Question

Help Me Please! Equilibrium Constants and Kc?

Consider the following reactions and their equilibrium constants at 301 K.
PCl5(g) ⇌1/4 P4(g) + 5/2 Cl2(g);
Kc = 3.70 ✕ 10−22

1/4 P4(g) + 3/2 Cl2(g) ⇌ PCl3(g);
Kc = 5.74 ✕ 1026

Calculate Kc for PCl5(g) ⇌ PCl3(g) + Cl2(g) at the same temperature.

Thank you for you Knowledge and teaching,
Sincerely, Ozzy

Answer 1

`K_c = 21.238 ✕ 10^(4)`

Explanation:

`PCl_5 (g) ⇌1/4 P_4 (g) + 5/2 Cl_2 (g)`
`K_(c1) = ([Cl_2]^(5/2)*[P]^(1/4))/[PCl_5] = 3.70 ✕ 10^(−22)`

`1/4 P_4 (g) + 3/2 Cl_2 (g) ⇌ PCl_3 (g)`
`K_(c2) = [PCl_3]/([P_4]^(1/4) * [Cl_2]^(3/2)) = 5.74 ✕ 10^(26)`

`PCl_5 (g) ⇌ PCl_3 (g) + Cl_2 (g)`
`K_c = ([PCl_3] * [Cl_2])/ [PCl_5]`
Multiplying the numerator and denominator by `[Cl_2]^(3/2) * [P_4]^(1/4)`
`K_c = [PCl_3]/ [PCl_5] * [Cl_2]^(5/2)/[Cl_2]^(3/2) * [P]^(1/4)/[P]^(1/4)`
Rearranging the terms,
`K_c = ([Cl_2]^(5/2)*[P]^(1/4))/[PCl_5] * [PCl_3]/([P_4]^(1/4) * [Cl_2]^(3/2))`
`K_c = K_(c1) * K_(c2)`

Therefore,
`K_c = 3.70 ✕ 10^(−22) * 5.74 ✕ 10^(26)`
`K_c = 21.238 ✕ 10^(4)`