Discuss the relation between $K_{p}$ and $K_{c}$ for this equilibrium:
$2 H_{2} O (g) + 2 {C l}_{2 (g)} ⇌ 4 H {C l}_{(g)} + O_{2 (g)}$

K_{p} = K_{c}

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K_{p} > K_{c}

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K_{p} < K_{c}

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K_{p} = \frac{1}{K_{c}}

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Solution

The correct option is B $K_{p} > K_{c}$
$Δ n =$ Number of moles of gaseous products $-$ number of moles of gaseous reactants.
$Δ n = 5 - 4 = 1$
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{p} = K_{c} (R T)^{1}$
$K_{p} = K_{c} (R T)$
As $R T > 1$
$K_{p} > K_{c}$

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Discuss the relation between Kp and Kc for this equilibrium:2H2O(g)+2Cl2(g)⇌4HCl(g)+O2(g)

The correct option is B Kp>KcΔn= Number of moles of gaseous products − number of moles of gaseous reactants.Δn=5−4=1Kp=Kc(RT)ΔnKp=Kc(RT)1Kp=Kc(RT)As RT>1Kp>Kc

Discuss the relation between $K_{p}$ and $K_{c}$ for this equilibrium:
$2 H_{2} O (g) + 2 {C l}_{2 (g)} ⇌ 4 H {C l}_{(g)} + O_{2 (g)}$

The correct option is B $K_{p} > K_{c}$
$Δ n =$ Number of moles of gaseous products $-$ number of moles of gaseous reactants.
$Δ n = 5 - 4 = 1$
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{p} = K_{c} (R T)^{1}$
$K_{p} = K_{c} (R T)$
As $R T > 1$
$K_{p} > K_{c}$