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Le Chatelier's Principle

In the follow...

Question

In the following reaction
${S O}_{2} (g) + O_{2} (g) ⇋ 2 {S O}_{3} (g)$
the equilibrium is not attained. The rate of forward reaction is greater than that of backward reaction. Thus, which of the following is the correct relation between $K_{p}$ and $Q_{p}$

K_{p} = Q_{p}

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K_{p} > Q_{p}

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K_{p} < Q_{p}

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K_{p} = Q_{p} = 1

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Solution

The correct option is B $K_{p} > Q_{p}$
In the following reaction
${S O}_{2} (g) + O_{2} (g) ⇋ 2 {S O}_{3} (g)$
the equilibrium is not attained. The rate of forward reaction is greater than that of backward reaction. Thus, the following is the correct relation between $K_{p}$ and $Q_{p}$
$K_{p} > Q_{p}$
The rate of forward reaction is greater than that of backward reaction. It means that the partial pressure of reactants is more than equilibrium value and the partial pressure of product is less than equilibrium value.

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2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

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2 N O C I (g) ⇌ 2 N O (g) + C l_{2} (g); K_{p} = 1.7 \times 10^{- 2}

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