Question

At a certain temperature the reaction CO(g) + H2O(g) CO2(g) + H2(g) has Kc = 0.400. Exactly 1.00 mol of each gas was placed in a 100.0 L vessel and the mixture underwent reaction. What was the equilibrium concentration of each gas?

Answer 1

M = 1/100 = 0.01

K = CO2*H2/(CO*H2O)

K = 0.4

find concnetrations

[CO2] = 0.01 -x

[H2]= 0.01 -x

[CO] = 0.01+x

[H2O] = 0.01+x

substitutein

K = CO2*H2/(CO*H2O)

0.4 = (0.01+x)(0.01+x)/((0.01 -x)(0.01 -x))

0.4 = (0.01+x)^2 / (0.01-x)^2

solve for x

sqrt(0.4) = (0.01+x)/(0.01-x)

(0.01-x)0.632 = 0.01+x

0.00632 - 0.632x = 0.01 + x

1.632x = -0.01+0.00632

x = (-0.01+0.00632 )/1.632 = -0.0022

[CO2] = 0.01 +x = 0.01--0.0022 = 0.0078

[H2]= 0.01 +x= 0.01-0.0022 = 0.0078

[CO] = 0.01-x = 0.01+0.0022 = 0.0122

[H2O] = 0.01+x = 0.01+0.0022 = 0.0122

proof

K = CO2*H2/(CO*H2O)

K = (0.0022*0.0022)/(0.0122*0.0122) = 0.4