13 Question 3 points a See page 72 The equilibrium constant

13 Question (3 points) a See page 72 The equilibrium constant K, for the reaction below is is 6.555x10 D+E 1st attempt Part 1 (1point) The initial composition of the reaction mixture is [C)]-ID]-[E]-1.9110 103 M What is the equilibrium concentration of C? See Periodic Table Part 2 (1 point) What is the equilibrium concentration of D? See Hint Part 3 (1 point) What is the equillbriumconcentration of E?

Solution

Suppose initial Concentration of C, D & E=a_o=1.9110 * 10^-3 M

Given reaction C ----------> D + E

Initial Concentration ao ao ao

Change in concentration. -x. +x. +x

Equilibrium concentration ao -x ao+ x. ao+x

Kc = [D][E] / [C]

Kc = (ao+ x)(ao+x) / (ao-x) = (ao+x)^2 / (ao - x)

Kc * (ao-x) = ao^2 + x^2 + 2 ao * x ------(1)

Put the values of ao in (1)

6.555 * 10^-5 (ao-x) = ao^2 + x^2 + 2 ao * x

6.555 * 10^-5 ao - 6.555 * 10^-5 x = ao^2 + x^2 + 2 ao * x

x^2 + (2ao + 6.555 * 10^-5)x + (ao^2 + 6.555 * 10^-5 ao) = 0

x^2 + (2*1.9110*10^-3 + 6.555 * 10^-5)x +[(1.911*10^-3)^2 + 6.555 * 10^-5 * 1.9110 * 10^-3] = 0

x^2 + 0.00388755 x + (0.000003652 + 0.0000001253) = 0

x^2 + 0.00388755 x + 0.00000378 = 0

Solve above quadratic equation and put the value of x and ao

Part 1

Equilibrium concentration of C = (ao - x) M

Part 2

Equilibrium concentration of D = (ao + x) M

Part 3

Equilibrium concentration of E = (ao + x) M