Question

For the following equilibrium, \(K_c = 6.3 × 10^{14}\) at \(1000\  K\)
\(NO (g) + O_3 (g) ⇋ NO_2 (g) + O_2 (g)\)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is \(K'_c\) for the reverse reaction?

Answer 1

Given:

Reaction:
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)

Equilibrium constant for forward reaction,
K_c = 6.3 × 10¹⁴ at 1000 K

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Step 1: Relationship between forward and reverse equilibrium constants

For a reversible reaction:

K_c(forward) × K_c(reverse) = 1

Therefore,

K_c^′ (reverse) = 1 / K_c

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Step 2: Calculate K_c^′

K_c^′ = 1 / (6.3 × 10¹⁴)

K_c^′ = 1.59 × 10⁻¹⁵

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Final Answer:

The equilibrium constant for the reverse reaction is
K_c^′ = 1.6 × 10⁻¹⁵