Question

A mixture of 0.5940 M CO and 0.3950 M Cl2 is enclosed in a vessel and heated to 1000 K . CO(g)+Cl2(g)↽−−⇀COCl2(g)c=255.0 at 1000 K Calculate the equilibrium concentration of each gas at 1000 K

Answer 1

Answer:

[CO] = 0.2064M

[Cl2] = 0.0074M

[COCl2] = 0.3876M

Explanation:

Based on the reaction, Kc at 1000K is:

Kc = 255.0 = [COCl2] / [CO] [Cl2]

Where [] are concentrations at equilibrium

As the reaction is 1:1, the concentrations at equilibrium are:

[CO] = 0.5940M - X

[Cl2] = 0.3950M - X

[COCl2] = X

Where X is reaction coordinate

Replacing:

255.0 = [X] / [0.5940 - X] [0.3950 - X]

255.0 = X / 0.23463 - 0.989 X + X²

59.8307 - 252.195 X + 255 X² = X

59.8307 - 253.195 X + 255 X² = 0

Solving for X:

X = 0.605M. False solution because produce negative answers

X = 0.3876M. Right solution.

Replacing:

[CO] = 0.5940M - 0.3876M

[Cl2] = 0.3950M - 0.3876M

[COCl2] = 0.3876M

[CO] = 0.2064M

[Cl2] = 0.0074M

[COCl2] = 0.3876M

Answer 2

The concentration at equilibrium of CO is 0.2064 M, $\rm Cl_2$ is 0.0074 M, and $\rm COCl_2$ is 0.3876 M.

The initial concentration of CO = 0.5940 M

$\rm Cl_2$ = 0.3950 M

At equilibrium,

Let the concentration of formed $\rm COCl_2$ = x

Concentration of CO = 0.5940 - x M

Concentration of $\rm Cl_2$ = 0.3950 - x M

$\rm K_c$ = 255

$\rm K_c$ = $\rm \dfrac{[COCl_2]}{[CO]\;[Cl_2]}$

255 = $\rm \dfrac{x}{[0.5940-x]\;[0.3950-x]}$

(151.47 - 255x) (100.727-255x) = x

15,256 - 65,025 $\rm x^2$ = x

0 =  65,025 $\rm x^2$ +x - 15,256

x = $\rm\dfrac{-b\;\pm\;\sqrt{b^2\;-\;4ac} }{2a}$

a = 65,025

b = 1

c = -15,256

x = 0.38 M

Thus, the concentration at equilibrium of CO is 0.2064 M, $\rm Cl_2$ is 0.0074 M, and $\rm COCl_2$ is 0.3876 M.

For more information about the equilibrium concentration, refer to the link:

https://brainly.com/question/16645766