8. Equilibrium Calculations cont’d

Type 3 (system changing to equilibrium – Calculating concentration of a species)

Example 1: 2A (g) + B (g) 2C (g)

An unknown amount of C was placed into a 3.0 L flask. When equilibrium was

reached, the concentration of A was 0.60 M. If Keq has a value of 34.0, how many

moles of C were placed into the flask originally?

Step 1:

2A / + B / / 2C
I 0 / 0 / x
C +0.6 / +0.3 / - 0.6
E 0.6
/ 0.3 / x – 0.6

Step 2: Step 3 (solve for x):

Keq = [C]2 34 = [x - 0.6]2 = 3.72 = [x - 0.6]2

[A]2[B] [0.6]2[0.3] = x – 0.6

1.91 = x – 0.6

x = 2.5 M

Step 4 (find moles):

Moles of C = (2.5 M)(3L) = 7.5 mol of C

Example 2: A2 (g) + B2 (g) 2AB (g)

If 6.0 mol of A2 and 6.0 mol of B2 are placed in a 1.0 L bulb and allowed to come to

equilibrium, what will be the concentration of all the species at equilibrium?

Keq = 2.5.

Step 1:

A2 / + B2 / /
2AB
I 6 / 6 / 0
C -x / -x / +2x
E 6 – x
/ 6 – x / 2x

Step 2: Step 3 (solve for x): Step 4 (find [ ]’s):

Keq = [AB]2 2.5 = [2x]2 1.58 = 2x [A2] = 6 – 2.65 = 3.4 M

[A2][B2] [6 – x][6 – x] 6 – x [B2] = 3.4 M

= x = 2.65 [AB] = 2(2.65) = 5.3 M

Example 3: A (g) + B (g) C (g) + D (g)

A 1.0 L flask contains 1.0 mole of A, 4.0 mol of B, 4.0 mol of C and 4.0 mol of D

at equilibrium. If 3.0 mol of A is added, what will be the new concentration of D

when equilibrium is re-established?

Step 1 (find Keq): Keq = [C][D] = (4)(4) = 4.0

[A][B] (1)(4)

Step 2:

A / + B / / C + D
I 1 + 3 / 4 / 4 4
C -x / -x / +x +x
E 4 - x
/ 4 - x / 4 + x 4 + x

(Hint: we know the sign (+ or -) for the change from the direction of the equilibrium shift. We added A so it shifts right or product side. Therefore, C and D are + x.)

Step 3 (solve for x): Step 4:

4.0 = [4 + x]2 2.0 = 4 + x x = 1.3 M [D] = 4 + x = 5.3 M

[4 – x]2 4 – x

Type 4 (reaction shift)

· we can also use the Keq to predict which direction a reaction will proceed for

any initial concentrations of reactants and products.

· suppose we mix certain amounts of H2, I2 and HI:

H2 (g) + I2 (g) 2HI (g)

· which way will the reaction proceed? Product or reactant side?

· we use Reaction Quotient (Q). It has the same form as Keq, but the

concentrations of reactants and products are not the equilibrium concentrations.

· “Reaction Quotient” is also known as “Trial Keq”

Q > Keq reaction will shift to reactants

(Q ratio is too big and reaction will shift to make more reactants)

Q < Keq reaction will shift to products

(Q ratio is too small and reaction will shift to make more products)

Q = Keq the system will not shift.

(the reaction is already at equilibrium!)

Example: H2 (g) + I2 (g) 2HI (g)

The Keq for the above reaction is 54.4. If we place 1.0 mol of H2 and I2 and 2.0

mol of HI in a 1.0 L flask, which way will the reaction shift?

Q = [HI]2 = [2]2 = 4.0 Keq = 54.4

[H2][I2] [1][1]

Keq > Q

54.4 > 4.0 Therefore, reaction will shift to products!

Example: 2SO2 (g) + O2 (g) 2SO3 (g)

The Keq is 0.14 for the above reaction. If a 1.0 L flask is filled with 0.10 mol of SO3

and with 0.20 mol of SO2 and O2, is the reaction at equilibrium? If not, in which

direction does it proceed?

Q = [SO3]2 = [0.1]2 = 1.25 Keq = 0.14

[SO2]2[O2] [0.2]2[0.2]

Q > Keq

1.25 > 0.14 Therefore, reaction will shift to reactants!

Do Questions: # 52, 53, 55-65 (type 3); # 50, 51, 54 (type 4)