Question

1. You mix 5.0 M A with 7.5 M B and allow the mixture to come to equilibrium. At equilibrium you determine there is 1.0 M A. What is the equilibrium constant? Construct an ICE table to aid you.(2 points)

                  2A(g) + B(g) ⇆ C(g) + D(g)

2. The value of the equilibrium constant is 50 for the reaction below. Which of the following set of concentration would be an equilibrium position? Explain your reasoning for each answer. (2 points)

                  A(g) + B(g) ⇆ C(g)

A. [A] = 5 M, [B] = 10 M, [C] = 50 M

B. [A] = .1 M, [B] = 20 M, [C] = 100 M

C. [A] = 1 M, [B] = 0.5 M, [C] = 25 M

D. [A] = 50 M, [B] = 50 M, [C] = 50 M

3. If Q is equal to 0, and K is equal to 0.10, explain how the following reaction will proceed. (1 point)

                  A(g) + B(g) ⇆ 2C(g)

Answer 1

3.

  A(g) + B(g) ⇆ 2C(g)

Q = reaction quotient = [C]² / { [A] [B] } = 0

Inititial concentration of [C] = 0

so,

Q = [0]² / { [A] [B] } = 0

Now,

K = 0.10 = [C]_equi² / { [A]_equi [B]_equi }

So, here, [C] > 0

So, some of C is formed. Hence the reaction will proceed towards forward direction for making C.

2.

A(g) + B(g) ⇆ C(g)

Kc = [C]_equi / { [A]_equi [B]_equi }

50 = (25) / { (1) (0.5)

So, answer is option (C); [A] = 1 M, [B] = 0.5 M, [C] = 25 M

1.

2A(g) + B(g) ⇆ C(g) + D(g)
IC: 5 7.5 0 0
C: 2x - x +x +x
EC:   5 - 2x 7.5 - x x x

But

[A]_equi = 1

So,

5 - 2x = 1

-2x = -4

x = 2

So,

[B]_equi = 7.5 - x = 7.5 - 2 = 5.5

[C]_equi = x = 2

[D]_equi = x = 2

Now,

Kc = { [C]_equi [D]_equi } / { [A]_equi [B]_equi }

= ( 2 x 2 ) / ( 1 x 5.5 )

= 4 / 5.5

= 0.73