Arrhenius Equation:



Chapter 13 Worksheet 5 (ws13.5)

The Arrhenius Equation for the Rate Constant and Catalysis

The Arrhenius equation (effect of temperature and catalyst on rate)

Recall that the rate law the reaction aA + bB = cC is:

Rate = k[A]x[B]y

This equation explicitly shows the effect of reactant concentrations on the rate but what about the effect of temperature and a catalyst on the rate of a reaction? They are hiding in the rate constant, k! The value of k increases as the temperature increases and in the presence of a catalyst.

The effect of temperature on rate constants is given quantitatively by the Arrhenius equation:

k = Ae-Ea/RT

A is the “frequency factor” (although I like to call it the “orientation factor”)

Ea is the activation energy.

R is the ideal gas constant

T is the absolute temperature

e-Ea/RT is the fraction of molecules whose kinetic energy is equal to or greater than Ea.

1. Give the unit for each of the following:

A

R = 8.314

T

Ea

Exponent

2. Sketch 2 graphs: k vs. T and the Maxwell-Boltzmann distribution of kinetic energies at two different temperatures (fraction of molecules vs. kinetic energy). Use the second graph to explain why k increases exponentially with temperature.

3. Fill in the blanks with “increases” or “decreases”. According to the Arrhenius equation:

The rate of a reaction increases as A _____________, T ________________, and as Ea ________________.

4. True or false.

a. The rate of a reaction increases as the concentration of reactants increases.

b. Assuming part a is true, the rate of a reaction doubles if the concentration of a reactant doubles.

c. The rate of a reaction doubles if the absolute temperature doubles.

d. The rate of a reaction doubles if the activation energy is halved.

5. If the activation energy for a reaction is 100 kJ/mol (a typical value), what fraction of the molecules have enough energy to get over the activation energy barrier at 300 K?

6. Consider the Arrhenius equation.

k = Ae-Ea/RT

a. According to the Arrhenius equation, temperature would have a greater effect on the rate of which type of reaction: one with a large activation energy or one with a small activation energy? (Use common sense!)

b. How can you use the Arrhenius equation to experimentally determine the activation energy and the frequency factor (A) for a reaction?

c. The rate constant for the following reaction was determined at two different temperatures: BH4-(aq) + NH4+(aq) → BH3NH3(aq) + H2(g) The data is shown in the table below. Calculate the activation energy and the frequency factor for the reaction.

|T (oC) |k (M-1s-1) |

|30.0 |1.94x10-4 |

|40.0 |1.49x10-3 |

d. Calculate the rate constant for the reaction at 100oC.

7. The Arrhenius equation can be written in a useful form by taking the natural logarithm (ln) of both sides. (Remember that ln xy = ln x + ln y and that ln ex = x.).

a. Derive the logarithmic form of the Arrhenius equation.

b. Using the logarithmic form of the Arrhenius equation, you can easily determine the activation energy and the frequency factor graphically:

A plot of _______________ vs. _______________ will produce a straight line with a slope of _______________ and a y-intercept of _______________.

c. Data for the temperature dependence of the rate constant for the following reaction are given below.

2 N2O5(g) → 4 NO2(g) + O2(g)

|Temperature (οC) |k (1/s) |

|25 |3.7 x 10-5 |

|45 |5.1 x 10-4 |

|55 |1.7 x 10-3 |

|65 |5.2 x 10-3 |

Make the appropriate graph of the data in excel and determine the activation energy (in kJ/mol) and frequency factor (in 1/s) for this reaction. (You must manipulate the data before making the graph.)

d. Use the activation energy and frequency factor determined from the graph and the Arrhenius equation to calculate the rate constant for the reaction at 85 οC.

Catalysis

Catalysts are substances that increase the rate of a chemical reaction but are not consumed in the reaction.

Homogeneous Catalysts – Catalysts that are in the same phase (usually liquid) as the reactants.

Heterogeneous Catalysts – Catalyst that are not in the same phase as the reactants. These are often metals which provide a surface on which the reaction occurs.

Enzymes – Biological catalysts (usually proteins) that evolved to catalyze specific reactions in cells. Proteins are very large polymers (MM from 10,000 to > 100,000 g/mol) that fold into intricate 3-D structures. They have properties of both homogeneous and heterogeneous catalysts. Enzymes are incredibly efficient catalysts and, unlike other types of catalysts, they catalyze only one specific reaction or a small class of related reactions.

Consider the rate law: Rate = Ae-Ea/RT[A]x[B]y

This equation tells you that the rate of reaction can be increased by:

1. Increasing the concentration of reactants

2. Increasing the temperature

3. Decreasing the activation energy

4. Increasing the frequency factor

(We will ignore the effect of increasing the reaction orders x and y because it can be complicated.)

1. Complete the following table:

|Type of catalyst |Increases reactant |Increases temperature? |Decreases activation |Increases A? |Exhibits reaction |

| |concentration? | |energy? | |specificity? |

|Homogeneous | | | | | |

|Heterogeneous | | | | | |

|Enzyme | | | | | |

2. Name 3 ways a catalyst can lower the activation energy. For each, state which type of catalyst can do it that way.

3. How can a catalyst increase A? (This is why I like to call A the “orientation factor”.)

An example of homogeneous catalysis: The decomposition of hydrogen peroxide

Uncatalyzed reaction:

2 H2O2 (aq) → 2 H2O (l) + O2 (g)

(Assume that it occurs in a single elementary step.)

Catalysis by HBr:

Step 1: H2O2 (aq) + 2 H+ (aq) + 2 Br- (aq) → 2 H2O (l) + Br2 (aq)

Step 2: H2O2 (aq) + Br2 (aq) → 2 H+ (aq) + 2 Br- (aq) + O2 (g)

4. The picture below shows the reaction occurring in the presence of the catalyst (HBr). Based on this picture, one can conclude that step 2 is rate-limiiting in the catalyzed pathway. Why? (Hint: Br2(aq) is red. I will project a color picture.)

[pic]

5. The energy diagram for the uncatalyzed and catalyzed pathways are shown below. How is HBr catalyzing this reaction?

[pic]

Demonstration: Catalysis of the same reaction by an enzyme called “catalase”. (Take notes.)

An example of heterogeneous catalysis: catalysis on the surface of a metal

H2C=CH2 (g) + H2(g) → CH3CH3 (g)

Catalyzed by platinum

[pic]

Heterogeneous catalysts resemble enzymes as follows:

1. Reactants bind to the surface. (Increases A because 3-dimensional search is changed to 2-dimensional search.)

2. Bonds in reactants are weakened. (Reactants are destabilized.)

Heterogenous catalyst differ from enzymes as follows:

1. They are not dissolved. (This is an advantage because it is easy to separate the catalyst from products!)

2. They are not specific (They can bind to many different substances and can catalyze a range of reactions.)

3. They do not stabilize the transition state.

4. They do not hold reactants in the optimal orientation.

6. Recall that a catalyst speeds up a chemical reaction without being consumed. A catalyst can be recognized as a substance that is consumed in one step in a reaction mechanism but regenerated in a subsequent step. In contrast, an intermediate is produced in one step and consumed in a subsequent step. For the mechanism below, write the overall reaction and identify all intermediates and catalysts.

step 1: Ce4+ + Mn2+ → Ce3+ + Mn3+ Intermediates:

step 2: Ce4+ + Mn3+ → Ce3+ + Mn4+

step 3: Mn4+ + Tl+ → Mn2+ + Tl3+ Catalysts:

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