What is the relationship between Kp and Kc for the reaction PCl5 in equilibrium with PCl3 and Cl2?

B. K_p = K_c(RT) — The relationship is given by Kp = Kc(RT)^delta n, where delta n is the change in the number of moles of gaseous species.

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$What is the relationship between K_{p} and K_{c} for the reaction PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g) ?$

A. $K_{p} = K_{c}$
B. $K_{p} = K_{c} (R T)$
C. $K_{p} = K_{c} / (R T)$
D. $K_{p} = K_{c} (R T)^{2}$

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Correct answer: B — $K_{p} = K_{c} (R T)$

Solution

$The formula is K_{p} = K_{c} (R T)^{Δ n} .$
$For the reaction PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g), the number of moles of gaseous products is $1 + 1 = 2$.$
$The number of moles of gaseous reactants is 1 .$
$Thus, Δ n = 2 - 1 = 1 .$
$Substituting Δ n = 1, we get K_{p} = K_{c} (R T)^{1} .$
$Hence the answer is (B).$

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What is the relationship between Kp​ and Kc​ for the reaction PCl5​(g)⇌PCl3​(g)+Cl2​(g)?

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$What is the relationship between K_{p} and K_{c} for the reaction PCl_{5} (g) ⇌ PCl_{3} (g) + Cl_{2} (g) ?$