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

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)

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Keq = [C][D] /([A]^2[B])

We need ICE table

initially

[A] = 5.0

[B] = 7.5

[C] = 0

[D] = 0

the change

[A] = -2x

[B] = - x

[C] = x

[D] = x

in equilibrium

[A] = 5.0-2x

[B] = 7.5 - x

[C] = 0 + x

[D] = 0 + x

and we know

[A] = 5.0-2x = 1

x = (1-5)/(-2) = 2

substitute

[A] = 5.0-2*2 = 1

[B] = 7.5 - 2 = 5.5

[C] = 0 + 2

[D] = 0 + 2

Solve in Keq

Keq = [C][D] /([A]^2[B])

Keq = (2*2)/(1*5.5) = 0.727

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)

we need

Q = [C]/([A]*[B]) = 50

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

Q = (50)/(5*10) = 1; not in equilbrium

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

Q = (100)/(0.1*20) = 50; in equilbirium

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

Q = (25)/(1*0.5) = 50

in equilibrium

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

Q = (50)/(50*50) = 0.02., not in equilbirium

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)

given K = 0.1

Q is defined as:

Q = [C]^c * [D]^d / ([A]^a * [B]^b)

In this Case, the concentrations are NOT in equilibrium

Therefore:

If Q < Keq; this has much more reactants than products, therefore expect reactants to form more product in order to achieve equilibrium

If Q > Keq; this has much more products than reactants, therefore expect products to form more reactants in order to achieve equilibrium

If Q = Keq; this has the same ratio in equilibrium for reactants and products. Expect no reaction. It is safe to assume this is already in equilibrium.

Since Q < Keq

choose:

If Q > Keq; this has much more products than reactants, therefore expect products to form more reactants in order to achieve equilibrium