In the chemical reaction: Br2 (g) + Cl2 (g) ↔ 2 BrCl (g) KP = 0.150 If there is initially 0.500 atm of BrCl and nothing else in a container, what direction does the reaction move: a) Q > K and the reaction moves to the right b) Q > K and the reaction moves to the left c) Q < K and the reaction moves to the left d) Q < K and the reaction moves to the right

Question

Kinley Lowery

Chemistry

12 August, 06:41

In the chemical reaction: Br2 (g) + Cl2 (g) ↔ 2 BrCl (g) KP = 0.150 If there is initially 0.500 atm of BrCl and nothing else in a container, what direction does the reaction move: a) Q > K and the reaction moves to the right b) Q > K and the reaction moves to the left c) Q < K and the reaction moves to the left d) Q < K and the reaction moves to the right

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Aiyana Alvarez · Answer 1

Where delta n = no. of moles of products - no. of moles of reactants = 2 - 2 = 0

so Kp = Kc (RT) ^0 = Kc X 1 = Kc

so Kp = Kc = 0.14

now Kc = [Cl2] [Br2] / [BrCl]^2

0.14 = 0.02 X 0.02 / [BrCl]^2

[BrCl]^2 = 4 X 10^-4 / 0.14 = 2.857 X 10^-3

[BrCl] = square root of (2.857 X 10^-3) = 0.0535

now you know that PV = nRT

P = n/VRT = CRT

so for BrCl ... C = 0.0535 M or 0.0535 moles/L

R = 0.0821 L atm/K/mole

T = 350 K

so P = 0.0535 X 0.0821 X 350 = 1.537 atm

so partial pressure of BrCl at equilibrium = 1.537 atm