The value of Kc = 4.24 at 800K for the reaction, CO (g) + H2O (g) ? CO2 (g) + H2 (g) Calculate equilibrium concentrations of CO2, H2, CO and H2O at 800 K, if only CO and H2O are present initially at concentrations of 0.10M each ?

For the reaction, CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g) Initial concentration: 0.1M 0.1 M 0 0 Let x mole per litre of each of the productbe formed. At equilibrium: 0.1 - x M 0.1 - x M x M x M where x is the amount of CO2 and H2 at equilibrium. . Hence, equilibrium constant can be written as, Kc= x2/(0.1-x)2 = 4.24 x2 = 4.24(0.01 + x2-0.2x) x2 = 0.0424 + 4.24x2-0.848x 3.24x2 – 0.848x + 0.0424 = 0 a = 3.24, b = – 0.848, c = 0.0424 for quadratic equation ax2 + bx + c = 0, Thus solving we get two values of x x1= 0.067 x2= 0.194 Neglecting x2= 0.194 becuase x could not be more than initial concentration. Hence the equilibrium concentrations are, [CO2] = [H2] = x = 0.067 M [CO] = [H2O] = 0.1 – 0.067 = 0.033 M