Question

The equilibrium constant, K, for the following reaction is 2.36x10-2 at 618 K. COCI,(E)=CO(g) +C1, An equilibrium mixture of the three gases in a 10.4 L container at 618 K contains 0.361 M COCI, 9.23x10-2M CO and 9.23 102 M CI. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 17.8 L? [COCI,] = [CO] = [CI] -

Answer 1

given, k =20364 10.4L K = 2:36X10-2 al traicion of K = 0.0236 ] for the reaction; * v= 10.4L coceg ħ Cocgtc₂c9) o coa = M.com XV =0.361 X 10.4 23.75 mol rco = MCoXV 30.0923. X10-4 -0.96mol... nel = Maxw = 0.0923 X 100 4 00.96 mol new volume of container vf = 1492 [code] =ņ 3-35 20:21M). [Co] – [42] = 0.96 = 0.054 Due to increase in volume equilibrium will 1 [co] [12) 0.054 xoosy -00014 Resetion quotient (Q) = . [ cocea ] oogi 1 stk so resction will proceed in forward direction. 17.8 178 a distoub

STATIST Cock (9) co(g) + Cl2 cg ) to 0.2l a-Ore oosa tate 0-21-2 0.054 exe o osuta so, at equilibrium ka [co] [c] - (0-0.54*2% (0.054 exe) [cocer (o.al-21) 0.0236 = 7?+ 0.104x + 0.0029 (0.21-2) or 0.00495 -0-0236x2 = 224 0-10uze p0-0029 er 42+ 0.1276 % - 0.002 20: : 1 = 0.014M So equilibrium concentrations are [code] = 0.21-poly = 0.196 M. 1 [co] =ooosy +0.014 20.0687 [chen = 0.054 +0.014 -0.068MY