Question

CH 15

± Using the Equilibrium Constant

1)

Part A

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

2)

Part A

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.700 M , [B] = 1.20 M , and [C] = 0.600 M . The following reaction occurs and equilibrium is established:

A+2B⇌C

At equilibrium, [A] = 0.600 M and [C] = 0.700 M . Calculate the value of the equilibrium constant, Kc.

Express your answer numerically.

Answer 1

Part (A)

Relationship between the initial and the equilibrium concentrations of all the species

A+ 2B ⇌ C
I....0.70..........1.20...............0.600
C...(-x).............(-2x)................(+x)
E.0.700-x........1.20-2x.............0.600+x

By definition, the equilibrium constant will be

Kc = (0.600+x ) / {(0.700−x)⋅(1.20−2x)2 }

ou know that 0.700−x=0.600 ⇒ x=0.100

Replace the value of x into the equation and you'll get

Kc = (0.600+ 0.10 ) / {(0.700−0.10)⋅(1.20−2 * 0.10)2 }

Kc = ( 0.7 ) / { 0.6 x ( 1 )}

Kc = 1.16

Part (A)
A: 2.00, X, 2.00-X
B: 2.00, X, 2.00-X
C: 0.00, X, X
D: 0.00, X, X
1.16 = X^2/(2.00-X)^2
take the square root of both sides
1.08 = X / (2.00-X)
2.16 -1.08X = X
2.16 = 2.08X
X = 1.04
[A] = 2.00-1.04 = 0.96 M

Part (B)
A: 1.00, X, 1.00-X
B: 2.00, X, 2.00-X
C: 0.00, X, X
D: 0.00, X, X
1.16 = X^2/(1.00-X)(2.00-X)
1.16(X^2-3X + 2) = X^2
1.16X^2 -3.48X + 2.32 =0
X^2 - 3X +2 = 0

(X-1) (X-2) =0
X = 1
[D] = 1 M or [D] = 2 M