Question

A mixture of 0.577 M H_2O , 0.314 M Cl_2O , and 0.666 M HClO are enclosed in a vessel at 25°C .
H_2O(g) + Cl_2O(g) <-------> 2 HOCl (g) Kc = 0.0900 at 25°C
1. Calculate the equilibrium concentrations of each gas at 25°C .

Answer 1

Answer:

Equilibrium Concentration of H₂O(g)  = 0.803

Equilibrium Concentration of Cl₂O(g)  = 0.540

Equilibrium Concentration of HOCl (g) = 0.214

Explanation:

Given;

                H₂O(g)  +   Cl₂O(g) <-------> 2HOCl (g)

I                 0.577        0.314                   0.666                

C               - x              -x                        +2x

E             0.577 - x     0.314 - x               0.666 +2x

$K_c = \frac{[HOCL]^2}{[H_2O][CL_2O]} \\\\0.09 = \frac{[0.666+2x]^2}{[0.577-x][0.314-x]}\\\\0.09(0.1812 -0.891x+x^2) = (0.666+2x)(0.666+2x)\\\\0.0163-0.0802x+0.09x^2 = 0.4436+2.664x+4x^2\\\\3.91x^2+2.7742x+0.4273 =0\\\\x = -0.226, or -0.483$

Equilibrium Concentration of H₂O(g) = 0.577 - (- 0.226) = 0.803

Equilibrium Concentration of Cl₂O(g) = 0.314 - (- 0.226) = 0.540

Equilibrium Concentration of HOCl (g)  = 0.666 +2(- 0.226) = 0.214

Thus, from the result it can be seen that at equilibrium, the reactants are favored.