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? [CO] = M [H2O] = M [H2] = M [CO2] =

Answer 1

K = 0.4

in equilibirum

CO2 = 0.01+x

H2 = 0.01+x

H2O = 0.01-x

CO = 0.01-x

substitute

K = CO2*H2 / (H2O*CO)

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

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

0.6325(0.01-x) = 0.01+x

0.006325 - 0.6325x = 0.01 + x

-1.6325x = 0.01-0.006325

x = (0.01-0.006325)/-1.6325 = -0.00225

substitute

CO2 = 0.01+x = 0.01+-0.00225 = 0.00775

H2 = 0.01+x = 0.01+-0.00225 = 0.00775

H2O = 0.01-x = 0.01+0.00225 = 0.01225

CO = 0.01-x= 0.01+0.00225 = 0.01225

proof

K = (0.00775^2)/(0.01225^2) = 0.4