K_p = K_c(RT)^Δn

For Δn = −1:

K_p = K_c / (RT)

∴ K_c = K_p × RT

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Step 3: Substitute the values

R = 0.08314 L·bar·mol⁻¹·K⁻¹

K_c = (2.0 × 10¹⁰) × (0.08314 × 450)

K_c = (2.0 × 10¹⁰) × 37.4

K_c = 7.48 × 10¹¹

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Final Answer:

The value of the equilibrium constant in terms of concentration at 450 K is
K_c = 7.5 × 10¹¹