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Question

Chemistry Question on Equilibrium

In the reaction, PCl3(g)+Cl2(g)<=>PCl5(g) {PCl_{3(g)} + Cl_{2(g)} <=> PCl_{5(g)}} at equilibrium

A

Kp=Kc(RT)−1K_p = K_c(RT)^{-1}

B

Kp=KcRTK_p = K_c RT

C

Kp=Kc(RT)−2K_p = K_c(RT)^{-2}

D

Kp=KcK_p = K_c

Answer

Kp=Kc(RT)−1K_p = K_c(RT)^{-1}

Explanation

Solution

PCl3(g)+Cl2(g)<=>PCl5(g) {PCl_{3(g)} + Cl_{2(g)} <=> PCl_{5(g)}}
Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}
Δng=1−2=−1\Delta n_g = 1 - 2 = -1
∴   Kp=Kc(RT)−1\therefore\:\:\: K_p = K_c (RT)^{-1}