Question

Consider the equilibrium between COCl2, CO and Cl2. COCl2(g) CO(g) + Cl2(g) K = 0.112 at 904 K The reaction is allowed to reach equilibrium in a 7.00-L flask. At equilibrium, [COCl2] = 0.258 M, [CO] = 0.170 M and [Cl2] = 0.170 M. (a) The equilibrium mixture is transferred to a 14.0-L flask. In which direction will the reaction proceed to reach equilibrium? (b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 14.0-L flask. [COCl2] = M [CO] = M [Cl2] = M

Answer 1

a) Equilibrium COCi, (g) C0(g) Cl2 (g) Increasein volume decreases the pressure. According to the Le-chatlier principle, decreasing the pres sure shifts the equilibrium towar ds the side where there will be more number of moles of gasses. Since there are more number(1) of moles of gasses on products side, increase in volume shifts the equilibrium towards products (right) side Thus, the reaction will proceed to right side in the forward directi on to re establish equilibrium (7.00 L) (0.258 M) Initi al concentrati on of COCI2,COCI = 0,129 14.0 L (7.00 L) (0.170 M) [Co]= Initi al concentrati on of CO = 0,085 14.0 L [a-7.00 L) (0.170 M) 14.0 L Initi al concentrati on of Cl2, = 0.085 b) The ICE table constructed is as follows cOC, (g) C (g) cCo(g) Initial (M) Change (M) Equilibrium (M): 0.129 0.085 0.085 +x 0129-x 0.085+x 0.085+x [coa COC,] Equilibrium constant, K (0.085+x) (0.085+ x) (0.129-x) 0.112 x 0.0236 According to the ICE table, Equilibrium concentration of COC12, [COC1,] = (0.129- x) M = (0.129-0.0236) M = 0.1054 M 0.105 M Equilibrium concentration of CO, [C0] = (0.085+x) M = (0.085+0.0236) M = 0.1086 M 0.109 M Equilibrium concentration of Cla, [a = (0.085x) M (0.085+0.0236) M = 0.1086 M 0.109 M