Practice Keq Questions for Unit Test



Practice Keq Questions for Unit Test

1. What features are necessary for two compounds (A and B) to be involved in a chemical equilibrium with a constant Keq?

2. Why do gases have a greater amount of entropy than solids?

3. Will the following reactions occur spontaneously? Briefly explain your answer.

a) 4HCl (g) + O2 (g) ( 2H2O (g) + 2Cl2 (g) + 114.4 KJ

b) N2O4 (g) + 58.9 KJ ( 2NO2 (g)

c) 2Hg (g) + O2 (g) ( 2HgO (s) ∆H = 10 KJ

4. Consider the following equilibrium reaction:

3H2 (g) + N2 (g) (( 2NH3 (g) + 92.4 KJ

Sketch graphs to show the concentration changes for ALL of the components when the following stresses are applied to the system:

a) Reducing the volume of the container.

b) Injecting ammonia into the system.

c) Adding a catalyst.

d) Reducing the temperature.

5. Write the equilibrium expressions for the following reactions:

a) COCl2 (g) (( CO (g) + Cl2 (g)

b) Zn (s) + 2Ag+ (aq) (( Zn2+ (aq) + 2Ag (s)

c) 2HgO (s) (( 2Hg (l) + O2 (g)

d) wW + xX (( uU + vV

6. In the equilibrium reaction:

4HCl (g) + O2 (g) (( 2H2O (g) + 2Cl2 (g) + 114.4 KJ

Predict the direction of the equilibrium shift if the following changes occur:

a) The pressure is increased.

b) Energy is added.

c) Oxygen is added.

d) HCl is removed.

e) A catalyst is added.

7. Consider the following equilibrium reaction:

2NOCl (g) (( 2NO (g) + Cl2 (g)

In one experiment, 2.00 moles of NOCl were placed in a 1.00 L flask, and the concentration of NO after equilibrium was achieved was 0.66 M. Calculate the equilibrium constant at 25°C for the reaction

8. At 800°C, the equilibrium constant for the following reaction is 0.279.

CO2 (g) + H2 (g) (( CO (g) + H2O (g) ∆H = 42.6 KJ

At a different temperature the equilibrium constant is 0.100. Is this different temperature higher or lower than 800°C? Give your reasoning.

9. In the following equilibrium reaction at 2000°C, Keq = 1.6 x 103.

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

The equilibrium concentration of NO is 0.13 mol / L. If the equilibrium concentration of nitrogen and oxygen are equal, calculate the concentration of nitrogen.

Practice Keq Questions for Unit Test Answers

1. You need the following conditions for constant Keq:

• Closed system

• Rate of forward reaction equal to rate of reverse reaction

• Constant temperature

• Tendencies of maximum entropy and minimum enthalpy favour opposite sides of reaction

2. Gases have a greater amount of entropy than solids because their molecules are more randomly arranged. Gas molecules are randomly dispersed with lots of spacing in between each other, whereas solid molecules are usually packed tightly together in a lattice.

3. a) Min enthalpy favours products; Max entropy favours reactants.

Reaction is spontaneous!

b) Min enthalpy favours reactants; Max entropy favours products.

Reaction is spontaneous!

c) Min enthalpy favours reactants; Max entropy favours reactants.

Reaction is not spontaneous!

4. a) Reducing the volume of container (increasing pressure!):

[pic]

There are 3 spikes initially to show increase in Pressure. Since equilibrium shifts right, there will be a decrease in the reactants [H2] and [N2] and an increase in products [NH3].

b) Injecting ammonia into the system:

[pic]

There is a spike for [NH3] initially to show addition of ammonia into the system. Since equilibrium shifts left, there will be an increase in the reactants [H2] and [N2] and a decrease in product [NH3].

c) Adding a catalyst:

[pic]

The addition of a catalyst only speeds up the forward and reverse reaction rates but does not affect the equilibrium concentrations of all species. Therefore, the concentration lines for all three chemical species in the system are unchanged!

d) Reducing the temperature:

[pic]

There are not any spikes for a temperature change in the graph. Since equilibrium shifts right, there will be a decrease in the reactants [H2] and [N2] and an increase in product [NH3].

5. a) Keq = [CO][Cl2]

[COCl2]

b) Keq = [Zn2+]

[Ag+]2

c) Keq = [O2]

d) Keq = [U]u [V]v

[W]w [X]x

6. a) If you increase the pressure, equilibrium will shift right to decrease pressure.

b) If you increase energy (heat), equilibrium will shift left to decrease the heat.

c) If you add oxygen, equilibrium will shift right to decrease the oxygen concentration.

d) If you remove HCl, equilibrium will shift left to increase the HCl concentration.

e) If you add a catalyst, equilibrium will not shift.

7. [NOCl] = 2.00mol / 1.00L = 2.00 M

2NOCl (g) (( 2NO (g) + Cl2 (g)

Initial [ ] 2.00 0 0

Change [ ] - 2x + 2x - x

__________________________________________________

Equil [ ] 2.00 – 2x 0.66 x

By looking at the NO column, you can figure out that 2x = 0.66. Therefore, x = 0.33M!

Equil [ ] 1.34M 0.66M 0.33M

Keq = [NO]2 [Cl2] = (0.66)2 (0.33) = 0.080

[NOCl]2 (1.34)2

8. For the following reaction:

CO2 (g) + H2 (g) + 42.6 KJ (( CO (g) + H2O (g)

Keq = 0.279 @ 800ºC

Keq = 0.100 @ New Temp

A smaller Keq means that the reactants are favoured at this new temperature. This also means that the equilibrium shifted left due to the change in temperature. To cause the equilibrium to shift left, you would have to decrease the temperature of the system. Therefore, the new temperature is lower than 800ºC.

9. For the following reaction:

2NO (( N2 + O2

Equil [ ] 0.13M x x

Keq = [N2] [O2] = x2 = 1.6 x 103 (square root both sides of equation)

[NO]2 (0.13)2

x = 40

0.13

x = 5.2

[N2] = x = 5.2 M

-----------------------

[H2]

[N2]

[NH3]

Time

3x

1x

2x

[H2]

[N2]

[NH3]

Time

3x

1x

2x

[H2]

[N2]

[NH3]

Time

[H2]

[N2]

[NH3]

Time

3x

1x

2x

................
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