Integrated rate laws and reaction order ln[A] t = -kt + ln[A] 0 Effect of Concentration
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Integrated rate laws and reaction order
ln[A]t = -kt + ln[A]0
Effect of Concentration
12-2
1/[A]t = kt + 1/[A]0
Effect of Concentration
12-3
[A]t = -kt + [A]0
Effect of Concentration
Graphical determination of the reaction order for the decomposition of N2O5.
Effect of Concentration
12-5
For the chemical reaction A C, a plot of 1/[A]t
versus time was found to give a straight line with a positive slope. What is the order of reaction?
Answer: second order
Effect of Concentration
12-6
Effect of Concentration
For the reaction X + Y Z, the reaction rate is found to depend only upon the concentration of X. A plot of 1/X verses time gives a straight line. What is the rate for this reaction?
Answer: rate = k [X]2
12-7
Effect of Concentration
The following data were obtained for the gas phase decomposition of nitrogen dioxide at 300 C
NO2 → NO + ½ O2
Time [NO2], M0 sec 0.010050 0.00787100 0.00649200 0.00481300 0.00380
Is the reaction first order or second order?Answer: second order (plot of 1/[NO2] vs. time is linear
12-8
Effect of Concentration
The thermal decomposition of acetaldehyde, CH3CHO CH4 + CO, is a second-order reaction. The following data were obtained at 518C.
time, s Pressure CH3CHO, mmHg0 364
42 330105 290720 132
Calculate the rate constant for the decomposition of acetaldehyde from the above data.
Answer: k = 6.7 106/mmHg·s
12-9
Worksheet # 10 - 5
2. In the gas phase at 500.°C, cyclopropane
reacts to form propene.
a. Explain how this plot confirms that the reaction
is first order.
b. Calculate the first-order rate constant, k.
c. Determine the initial concentration of
cyclopropane in this experiment [y-intercept = -
2.30]
1. The gas-phase conversion of 1,3-butadiene to
1,5-cyclooctadiene, 2C4H6 → C8H12 was studied,
providing data for the plot shown at the right.
a. Explain how this plot confirms that the reaction
is second order.
b. Calculate the second-order rate constant, k.
c. Determine the initial concentration of 1,3-
butadiene in this experiment.
12-10
Worksheet # 10 – 5: Answers
12-1012-10
2. In the gas phase at 500.°C, cyclopropane reacts to
form propene.
a. Explain how this plot confirms that the reaction is first
order. Plot of ln [A] vs time
b. Calculate the first-order rate constant, k.
Answer = 0.00175 /s
c. Determine the initial concentration of cyclopropane in
this experiment. Answer = 0.10 M
1. The gas-phase conversion of 1,3-butadiene to 1,5-cyclooctadiene, 2C4H6 → C8H12 was studied, providing data for the plot shown at the right.a. Explain how this plot confirms that the reaction is second order. Plot of 1/[A] vs timeb. Calculate the second-order rate constant, k.
Answer = 0.04 /Msc. Determine the initial concentration of 1,3-butadiene in this experiment. Answer = 0.025 M
12-11
An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions
Zero Order First Order Second Order
Plot for straight line
Slope, y-intercept
Half-life
Rate law rate = k rate = k [A] rate = k [A]2
Units for k mol/L*s 1/s L/mol*s
Integrated rate law in straight-line form
[A]t =
k t + [A]0
ln[A]t =
-k t + ln[A]0
1/[A]t =
k t + 1/[A]0
[A]t vs. t ln[A]t vs. t 1/[A]t = t
k, [A]0 -k, ln[A]0k, 1/[A]0
[A]0/2k ln 2/k 1/k [A]0
Effect of Concentration
Effect of Temperature
• Increasing the temperatureof the reaction increases the number of effective collisions by both increasing the total number of collisions (because molecules move faster) and increasing the fraction of collisions that are effective (because the average kinetic energy is higher).
12-12
Figure 12.9
12-13Figure 12.9
Effect of Temperature
The dependence of possible collisions on the product
of reactant concentrations.
A
A
B
B
A
A
B
B
A
4 collisions
Add another molecule of A
6 collisions
Add another molecule of B
A
A
B
B
A B
Effect of Temperature
The Effect of Ea and T on the Fraction (f) of Collisions
with Sufficient Energy to Allow Reaction
Ea (kJ/mol) f (at T = 298 K)
50 1.70x10-9
75 7.03x10-14
100 2.90x10-18
T f (at Ea = 50 kJ/mol)
250C(298K) 1.70x10-9
350C(308K) 3.29x10-9
450C(318K) 6.12x10-9
Effect of Temperature
The effect of temperature on the distribution of collision energies
Effect of Temperature
Energy-level diagram for a reaction
REACTANTS
PRODUCTS
ACTIVATED STATE
Co
llisi
on
En
ergy
Co
llisi
on
En
ergy
Ea (forward)
Ea (reverse)
The forward reaction is exothermic because the reactants have more energy than the products.
Effect of Temperature
An energy-level diagram of the fraction of collisions exceeding Ea.
Effect of Temperature
Suppose the collision rate between molecules A and B at 25°C is 10,000 collisions per second. The number of effective collisions at this same temperature is 100 collisions per second. How will each of the following affect the total number of collisions and the fraction of effective collisions between molecules A and B?
– The temperature decreases
– The concentration of reactant B decreases
12-19
Sample Problem
When the temperature decreases• Average kinetic energy decreases• Average velocity of the molecules decreases
Should decrease the number of collisions and the fraction of effective collisions!
Therefore, we expect the total number of collisions to be less than 10,000 and the fraction of effective collisions to be less than 100.
12-20
Sample Problem
When the concentration of reactant B decreases– The number of collisions between A and B decreases
Less B molecules can make contact with A molecules.
– The fraction of effective collisions should remain the same
The average kinetic energy and temperature of the molecules emains the same.
Therefore, we expect the total number of collisions to be less than 10,000 and the number of effective collisions to be less than 100, but still one-tenth of the total number of collisions (same fraction as before).
12-21
Sample Problem
Figure 16.10
Dependence of the rate constant on temperature
Effect of Temperature
The Arrhenius Equation
k AeEa
RT
ln k = ln A - Ea/RT
lnk2
k1
=Ea
R-
1
T2
1
T1
-
where k is the kinetic rate constant at T
Ea is the activation energy
R is the energy gas constant
T is the Kelvin temperature
A is the collision frequency factor
Effect of Temperature
R = 8.314 J/mol*K
Graphical determination of the activation energy
ln k = -Ea/R (1/T) + ln A
Effect of Temperature
SOLUTION:
The decomposition of hydrogen iodide,
2HI(g) H2(g) + I2(g)
has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea.
lnk2
k1
=Ea
-R
1
T2
1
T1
- Ea = - R lnk2
k1
1
T2
1
T1
-
-1
1
600K
1
500K-ln
1.10x10-5L/mol*s
9..51x10-9L/mol*s
Ea = - (8.314J/mol*K)
Ea = 1.76x105 J/mol = 176 kJ/mol
Sample Problem
12-26
Sample Problem
The activation energy for the reaction CH3CO CH3 + CO is 71 kJ/mol. How many times greater is the rate constant for this reaction at 170C than at 150C?
Answer: 2.5
12-27
The activation energy for the following first-order reaction is 102 kJ/mol.
N2O5(g) 2NO2(g) + (1/2)O2(g)
The value of the rate constant (k) is 1.35 104 s1 at 35C. What is the value of k at 0C?
Answer: 8.2 107 s1
Sample Problem
12-28
Worksheet # 10 - 6
1. The rate constant for the gas phase decomposition of N2O5,N2O5 → 2NO2 + ½ O2
has the following temperature dependence:
Determine the activation energy for this reaction. 2. A first order reaction has rate constants of 0.046 /second and 0.081
/second at 0 C and 20.0 C, respectively. What is the value of the activation energy?
3. The activation energy for the decomposition of HI to H2 and I2 is 186 kJ/mol. The rate constant at 555 K is 3.52 x 10 -7 L/mol sec. What is the rate constant of 645 K?
Temperature (K) Rate constant, k ,(1/second)
338 0.0049
318 0.00050
298 0.000035
12-29
Worksheet # 10 - 6: Answers
12-2912-29
Worksheet # 10 - 6
1. The rate constant for the gas phase decomposition of N2O5,N2O5 → 2NO2 + ½ O2
has the following temperature dependence:
Determine the activation energy for this reaction. Ea = 100 kJ/mol2. A first order reaction has rate constants of 0.046 /second and 0.081
/second at 0 C and 20.0 C, respectively. What is the value of the activation energy? Ea = 18.8 kJ/mol
3. The activation energy for the decomposition of HI to H2 and I2 is 186 kJ/mol. The rate constant at 555 K is 3.52 x 10 -7 L/mol sec. What is the rate constant of 645 K? k = 9.5 x 10 -5 L/mol sec
Temperature (K) Rate constant, k ,(1/second)
338 0.0049
318 0.00050
298 0.000035
Effect of Surface Area
12-30Iron Nail Steel wool
Figure 12.4
Steel wool has more iron atoms exposed on the surface than the same mass of iron nail.
12-31
• The catalyst called catalase in this piece of liver causes the decomposition of H2O2 to occur faster.
Figure 12.5
Effect of Catalysts
• Adding an appropriate catalyst increases the number of effective collisions by lowering the activation energy. This also increases the fraction of collisions that are effective.
12-32
Figure 12.10
Effect of Catalysts
• Catalytic converters dramatically speed up the reactions of toxic gases to form harmless products:
– CO to CO2
– NO to N2 and O2
12-33
Catalyst is a palladium/platinum metal surfaceFigure 12.11
Effect of Catalysts
The thousands of enzymes in our bodies act to catalyze specific biological processes.
12-34
Figure 12.12
Effect of Catalysts
The enzyme sucrase catalyzes the decomposition of sucrose by making bond-breaking easier:
12-35
Figure 12.13
Effect of Catalysts
The metal-catalyzed hydrogenation of ethylene
H2C CH2 (g) + H2 (g) H3C CH3 (g)
Effect of Catalysts
Effect of Catalysts
• This catalyzed
reaction has two
transition states,
and an
intermediate that is
lower in energy
than the two
transition states.
12-38
Effect of Catalysts
Reaction energy diagram of a catalyzed and an uncatalyzed process.
Effect of Catalysts
• Each catalyst has its own specific way of functioning.
• In general a catalyst lowers the energy of activation.
• Lowering the Ea increases the rate constant, k, and thereby increases the rate of the reaction
Effect of Catalysts
12-41
• A catalyst increases the rate of the forward AND the
reverse reactions.
• A catalyzed reaction yields the products more quickly,
but does not yield more product than the uncatalyzed
reaction.
• A catalyst lowers Ea by providing a different mechanism, for the reaction through a new, lower energy pathway.
Effect of Catalysts
12-42
• Chlorine atoms from CF2Cl2 catalyze the decomposition of ozone in the stratosphere:
O3(g) + Cl(g) ClO(g) + O2(g)
ClO(g) + O3(g) Cl(g) + 2O2(g)
• The ClO(g) formed in step 1 is an intermediate that is formed temporarily.
• The Cl (g) is a catalyst as it is used in the first reaction and formed in the second reaction
Effect of Catalysts
12-43
• A catalyst is not a reactant or product. It interacts with the reactants, but is not permanently changed during the reaction.
• Since catalysts are “recycled,” small amounts are needed and last a long time.
Effect of Catalysts
• The decomposition of HI is an exothermic reaction:
2 HI(g) → H2(g) + I2(g)
Draw an energy diagram for the uncatalyzed reaction. Label reactants and products. Sketch a possible activated complex. Use a dotted line to show the energy changes when platinum metal, a catalyst that increases the rate of the reaction, is added to the system.
12-44
Sample Problem
12-45
Sample Problem
Ethene (H2C=CH2) can be converted to ethanol (CH3CH2OH) by a three-step process. Identify any intermediates or catalysts.
H2C=CH2 + H3O+ → H3C-CH2+ + H2O
H3C-CH2+ + H2O → CH3CH2OH2
+
CH3CH2OH2+ + H2O → CH3CH2OH + H3O+
12-46
Sample Problem
Identify any intermediates or catalysts.
H2C=CH2 + H3O+ → H3C-CH2+ + H2O
H3C-CH2+ + H2O → CH3CH2OH2
+
CH3CH2OH2+ + H2O → CH3CH2OH + H3O+
Intermediates in this process include H3C-CH2+ and
CH3CH2OH2+.
– H3C-CH2+ is formed in step 1, then used in step 2.
– CH3CH2OH2+ is formed in step 2, then used in step 3.
– H3O+ is used in step 1, then regenerated in step 3, so it is a catalyst.
12-47
Sample Problem
Effect of Catalysts
12-49
Worksheet # 10 - 5
1. One mechanism for the destruction of ozone in the upper atmosphere is:O3 (g) + NO (g) → NO2 (g) + O2 (g)NO2 (g) + O (g) → NO (g) + O2 (g)
Overall reaction O3 (g) + O (g) → 2O2 (g)
a) Which species is the catalyst?b) Which species is the intermediate?c) The energy of activation of the overall reaction is 14.0 kJ. Ea for the same reaction when catalyzed is 11.9 kJ. Draw the energy diagram of both the catalyzed and uncatalyzed reaction.
2. Chlorine atoms (from the decomposition of freons in the upper atmosphere) can act as catalyst for the destruction of ozone. The activation energy for the reaction
O3 + Cl → ClO + O2is 2.1 kJ/mol. Which is the more effective catalyst for the destruction of ozone, Cl or NO?
12-5012-50
Worksheet # 10 – 5: Answers
1. One mechanism for the destruction of ozone in the upper atmosphere is:O3 (g) + NO (g) → NO2 (g) + O2 (g)NO2 (g) + O (g) → NO (g) + O2 (g)
Overall reaction O3 (g) + O (g) → 2O2 (g)
a) Which species is the catalyst? NOb) Which species is the intermediate? NO2
c) The energy of activation of the overall reaction is 14.0 kJ. Ea for the same reaction when catalyzed is 11.9 kJ. Draw the energy diagram of both the catalyzed and uncatalyzed reaction.
2. Chlorine atoms (from the decomposition of freons in the upper atmosphere) can act as catalyst for the destruction of ozone. The activation energy for the reaction
O3 + Cl → ClO + O2is 2.1 kJ/mol. Which is the more effective catalyst for the destruction of ozone, Cl or NO? Cl
Spectrophotometric monitoring of a reaction.
Chemical Instruments
Conductometric monitoring of a reaction
Manometric monitoring of a reaction
Chemical Instruments
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