Equilibrium Practice Problems - LCPS



Equilibrium Practice Problems

1. Write the equilibrium expression for each of the following reactions:

N2 (g) + 3 H2 (g) ( 2 NH3 (g)

K = [NH3]2

[N2] [H2]3

I2 (s) + Cl2 (g) ( 2 ICl (g)

K = [ICl]2

[Cl2]

NO2 (g) ( NO (g) + ½ O2 (g)

K = [NO] [O2]1/2

[NO2]

2. The dissociation of acetic acid, CH3COOH, has an equilibrium constant at

25(C of 1.8 x 10-5. The reaction is CH3COOH (aq) ( CH3COO- (aq) + H+ (aq)

If the equilibrium concentration of CH3COOH is 0.46 moles in 0.500 L of

water and that of CH3COO- is 8.1 x 10-3 moles in the same 0.500 L,

calculate [H+] for the reaction.

K = [CH3COO-] [H+] Plug in known values and solve.

[CH3COOH]

1.8 x 10-5 = [8.1 x 10-3 moles/0.500 L] [H+]

[0.46 moles/0.500 L]

[H+] = 1.0 x 10-3 M

3. Indicate the effect of a catalyst, pressure, temperature and concentration on each of the following on:

catalyst pressure temperature concentration

a. speed presence direct direct direct

increase relationship relationship relationship

b. posn nothing based on moles based on inverse

enthalpy of rxn relationship

4. Given the initial partial pressures of (PPCl5) = 0.0500 atm, (PPCl3) = 0.150 atm, and (PCl2) = 0.250 atm at 250(C for the following reaction, what must each equilibrium partial pressure be?

PCl5 (g) ( PCl3 (g) + Cl2 (g) Kp = 2.15

Determine reaction quotient, Q = 0.150 x 0.250 = 0.75

0.0500 Shift to products

Make a chart describing relationships in change.

PCl5 (g) ( PCl3 (g) + Cl2 (g)

Initial 0.0500 0.150 0.250

Change -x +x +x

After 0.0500 -x 0.150 +x 0.250 + x

Plug into equilibrium expression and solve for x.

2.15 = (.150 +x)(0.250 + x) so x = 0.0272

(0.0500 –x)

Determine Concentrations.

PCl5 (g) ( PCl3 (g) + Cl2 (g)

After 0.0500 -x 0.150 +x 0.250 + x

Substitute 0.0228 .177 .277

Confirm value of K with these values to check your answer.

5. Calculate Kp for the following equilibrium at 250(C if K = 0.042.

PCl5 (g) ( PCl3 (g) + Cl2 (g)

K = Kp(RT)-( n SO 0.042 = Kp (0.08206x523)-(2-1)

Therefore Kp = 1.8

6. The reaction 2 NO (g) ( N2 (g) + O2 (g) has a value of

K= 2400 at 2000 K. If 0.61 g of NO are put in a previously empty 3.00 L

vessel, calculate the equilibrium concentrations of NO, N2, and O2.

Make a chart describing relationships in change.

2 NO (g) ( N2 (g) + O2 (g)

Initial 6.8 x 10-3 0 0

Change -2x +x +x

After 6.8 x 10-3-2x x x

Plug into equilibrium expression and solve for x.

2400 = [x] [x] so x = 0.0034

[6.8 x 10-3-2x]2

Determine Concentrations.

2 NO (g) ( N2 (g) + O2 (g)

After 6.8 x 10-3- 2X X X

Substitute 0 0.0034 0.0034

Does this make sense considering the value of K.

7. Using the same reaction as in #6, calculate the equilibrium concentrations of NO, N2, and O2 if the initial concentrations of each species are:

[NO] = 0 M, [N2] = 0.850 M, [O2] = 0.560 M.

Make a chart describing relationships in change.

2 NO (g) ( N2 (g) + O2 (g)

Initial 0 0.850 M 0.560 M

Change +2x -x -x

After 2x 0.850 M - x 0.560 M - x

Plug into equilibrium expression and solve for x.

2400 = [0.850 M - x] [0.560 M - x] so x = 0.007

[2x]2

Determine Concentrations.

2 NO (g) ( N2 (g) + O2 (g)

After 2x 0.850 M - x 0.560 M - x

Substitute 0.014 0.843 0.553

Does this make sense considering the value of K.

8. The following reaction has an equilibrium constant of 620 at a certain temperature. Calculate the equilibrium concentrations of all species if 4.5 mol of each component were added to a 3.0 L flask.

H2 (g) + F2 (g) ( 2 HF (g)

Determine molarity of solutions [4.5 mol / 3.0L ] = 1.5 M of all 3 solutions

Make a chart describing relationships in change.

H2 (g) + F2 (g) ( 2 HF (g)

Initial 1.5 M 1.5 M 1.5 M

Change -x -x +2x

After 1.5 –x 1.5 – x 1.5 +2x

Plug into equilibrium expression and solve for x.

620 = [1.5 +2x]2 so x = 1.33

[1.5 – x]2

Determine Concentrations.

After 1.5 –x 1.5 – x 1.5 +2x

Substitute 0.17 M 0.17 M 4.13 M

9. Ammonia undergoes hydrolysis according to the following reaction:

NH3 (aq) + H2O ( NH4+ (aq) + OH- (aq) K = 1.8 x 10-5

Calculate [NH3], [NH4+] and [OH-] in a solution originally 0.200 M NH3.

NH3 (aq) + H2O ( NH4+ (aq) + OH- (aq

Initial 0.200 M 0 0

Change -x +x +x

After 0.2 – x x x

Plug into equilibrium expression and solve for x.

1.8 x 10-5 = [x]2 so x = .0018

[0.2 – x]

Determine Concentrations.

After 0.2 – x x x

Substitute 0.2 M .0018 M .0018M

10. The equilibrium constant for the following reaction is 600(C is 4.0. Initially, two moles of CO and one mole of H2O were mixed in a 1.0 liter container. Determine the concentration of all species at equilibrium.

CO(g) + H2O(g) ( CO2 (g) + H2 (g)

Initial 2.0 M 1.0 M 0 0

Change -x -x +x +x

After 2.0 – x 1.0 - x x x

Plug into equilibrium expression and solve for x.

4.0 = [x]2 so x = 0.85

[1.0 – x][2.0 – x]

Determine Concentrations.

After 2.0 – 0.85 1.0 - 0.85 0.85 0.85

Substitute 1.15 0.15 0.85 0.85

11. If 5.0 moles of O2 and 4.0 moles of NO were entered into an empty 1.0 liter flask, calculate the equilibrium constant if the amount of NO2 found at equilibrium was 1.5 moles.

2 NO (g) + O2 (g) ( 2 NO2 (g)

Initial 4.0 M 5.0 M 0

Change -2x -x +2x

After 4.0 – 2x 5.0 - x 2x

Using 1.5 = 2x and therefore x = 0.75 M

After 4.0 – 1.5 5.0 – 0.75 1.5

Substitute 2.5 4.25 1.5

K = (1.5)2 = 0.085

(2.5)2(4.25)

12. Two moles of NH3 were entered in a 1.0 liter container at 650(C. At equilibrium only 71% of the original NH3 was found. Determine the equilibrium constant of the following reaction.

2 NH3 (g) ( N2 (g) + 3 H2 (g)

Initial 2.0 M 0 0

Change -2x +x +3x

After 2.0 – 2x +x +3x

Using 2.0 – 2x = 2.0(71%) so x = 0.29

After 2.0 – 2(.29) 0.29 +3(0.29)

Substitute 1.42 0.29 0.87

K = (0.29)(0.87)3 = 0.095

(1.42)2

13. If the equilibrium constant for the following reaction is 0.10, determine the final concentration of ICl, if 4.0 moles of I2 and Cl2 were entered initially into an empty 1.0 liter flask.

I2 (g) + Cl2 (g) ( 2 ICl (g)

Initial 4.0 M 4.0 M 0

Change -x -x +2x

After 4.0 – x 4.0 - x 2x

Plug into equilibrium expression and solve for x.

0.10 = [2x]2 so x = 0.546

[4.0 – x]2

Determine Concentrations.

After 4.0 – 0.546 4.0 - 0.546 2(0.546)

Substitute 3.5 3.5 1.1

14. The reaction of carbon disulfide with chlorine is as follows:

CS2 (g) + 3 Cl2 (g) ( CCl4 (g) + S2Cl2 (g) (H = -238 kJ

Predict the effect of the following changes to the system on the direction

of equilibrium.

a. The pressure on the system is doubled by halving the volume.

b. CCl4 is removed as it is generated.

c. Heat is added to the system.

15. The reaction of nitrogen gas with hydloric acid is as follows:

N2 (g) + 6 HCl (g) ( 2 NH3 (g) + 3 Cl2 (g) (H = 461 kJ

Predict the effect of the following changes to the system on the direction

of equilibrium

a. Triple the volume of the system.

b. The amount of nitrogen is doubled.

c. Heat is added to the system.

Sometimes it’s our fault that things are out of balance.

(See below.)

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