Question

Consider the following reaction and associated equilibrium constant:
aA(g)⇌bB(g), Kc = 1.5

Part A

Find the equilibrium concentrations of A and B for a=1 and b=1. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction.

Express your answers using two significant figures separated by a comma.

[A], [B] = ?M

Part B

Find the equilibrium concentrations of A and B for a=2 and b=2. Assume that the initial concentration of A is 1.0 M and that no B is present at the beginning of the reaction.

Express your answers using two significant figures separated by a comma.

[A], [B] = ?M

Part C

Find the equilibrium concentrations of A and B for a = 2 and b = 1. Assume that the initial concentration of Ais 1.0 M and that no B is present at the beginning of the reaction.

Express your answers using two significant figures separated by a comma.

[A], [B] = ?M

Answer 1

Part. A :-

ICE table is :-

..............................A <-------------------> B

Initial (I)................1.0 M.......................0.0 M

Change (C).......... - Y............................+ Y

Equilibrium (E).....(1.0-Y) M...................Y M

Y = Amount dissociated

Expression of K_c is :

K_c = [B] / [A]

1.5 = Y / (1.0 - Y)

1.5 - 1.5 Y = Y

Y = 0.6

Hence, Equilibrium concentration of A = [A] = 1.0 - Y = 1.0 - 0.6 = 0.4 M and

Equilibrium concentration of B = [ B] = Y = 0.6 M

Part. B :-

ICE table is :-

..............................2A <-------------------> 2B

Initial (I)................1.0 M.......................0.0 M

Change (C).......... - 2Y............................+ 2Y

Equilibrium (E).....(1.0-2Y) M...................2Y M

Y = Amount dissociated

Expression of K_c is :

K_c = [B]² / [A]²

1.5 = (2Y)² / (1.0 - 2Y)²

2Y / (1.0-2Y) = (1.5)^1/2

2Y = 1.225 - 2.45 Y

Y = 0.275

Hence, Equilibrium concentration of A = [A] = 1.0 - 2Y = 1.0 - 2x0.275 = 0.45 M and

Equilibrium concentration of B = [ B] = Y = 0.275 M

Part. C :-

ICE table is :-

..............................2A <-------------------> B

Initial (I)................1.0 M.......................0.0 M

Change (C).......... - Y............................+ Y/2

Equilibrium (E).....(1.0-Y) M...................Y/2 M

Y = Amount dissociated

Expression of K_c is :

K_c = [B] / [A]²

1.5 = (Y/2) / (1.0 - Y)²

1.5 +1.5 Y² - 3 Y = 0.5 Y

1.5 Y² - 3.5 Y +1.5 = 0

Y² - 2.33 Y + 1 = 0

On solving, we have

Y = 0.567 (Possible value)

Hence, Equilibrium concentration of A = [A] = 1.0 - Y = 1.0 - 0.567 = 0.43 M and

Equilibrium concentration of B = [ B] = Y/2 = 0.567 / 2 = 0.28 M