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 for 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
Entire Exponent
Same as k (e.g. s-1, M-1s-1, etc.)
J/mol-K
K
J/mol
None
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.
k increases exponentially with temperature because the Maxwell-Boltzmann distribution is an
exponential function. A small increase in temperature greatly increases the fraction of
molecules with sufficient energy to react. (Those with kinetic energy greater than Ea.)
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3. Fill in the blanks with “increases” or “decreases”. According to the Arrhenius equation:
The rate of a reaction increases as A _____increases______, T _______increases_______, and as
Ea ______decreases__________.
4. True or false.
a. The rate of a reaction increases as the concentration of reactants increases.
True unless the reaction is zeroth order with respect to the reactant.
b. Assuming part a is true, the rate of a reaction doubles if the concentration of a reactant
doubles.
True only if the reaction is first order with respect to the reactant.
c. The rate of a reaction doubles if the absolute temperature doubles.
False, the rate increases exponentially.
d. The rate of a reaction doubles is the activation energy is halved.
False, the rate increases exponentially.
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?
 (100kJ / mol )
f=
e-Ea/RT
=e
(8.314 x10 3 kJ / mol  K )( 300K )
 3.87 x10 18 (a tiny fraction!!)
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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!)
Temperature has a greater effect on the rate of a reaction with a large activation energy. Think
about the effect of running speed when trying to get over a wall. Your speed will have a much
greater effect on your ability to scale a high wall than a low wall!
b. How can you use the Arrhenius equation to experimentally determine the activation
energy and the frequency factor (A) for a reaction?
Measure the rate constant at 2 (or more) different temperatures.
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 for the reaction.
T (oC)
30.0
40.0
100.0
k (M-1s-1)
1.94x10-4
1.49x10-3
?
T (K)
303
313
373
1/T (K-1)
3.30003 x 10-3
3.19489 x 10-3
2.68097 x 10-3
lnk (n unit)
-8.5477
-6.5090
?
Calculate the slope of the line produced by plotting lnk vs 1/T and set this equal to –Ea/R
Slope =
Δy
( 6.5090)  ( 8.5477)

 1.9398x104 K = -Ea/R
-3
1
3
1
Δx 3.1949x10 K  3.3000x10 K
Ea = -R(slope) =( -8.314 J/mol-K)(-1.9398x104K) = 1.61 x 105 J/mol = 161 kJ/mol
d. Calculate the rate constant for the reaction at 100oC.
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lnk  ( 6.5090)
  1.9398x10 K
-3
1
3
1
2.6810x10 K  3.1949x10 K
k = 31.2 M-1s-1 (Notice that this is MUCH larger than k at 40oC!)
Note: In these calculations, you will get significantly different answers if you round the numbers to
3 sig figs before you get the final answers. This could be important on a multiple choice exam when
there are two similar choices.
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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.).
e.
Derive the logarithmic form of the Arrhenius equation.
ln k  
Ea 1
 ln A
R T
f. Using the logarithmic form of the Arrhenius equation, you can easily determine the
activation energy and the frequency factor graphically:
A plot of _____ln k__________ vs. ____1/T___________ will produce a straight line
with a slope of ______-Ea/R_________ and a y-intercept of _______ln A________.
g. 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 (C)
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.) See excel file on my website.
h. 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 C.
slope = -Ea/R
-12459 K = -Ea/8.314 J/mol-K
Ea = 103584 J/mol = 104 kJ/mol
y-intercept = ln A = 31.6
A = 5.29E13 s-1
k = Ae-Ea/RT = (5.29E13)e-(103584)/(8.314)(358) = 0.041 s-1
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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:
Increases
reactant
concentration?
Homogeneous
No
Heterogeneous
No
Enzyme
No
Type of
catalyst
Increases
temperature?
No
No
No
Decreases
activation
energy?
Yes
Yes
Yes
Increases A?
No
Yes
Yes
Exhibits
reaction
specificity?
No
No
Yes
2. Name 3 ways a catalyst can lower the activation energy. For each, state which type of catalyst
can do it that way.
a. Provide new pathway (mechanism) with more stable (lower energy) transition state. All
catalysts can do this.
b. Stabilize (decrease the energy of) the transition state through binding to catalyst. Enzymes do
this.
c. Destabilize (increase the energy of) the reactants (weaken or add stress to bonds).
Heterogeneous catalysts and enzymes can do this.
3. How can a catalyst increase A? (This is why I like to call A the “orientation factor”.)
Hold reactants close together. (Enzymes and heterogeneous catalysts do this.)
Hold reactants in optimal orientation for reaction to occur. (Enzymes do this.)
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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.)
If step 1 was rate-limiting, then Br2 would not accumulate and the solution would never turn
red.
5. The energy diagram for the uncatalyzed and catalyzed pathways are shown below. How is
HBr catalyzing this reaction?
The catalyst provides a new pathway with a
lower energy transition state. HBr first
reacts with one molecule of hydrogen
peroxide to produce water and Br2 (fast).
The Br2 intermediate then reacts with the
second molecule of hydrogen peroxide in
the rate-limiting step to produce oxygen
and regenerate the catalyst. The transition
state for the second step is different and
more stable than the transition state for the
uncatalyzed reaction.
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Demonstration: Catalysis of the same reaction by an enzyme called “catalase”. (Take notes.)
Main points:
Emzymes are proteins that act as very efficient and specific catlalysts inside living cells. The
potato cells contain the enzyme called catalase that catalyzes the decomposition of hydrogen
peroxide. (Hydrogen peroxide can damage cells.)
Catalase increases the rate of the reaction by a factor of ~1018! In contrast, HBr increases the
rate by a factor of a few thousand!
Hydrogen peroxide binds to the surface of the enzyme (in the “active site”). A conformational
change in the protein stresses the bonds in the hydrogen peroxide (destabilizes the reactant).
Reactive amino acid side chains react with hydrogen peroxide to provide the new pathway.
The transition state is greatly stabilized because it binds very tightly to the enzyme (much more
tightly than does hydrogen peroxide itself.
Enzymes (proteins) fold into a specific active conformation. Heat can “denature” a protein by
causing it to unfold and destroy its function. That is why the cooked potato didn’t work.
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An example of heterogeneous catalysis: catalysis on the surface of a metal
H2C=CH2 (g) + H2(g) → CH3CH3 (g)
Catalyzed by platinum
Heterogeneous catalysts resemble enzymes as follows:
1. Reactants bind to the surface. (Increases A because 3-dimensional search is changed to 2dimensional 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.
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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 mechanisms below, write the overall reaction and identify all intermediates
and catalysts.
step 1: Ce4+ + Mn2+ → Ce3+ + Mn3+
step 2: Ce4+ + Mn3+ → Ce3+ + Mn4+
step 3: Mn4+ + Tl+ → Mn2+ + Tl3+
2Ce4+ + Tl+ → 2Ce3+ + Tl3+
Intermediates: Mn3+, Mn4+
Catalysts: Mn2+
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