For the following reaction, K c = 255 at 1000 K. CO (g) + Cl2 (g) COCl2 (g) A re

Question

For the following reaction, Kc = 255 at 1000 K.
CO (g) + Cl2 (g) COCl2 (g)
A reaction mixture initially contains a CO concentration of 0.1470 M and a Cl2 concentration of 0.172 M at 1000 K.

Part A

What is the equilibrium concentration of CO at 1000 K?

Express your answer in molarity to three significant figures.

Part B

What is the equilibrium concentration of Cl2 at 1000 K?

Express your answer in molarity to three significant figures.

Part C

What is the equilibrium concentration of COCl2 at 1000 K?

Explanation / Answer

Given,

Kc = 255 at 1000 K

Initial concentrations, [CO]i = 0.1470 M, [Cl2]i = 0.172 M

At equilibrium, Kc = [COCl2] / [CO] [Cl2]

Kc = x / (0.1470-x)(0.172-x)

or 255 = x / (x2 - 0.319x + 0.0253 )

   255x2 - 82.345 x + 6.452 = 0

solving for x,

x = [b ± (b24ac)] / 2a

   = [- (-82.345) ± {(-82.345)2 - 4(255)(6.452)}] / 2(255)

   = [82.345 ± 14.13] / 510

now, x = 0.189 or 0.133

Because there are two possible solutions, we have to determine which is the real solution.

[CO]i - 0.189 = 0.1470 - 0.189 = - 0.042

[Cl2]i - 0.189 = 0.172 - 0.189 = - 0.017

So, possible value of x is 0.133

Thus at equilibrium (at 1000 k),  

Part A : [CO] = 0.01470 - x = 0.1470 - 0.133 = 0.014 M

Part B : [Cl2] = 0.172 - 0.133 = 0.039 M

Part C : [COCl2] = 0.133 M

Condition [CO] [CO] [COCl2] Initial 0.1470 0.172 0 Change -x -x +x Equilibrium 0.1470-x 0.172-x x

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