In which of the following equilibrium equation $K_{p} > K_{c}$ ?

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

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P C l_{3} (g) + C l_{2} ⇌ P C l_{5} (g)

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H_{2} (g) + I_{2} (g) ⇌ 2 H l (g)

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N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

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Solution

The correct option is A $2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
$K_{p} = K_{c} (R T)^{Δ n}$
where $Δ n =$ (number of gaseous products) $-$ (No. of gaseous reactants)
So, for $K_{P} > K_{C}$
$Δ n \geq 1$
Here, $2 S O_{3} ⇋ 2 S O_{2} + O_{2}$
Has $Δ n = 3 - 2 = 1$
So, for this reaction $K_{P} > K_{C}$

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Similar questions

K_{p}

K_{c}

log \frac{K_{p}}{K_{c}} + log R T = 0

log \frac{K_{P}}{K_{C}} + log R T = 0

2 H I_{(g)} ⟶ H_{2 (g)} + I_{2 (g)}

2 N H_{3 (g)} ⟶ N_{2 (g)} + 3 H_{2 (g)}

2 N O_{(g)} ⟶ N_{2 (g)} + O_{2 (g)}

P C l_{5 (g)} ⟶ P C l_{3 (g)} + C l_{2 (g)}

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

2 H I (g) ⇌ H_{2} (g) + I_{2} (g)

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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