Chapter 10 Reaction Rates and Chemical...

Chapter 10 – Reaction Rates and Chemical Equilibrium

Section 10. – Rates of Reactions

Goal: Learn how temperature, concentration, and catalysts affect the rate of reaction.

Summary

• The rate of a reaction is the speed at which the reactants are converted to products.

• Activation energy: The energy that must be provided by a collision to break apart the bonds of the

reacting molecules (and for the reaction to proceed and produce products).

• Factors that can increase the rate of reaction:

o Increasing temperature – increases the changes of collisions, and provides more energy to the

molecules that can be used as activation energy.

o Increasing reactant concentration – more molecules means more chances of collisions.

o Add a catalyst – lowers the activation energy

Practice Problems

1. What is meant by rates of reaction?

2. Why does bread mold grow more quickly at room temperature?

3. How does a catalyst affect the activation energy?

a. Lowers the activation energy.

b. Raise the activation energy.

c. Doesn’t affect the activation energy.

4. In the reaction, H2(g) + Br2(g) → 2HBr(g), what happens to the number of collisions when more Br2

molecules are added?

a. Increases the number of collisions.

b. Decreases the number of collisions.

c. Has no effect on the number of collisions.

5. In the reaction, 2H2(g) + CO(g) → CH3OH(g), what happens to the number of collisions when the

temperature of the reaction is decreased?

a. Increases the number of collisions.

b. Decreases the number of collisions.

c. Has no effect on the number of collisions.

6. How would adding some SO2 change the rate of reaction for 2SO2(g) + O2(g) → N2(g) +2H2O(g)?

a. Increases the rate of reaction.

b. Decreases the rate of reaction.

c. Has no effect on the rate of reaction.

7. How would decreasing the temperature change the rate of reaction for:

2SO2 (g) + O2 (g) → N2 (g) +2H2O (g)?




8. How would adding a catalyst change the rate of reaction for

2SO2 (g) + O2 (g) → N2 (g) +2H2O (g)?




9. How would removing O2(g) change the rate of reaction for

2SO2 (g) + O2 (g) → N2 (g) +2H2O (g)?




10. How would adding some NO(g) change the rate of reaction for:

2NO (g) + 2H2 (g) → N2 (g) +2 H2O (g)?




11. How would decreasing the temperature change the rate of reaction for

2NO (g) + 2H2 (g) → N2 (g) +2 H2O (g)?




12. How would removing some H2 (g) change the rate of reaction for

2NO (g) + 2H2 (g) → N2 (g) +2 H2O (g)?




13. How would adding a catalyst change the rate of reaction for

2NO (g) + 2H2 (g) → N2 (g) +2 H2O (g)?




Section 10.2 – Chemical Equilibrium

Goal: Use the concept of reversible reactions to explain chemical equilibrium

Summary

• Chemical equilibrium occurs in a reversible reaction when

the rate of the forward reaction becomes equal to the rate

of the reverse reaction.

• At equilibrium, no further change occurs in the

concentrations of the reactants and products as the forward

and reverse reactions continue.

Understanding the Concepts

The following diagrams show the chemical reaction with time:

If A is blue and B is orange, state whether or not the reaction has reached equilibrium in this time period and

why.

The following diagrams show the chemical reaction with time:

If C is blue and D is yellow, state whether or not the reaction has reached equilibrium in this time period and

explain why.

Practice Problems

14. When does a reversible reaction reach equilibrium?

a. Forward and reverse reaction rates are equal; concentration of products and reactants is constant.

b. Forward rate triples and reverse reaction rate doubles; concentration of products and reactants is

constant.

c. Forward and reverse reaction rates are equal; concentration of products fluctuates while the reactants

stays constant.

15. Which of the following is at equilibrium?

a. The rate of the forward reaction is twice as fast as the rate of the reverse reaction.

b. The concentration of the reactant and the products do not change.

c. The rate of the reverse reaction changes.

16. Which of the following are not at equilibrium?

a. The rates of the forward and reverse reactions are equal.

b. The rate of the forward reaction doubles and the rate of the reverse reaction is halved.

c. The concentration of reactants and the products are constant.

Section 10.3 – Equilibrium Constants

Goal: Calculate the equilibrium constant for a reversible reaction given the concentrations of reactants and

products at equilibrium.

Summary

• An equilibrium constant, Kc, is the numerical value obtained by substituting

experimentally measured molar concentrations at equilibrium into the

equilibrium constant expression.

• An equilibrium constant expression for a reversible reaction is written by

multiplying the concentration of the products in the numerator and dividing

by the product of the concentrations of the reactants in the denominator.

• Each concentration is raised to a power equal to its coefficient in the

balanced chemical equation.

Writing the equilibrium constant expression

aA + bB cC + dD

Example: Write the equilibrium constant expression for the following chemical reaction:

2NO2(g) N2O4(g)

Answer:

Kc = [N2O4]

[NO2]2

Calculating an equilibrium constant

Example: Calculate the numerical value of Kc for the following reaction when the equilibrium mixture contains 0.025M NO2 and 0.087M N2O4.

2NO2(g) N2O4(g)

Answer: Write the equilibrium constant expression, substitute the molar concentrations and calculate.


Write the equilibrium constant expression for the following reactions:

CH4 (g) + 2O2 (g) ⇄ CO2 (g) + 2H2O(g)

4NH3 (g) +3O2 (g) ⇄ 2N2 (g) +6H2O (g)

C (s) +2H2 (g) ⇄ CH4 (g)

2C2H6 (g) + 7O2 (g) ⇄ 4C)2 (g) +6H2O (g)

2KHCO3 (s) ⇄ K2CO3 (s) +CO2 (g) + H2O (g)

4NH3 (g) + 5O2 (g) ⇄ 4NO (g) + 6H2O (g)

Practice Problems

17. What is the numerical value of Kc for the reaction, N2O4 (g) ⇄ 2NO2 (g), of the equilibrium mixture that contains 0.030 M N2O4 ⇄ and 0.21 M NO2?

a. 1.47

b. 0.68

c. 0.013

d. 7.00

e. 0.14

18. What is the numerical value of Kc for the reaction, CO2 (g) + H2 ⇄ CO (g) + H2O (g) if the equilibrium

mixture contains 0.3 M CO2, 0.33 M H2, 0.2 M CO, and 0.30 M H2O?

a. 1.65

b. 0.61

c. 0.15

d. 1.00

e. 0.10

19. What is the numerical value of Kc for the reaction, CO2 (g) + 3H2 ⇄ CH4 (g) + H2O (g) if the equilibrium

mixture contains 0.51 M CO, 0.30 M H2 1.8 M CH4, and 2.0 M H2O?

a. 2

b. 200

c. 260

d. 1800

e. 52

20. What is the numerical value of Kc for the reaction, N2 (g) + 3H2 (g) ⇄ 2NH3 (g), if the equilibrium

mixture contains 0.44 M N2, 0.40 M H2, and 2.2 M NH3?

a. 100

b. 170

c. 220

d. 2.2

e. 0.17

Section 10.4 – Using Equilibrium Constants

Goal: Use an equilibrium constant to predict the extent of reaction and to calculate equilibrium concentrations.

Summary

• A large value of Kc indicates that an equilibrium mixture contains mostly products and few reactants,

whereas a small value of Kc indicates that the equilibrium mixture contains mostly reactants.

• Equilibrium constants can be used to calculate the concentration of a component in the equilibrium

mixture.

Calculating equilibrium concentrations

To determine the concentration of a product or a reactant at equilibrium, we use the equilibrium constant

expression to solve for the unknown concentration.

Example: Calculate the equilibrium concentration for CF4 if Kc = 2.0 and the equilibrium mixture contains

0.10M COF2 and 0.050M CO2.

2COF2(g) ⇄ CO2(g) + CF4(g)

Answer: Write the equilibrium constant expression.

Kc = [CO2][CF4]

[COF2]2

Rearrange the equation, substitute the molar concentrations and calculate [CF4].

[CF4] = [COF2]2𝐾𝑐

[CO2]=

2.0(0.10)2

(0.050)= 0.40𝑀


Would the equilibrium constant, Kc, for the reactions in the diagrams have a large or a small value?

Practice Problems:

21. Does the equilibrium mixture contain mostly products or reactants?






27. The equilibrium constant, Kc, for H2(g) + I2(g) ⇄ 2HI(g) is 54. If the equilibrium mixture contains 0.15

M I2 and 0.030 M HI, what is the molar concentration of H2?

a. 0.030 M

b. 1.1x10-4 M

c. 0.15 M

d. 9.0x103 M

e. 3.7x10-3 M

28. The equilibrium constant, Kc, for the reaction, N2O4 (g) ⇄ 2NO2 (g), is 4.6 x 10-3. If the equilibrium

mixture contains 0.050 M NO2, what is the molar concentration of N2O4?

a. 0.54 M

b. 0.49 M

c. 11 M

d. 1.5 M

e. 5.4 M

29. The Kc for the reaction, 2NOBr(g) ⇄ 2NO(g) + Br2(g), at 100 oC is 2.0. If the equilibrium mixture

contains 2.0 M NO and 1.0 M Br2, what is the molar concentration of NOBr

a. 0.14 M

b. 0.71 M

c. 1.0 M

d. 1.4 M

e. 4.0 M

30. The Kc for the reaction, 3H2 + N2 ⇄ 2NH3 at 225oC is 1.7x102. If the equilibrium mixture contains 0.18

M H2 and 0.020 M N2 what is the molar concentration of NH3?

a. 50 M

b. 0.61 M

c. 0.020 M

d. 1.6 M

e. 0.20 M

Section 10.5 – Changing Equilibrium Conditions: Le Châtelier’s Principle

Goal: Use Le Châtelier’s principle to describe the changes made in equilibrium concentrations when reaction

conditions change.

Summary:

Le Châtelier’s principle states that when a system at equilibrium is disturbed by changes in concentration,

volume, or temperature, the system will shift in the direction that will reduce that stress.

• When reactants are removed or products are added to an equilibrium mixture, the system shifts in the

direction of the reactants.

• When reactants are added or products are removed from an equilibrium mixture, the system shifts in the

direction of the products.

• A decrease in the volume of a reaction container causes a shift in the direction of the smaller number of

moles of gas.

• An increase in the volume of a reaction container causes a shift in the direction of the greater number of

moles of gas.

• Increasing temperature of an endothermic reaction or decreasing the temperature of an exothermic

reaction will cause the system to shift in the direction of the products.

• Decreasing the temperature of an endothermic reaction or increasing the temperature of an exothermic

reaction will cause the system to shift in the direction of reactants.

Understanding the Concepts:

Would T2 be higher or lower than T1 for the reaction shown in the diagram?

Would the reaction shown in the diagrams be exothermic or endothermic?

Practice Problems:

31. Given 3O2 (g) + heat ⇄ 2O3 (g), what effect does adding more O2 (g) have?

a. Equilibrium shifts in the direction of the products.

b. Equilibrium shifts in the direction of the reactants.

c. Equilibrium does not change.

32. Given 3O2 (g) + heat ⇄ 2O3 (g), what effect does adding more O3 (g) have?




33. Given 3O2 (g) + heat ⇄ 2O3 (g), what effect does increasing the temperature have?




34. Given 3O2 (g) + heat ⇄ 2O3 (g), what effect does increasing the volume of the container have?




35. Given 3O2 (g) + heat ⇄ 2O3 (g), what effect does adding a catalyst have?




36. Given N2 (g) + 3H2 ⇄ 2NH3 (g) +92 kJ, what effect does removing some N2 (g) have?




37. Given N2 (g) + 3H2 ⇄ 2NH3 (g) +92 kJ, what effect does temperature have?




38. Given N2 (g) + 3H2 ⇄ 2NH3 (g) +92 kJ, what effect does adding more NH3 (g) have?




39. Given N2 (g) + 3H2 ⇄ 2NH3 (g) +92 kJ, what effect does adding more H2 (g) have?




40. Given N2 (g) + 3H2 ⇄ 2NH3 (g) +92 kJ, what effect does increasing the volume of the container have?




41. Given H2 (g) + Cl2 (g) + heat ⇄ 2HCl (g), what effect does adding more H2 (g) have?




42. Given H2 (g) + Cl2 (g) + heat ⇄ 2HCl (g), what effect does increasing the temperature have?




43. Given H2 (g) + Cl2 (g) + heat ⇄ 2HCl (g), what effect does removing some Cl2 (g) have?




44. Given H2 (g) + Cl2 (g) + heat ⇄ 2HCl (g), what effect does removing some HCl (g) have?




45. Given H2 (g) + Cl2 (g) + heat ⇄ 2HCl (g), what effect does adding a catalyst have?




46. Given CO (g) + H2O (g) ⇄ CO2 + H2 (g) + heat, what effect does decreasing the temperature have?




47. Given CO (g) + H2O (g) ⇄ CO2 + H2 (g) + heat, what effect does adding more H2 (g) have?




48. Given CO (g) + H2O (g) ⇄ CO2 + H2 (g) + heat, what effect does removing CO2 (g) as it forms have?




49. Given CO (g) + H2O (g) ⇄ CO2 + H2 (g) + heat, what effect does adding more H2O (g) have?




50. Given CO (g) + H2O (g) ⇄ CO2 + H2 (g) + heat, what effect does decreasing the volume of the container

have?




Challenge Problems

51. The Kc = 2.0 at 100 oC for the decomposition reaction of NOBr.

In an experiment, 1.0 mole of NOBr, 1.0 mole NO, and 1.0 mole of Br2 where placed in a 1.0 L container.

a. Write the equilibrium constant expression for the reaction.

b. Is the system at equilibrium?

c. If not, will the rate of the forward or reverse reaction initially speed up?

d. At equilibrium, which concentration(s) will be greater than 1.0 M, and which will be less than 1.0 M.

52. Consider the following reaction:

a. Write the equilibrium constant expression for the reaction.

b. Initially, 0.60 mole of PCl5 is placed in a 1.0 L flask. At equilibrium, there is 0.16 mole of PCl3 in the

flask. What are the equilibrium concentrations of PCl5 and Cl2?

c. What is the numerical value of the equilibrium constant, Kc, for the reaction?

d. If 0.20 mole of Cl2 is added to the equilibrium mixture, will the concentration of PCl5 increase or

decrease.

53. Indicate if you would increase or decrease the volume of the container to increase the yield of the products

in each of the following:

54. Indicate if you would increase or decrease the volume of the container to increase the yield of the products

in each of the following:

Date post:	06-Mar-2018
Category:	Documents
Upload:	vandat
View:	233 times
Download:	2 times

Chapter 10 Reaction Rates and Chemical...

Documents