A mixture of 0.430 M H 2 O , 0.398 M Cl 2 O , and 0.732 M HClO are enclosed in a vessel at 25 ° C . H 2 O ( g ) + Cl 2 O ( g ) − ⇀ ↽ − 2 HOCl ( g ) K c = 0.0900 at 25 ° C Calculate the equilibrium

Question

A mixture of 0.430 M H 2 O , 0.398 M Cl 2 O , and 0.732 M HClO are enclosed in a vessel at 25 ° C . H 2 O ( g ) + Cl 2 O ( g ) − ⇀ ↽ − 2 HOCl ( g ) K c = 0.0900 at 25 ° C Calculate the equilibrium concentrations of each gas at 25 ° C .

[H_2O]=

[Cl_2O]=

[HOCl]=

Answer 1

H₂O(g) + Cl₂O(g) <==> 2HOCl(g) ... Kc = 0.0900

Write the equilibrium expression:

Kc = [HOCl]² / [Cl₂O]

Set up an ICE table:

H₂O(g) + Cl₂O(g) <==> 2HOCl(g)

0.430...........0.398...............0.732.............Initial

+x.................+x...................-2x.................Change (NOTE: Kc < 1, and [HOCl] is greater than [H₂O] & [Cl₂O]

0.430+x.....0.398+x...........0.732-2x..........Equilibrium

Plug into equilibrium expression and solve for x:

0.0900 = (0.732-2x)² / (0.430+x)(0.398+x)

To simplify, let 0.430+x and 0.398+x be equal to each other. We can then take the square root of both sides.

0.0900 = (0.732 - 2x)² / (0.430 - x)² and taking √ of both sides we get...

0.300 = 0.732- 2x / 0.430 - x

0.732 - 2x = 0.129 - 0.300x

1.70x = 0.603

x = 0.355 M

At equilibrium,...

[H₂O] = 0.430 + 0.355 = 0.785 M

[Cl₂O] = 0.398 + 0.355 = 0.753 M

[HOCl] = 0.732 - (2x0.355) = 0.732 - 0.710 = 0.022 M

Again, please check all the math. Also, you may be required to solve the quadratic without making the assumptions that have been made in the above solution.