kinetics: the differential and integrated rate laws β¦...1 dr. laurence lavelle kinetics: the...
TRANSCRIPT
1
Dr. Laurence Lavelle
Kinetics: The Differential and Integrated Rate Laws in Chemistry (and Physics, Biology, etc.)
In general, for all reactions: aA β bB + cC
Rate = β1ππππ[π΄π΄]ππππ
= 1ππππ[π΅π΅]ππππ
= 1ππππ[πΆπΆ]ππππ
*Notice for the reactants, there is a negative sign in front. This is because as time progresses, the concentration of reactant decreases. For products, their concentration over time increases which is why theirs is positive (no negative sign in front).
As reactant A goes to product B: A β B
Rate = β1ππππ[π΄π΄]ππππ
= ππ[π΄π΄]ππ
*In the following derivations of the three integrated rate laws, we assume the stoichiometric coefficient "a" in front of reactant A is 1.
1st order reaction Differential Rate Law ππππππππ = βπ π [π¨π¨]
π π π π = ππ[π¨π¨]ππ
Separate the variables ππ[π΄π΄][π΄π΄]
= βππ ππππ
Integrate both sides of the equation β« ππ[π΄π΄][π΄π΄]
= β«βππ ππππ = βππ β«ππππ
Indefinite integral constant, b ln[π΄π΄] = βππππ + ππ
At time t = 0, ππ = ln[π΄π΄]0 where [A]0 is the initial reactant concentration
Integrated Rate Law (linear form) π₯π₯π₯π₯[π¨π¨] = βπππ π + π₯π₯π₯π₯[π¨π¨]ππ
To more clearly see the exponential relationship between time, t, and reactant concentration, [A], for a first-order reaction we can convert the integrated first-order rate-law (linear form) to its non-linear exponential form:
2
Integrated Rate Law (linear form) ln [π΄π΄] = βππππ + ln[π΄π΄]0
Raise both sides to the power of e ππln [π΄π΄] = ππβππππ ππln [π΄π΄]0
Simplify, and obtain non-linear form [π΄π΄] = [π΄π΄]0 ππβππππ
One can also then write the non-linear differential rate law:
Rate = βππ[π΄π΄]ππππ
= ππ[π΄π΄] = ππ[π΄π΄]0 ππβππππ
2nd order reaction
Differential Rate Law ππππππππ = βπ π [π¨π¨]π π π π
= ππ[π¨π¨]ππ
Separate the variables βππ[π΄π΄][π΄π΄]2
= ππ ππππ
Integrate both sides of the equation β«β ππ[π΄π΄][π΄π΄]2
= β«ππ ππππ = ππ β«ππππ
Indefinite integral constant, b βοΏ½β 1[π΄π΄]οΏ½ = ππππ + ππ
At time t = 0, ππ = οΏ½ 1[π΄π΄]0
οΏ½ where [A]0 is the initial reactant concentration
Integrated Rate Law ππ[π¨π¨]
= πππ π + ππ[π¨π¨]ππ
Zero-order reaction
Differential Rate Law ππππππππ = βπ π [π¨π¨]
π π π π = ππ[π¨π¨]ππ
Simplify βππ[π΄π΄]ππππ
= ππ
Separate the variables ππ[π΄π΄] = βππππππ
Integrate both sides of the equation β«ππ[π΄π΄] = β«βππ ππππ = βππ β«ππππ
Indefinite integral constant, b [π΄π΄] = βππππ + ππ
At time t = 0, ππ = [π΄π΄]0 where [A]0 is the initial reactant concentration
Integrated Rate Law [π¨π¨] = βπππ π + [π¨π¨]ππ
3
First, Second, and Zero Order Plots for Chemical Reactions
1st order reaction
ln [π΄π΄] = βππππ + ln[A]0
2nd order reaction
1[π΄π΄]
= ππππ + 1
[π΄π΄]0
Zero-order reaction
[π΄π΄] = βππππ + [π΄π΄]0
4
Integrated Rate Law (continued)
- As reactant A goes to product B: aA β bB
Rate = β1ππππ[π΄π΄]ππππ
= ππ[π΄π΄]ππ
*In the following derivations of the three integrated rate laws, we assume the stoichiometric coefficient "a" in front of reactant A is no longer 1.
β’ Therefore, with each new integrated rate law, the proportionality constant is now written as ak'. However, when a rate constant is given, we treat it as a generic proportionality constant and call it k. This is because when an experiment is first performed in lab, the overall balanced reaction with correct coefficients is unknown. To figure out the order of the reaction, the data points are plotted and the slope is calculated and just called k.
1st order reaction: 2A β B Differential Rate Law Rate = β1
2ππ[π΄π΄]ππππ
= ππ[π΄π΄]1
Separate the variables ππ[π΄π΄][π΄π΄]
= β2ππ ππππ
Integrate both sides of the equation β« ππ[π΄π΄][π΄π΄]
= β«β2ππ ππππ = β2ππ β«ππππ
Indefinite integral constant, b ln[π΄π΄] = β2ππππ + ππ
At time t = 0, ππ = ππππ[π΄π΄]0 where [A]0 is the initial reactant concentration
Integrated Rate Law (linear form) ln [π΄π΄] = β2ππβ²ππ + ln[A]0
5
2nd order reaction: 3A β B
Differential Rate Law Rate = β13ππ[π΄π΄]ππππ
= ππ[π΄π΄]2
Separate the variables ππ[π΄π΄][π΄π΄]2
= β3ππ ππππ
Integrate both sides of the equation β«β ππ[π΄π΄][π΄π΄]2
= β«3ππ ππππ = 3ππ β«ππππ
Indefinite integral constant, b βοΏ½β 1[π΄π΄]οΏ½ = 3ππππ + ππ
At time t = 0, ππ = οΏ½ 1[π΄π΄]0
οΏ½ where [A]0 is the initial reactant concentration
Integrated Rate Law 1[π΄π΄]
= 3ππβ²ππ + 1[π΄π΄]0
Zero-order reaction: 4A β B
Differential Rate Law Rate = β14ππ[π΄π΄]ππππ
= ππ[π΄π΄]0
Simplify β14ππ[π΄π΄]ππππ
= ππ
Separate the variables ππ[π΄π΄] = β4ππππ
Integrate both sides of the equation β«ππ[π΄π΄] = β«β4ππ ππππ = β4ππ β«ππππ
Indefinite integral constant, b [π΄π΄] = β4ππππ + ππ
At time t = 0, ππ = [π΄π΄]0 where [A]0 is the initial reactant concentration
Integrated Rate Law [π΄π΄] = β4ππβ²ππ + [π΄π΄]0
6
Half-Life
- The half-life of a reaction (t1/2) is defined as the time it takes for the concentration of the reactant to decrease to half its original concentration.
-The shorter the half-life, the faster the reaction...the faster the reaction, the larger the rate constant.
1st order reaction
Integrated Rate Law ππππ[π΄π΄] = βππππ + ππππ[π΄π΄]0
At time t = t1/2 [π΄π΄] = 12
[π΄π΄]0
Substitute for [A] ln 12
[π΄π΄]0 = βππππ1 2β + ln[π΄π΄]0
Rearrange the terms ln 12
[π΄π΄]0 β ln[π΄π΄]0 = βππππ1 2β
ln 12
[π΄π΄]0[π΄π΄]0
= βππππ1 2β
β0.693 = βππππ1 2β
Half-life equation π π ππ ππβ = ππ.ππππππππ
7
2nd order reaction
Integrated Rate Law 1[π΄π΄] = ππππ + 1
[π΄π΄]0
At time t = t1/2 [π΄π΄] = 12
[π΄π΄]0 Therefore 1[π΄π΄]
= 2 β 1[π΄π΄]0
Substitute for [A] 2 β 1[π΄π΄]0
= ππππ1 2β + 1[π΄π΄]0
Rearrange the terms 2 β 1[π΄π΄]0
β 1[π΄π΄]0
= ππππ1 2β
1[π΄π΄]0
= ππππ1 2β
Half-life equation π π ππ ππβ = ππππ[π¨π¨]ππ
*Note that the half-life for a 2nd order reaction depends on the initial concentration of reactant; it is inversely proportional. Thus, the larger the initial concentration, the smaller the half-life.
Zero-order reaction
Integrated Rate Law [π΄π΄] = βππππ + [π΄π΄]0
At time t = t1/2 [π΄π΄] = 12
[π΄π΄]0
Substitute for [A] 12
[π΄π΄]0 = βππππ1 2β + [π΄π΄]0
Rearrange the terms ππππ1 2β = [π΄π΄]0 β12
[π΄π΄]0
ππππ1 2β = 12
[π΄π΄]0
Half-life equation π π ππ ππβ = [π¨π¨]ππππππ
*Note that the half-life for a zero-order reaction depends on the initial concentration of reactant; it is directly proportional. Thus, the larger the initial concentration, the larger the half-life.
8
Examples Using Integrated Rate Laws and
Half-Lives in Chemistry
1st order reaction
2N2O5(g) β 4NO2(g) + O2(g)
β’ Write the overall rate expression for this first order reaction.
πππππ π ππ = ππ[π΅π΅πππΆπΆππ]ππ
*Note that the exponent does not match the coefficient in front of N2O5. By just looking at the overall balanced equation, you cannot tell the order of the reaction.
β’ The half-life of N2O5 at 25ΛC is 1.87 x 104 seconds. Determine the rate constant for the above reaction.
t1 2β = 0.693
k
1.87 Γ 104 = 0.693
k
ππ = ππ.ππ Γ ππππβππ π¬π¬βππ
β’ Using the information above, if you were to start with 0.250 M N2O5, how much is left after 10.0 hours?
ππππ[π΄π΄] = βππππ + ππππ[π΄π΄]0
ππππ[π΄π΄] = β(3.7 Γ 10β5 sβ1)(3.6 Γ 104s) + ln[0.250]
[π¨π¨] = ππ.ππππππππ ππ
9
1st order reaction
2N2O(g) β 2NO(g) + O2(g)
β’ Write the overall rate expression for this first order reaction.
πππππ π ππ = ππ[π΅π΅πππΆπΆ]ππ
*Note that the exponent does not match the coefficient in front of N2O. By just looking at the overall balanced equation, you cannot tell the order of the reaction.
β’ The half-life of N2O at 1000 K is 0.912 seconds. Determine the rate constant for the above reaction.
t1 2β = 0.693
k
0.912 = 0.693
k
ππ = ππ.ππππ π¬π¬βππ
β’ Using the information above, if you were to start with 1.50 M N2O5, how much is left after 20 seconds?
ππππ[π΄π΄] = βππππ + ππππ[π΄π΄]0
ππππ[π΄π΄] = β(0.76 sβ1)(20 s) + ln[1.50]
[π¨π¨] = ππ.ππππ Γ ππππβππ ππ
β’ Using the equation ππππ[π΄π΄] = β2ππβ²ππ + ππππ[π΄π΄]0 and the information above, calculate kβ.
ππ = 2ππβ²
0.76 sβ1 = 2ππβ²
ππ.ππππ π¬π¬βππ = ππβ²
10
2nd order reaction
2C4H6(g) β C8H12(g)
β’ Write the overall rate expression given that the experimentally determined reaction of the dimerization of butadiene is second order.
πππππ π ππ = ππ[πͺπͺπππ―π―ππ]ππ
*Note that in this example the exponent does match the coefficient in front of C4H6. However, this is just a coincidence. By just looking at the overall balanced equation, you cannot tell the order of the reaction.
β’ The dimerization reaction of butadiene is second order and has a rate
constant of 0.0140M-1Β·s-1 at 500ΛC. Determine the C4H6 concentration after 115.0 seconds if the initial concentration of C4H6 is 0.0500 M.
1[π΄π΄] = ππππ +
1[π΄π΄]0
1[π΄π΄] = (0.0140 Mβ1 Β· sβ1 )(115.0 s) +
1[0.0500]
1[π΄π΄] = 21.6
[π¨π¨] = ππ.ππππππππ ππ
β’ Using the information above, calculate the half-life for the dimerization reaction of butadiene when the initial concentration of C4H6 is 0.0500M.
ππ1 2β = 1
ππ[π΄π΄]0
ππ1 2β =1
(0.0140Mβ1. sβ1)[0.0500]
π π ππ ππβ = ππππππππ π¬π¬
11
2nd order reaction
2HI(g) β H2(g) + I2(g)
β’ Write the overall rate expression for this experimentally determined second order reaction.
πππππ π ππ = ππ[π―π―π―π―]ππ
β’ This second order reaction has a rate constant of 6.4 x 10-9 M-1Β·s-1 at 500 K. Determine the concentration of HI after 90 days if the initial concentration of C4H6 is 0.850 M.
1[π΄π΄] = ππππ + 1[π΄π΄]0
1[π΄π΄] = (6.4 Γ 10β9 Mβ1 Β· sβ1 )(7776000 s) + 1[0.850]
1[π΄π΄] = 1.23
[π¨π¨] = ππ.ππππππ ππ
β’ Using the information above, calculate the half-life for the decomposition of HI when the initial concentration of HI is 2.00 M
ππ1 2β = 1ππ[π΄π΄]0
ππ1 2β =1
(6.4 Γ 10β9 Mβ1 Β· sβ1)[2.00]
π π ππ ππβ = ππ.ππ Γ ππππππ π¬π¬
β’ Using the equation 1[π΄π΄] = 2ππβ²ππ + 1[π΄π΄]0
and the information above,
calculate kβ and discuss whether this reaction is kinetically favorable.
ππ = 2ππβ²
6.4 Γ 10β9 Mβ1 Β· sβ1 = 2ππβ²
ππ.ππ Γ ππππβππ ππβππ Β· π¬π¬βππ = ππβ²
This reaction is kinetically unfavorable and will take a long time to occur as the rate constant is ~10-9: the smaller the rate constant, the slower the reaction.
12
Zero-order reaction
2NH3(g) β N2(g) + 3H2(g)
β’ The above reaction, known as the reverse Haber Bosh process takes place on a hot platinum surface. Write the overall rate expression for this zero order reaction.
πππππ π ππ = ππ[π΅π΅π―π―ππ]ππ
*Note that the exponent does not match the coefficient in front of NH3. By just looking at the overall balanced equation, you cannot tell that this is a zero order reaction.
β’ At a given temperature, the rate constant for the decomposition of
ammonia is 5.63 x 10-4 MΒ·s-1. Determine the initial concentration of NH3, if after 90.0 seconds the remaining concentration was 0.599 M.
[π΄π΄] = βππππ + [π΄π΄]0
[0.599] = β(5.63 Γ 10β4M β sβ1)(90.0s) + [π΄π΄]0
[π¨π¨]ππ = ππ.ππππππ ππ
β’ Using the information above, calculate the half-life for the decomposition of ammonia in the reverse Haber Bosh process if the initial concentration of ammonia was 0.350 M.
ππ1 2β = [π΄π΄]02ππ
ππ1 2β = 0.350 M
2(5.63 Γ 10β4M β sβ1)
π π ππ ππβ = ππππππ π¬π¬
13
Radioactive Decay
- In microbiology and physics labs, we frequently solve problems about carbon dating and radioactive decay. Radioactive decay is a first order reaction. In these types of problems, the rate constant k is referred to as the decay constant, and N is the number of radioactive nuclei.
Differential Rate Law πΉπΉπππ π ππ = βπ π π΅π΅π π π π
= πππ΅π΅
Separate the variables ππππππ
= βππ ππππ
Integrate both sides of the equation β« ππππππ
= β«βππ ππππ = βππ β«ππππ
Indefinite integral constant, b lnππ = βππππ + ππ
At time t = 0, ππ = ln ππ0 where N0 is the initial number of radioactive nuclei.
Integrated Rate Law (linear form) π₯π₯π₯π₯ π΅π΅ = βπππ π + π₯π₯π₯π₯π΅π΅ππ
To more clearly see the exponential relationship between time, t, and the number of radioactive nuclei, N, we can convert the integrated first-order rate-law (linear form) to its non-linear exponential form:
Raise both sides to the power of e ππln ππ = ππβππππ ππln ππ0
Simplify, and obtain non-linear form ππ = ππ0 ππβππππ
One can also then write the non-linear differential rate law: π π ππππππ = βππππ
ππππ= ππππ = ππ[ππ]0 ππβππππ
β’ Carbon-14 dating is used to determine the age of various objects and has a half-life of 5730 years. If a sample today contains 0.096 mg of 14C, how much 14C was present in this sample 12,530 years ago? Hint: To simplify, let mass of 14C = number of radioactive nuclei, N
t1 2β = 0.693k
= 5730 years
ππ = 1.21 Γ 10β4 π¦π¦πππππ¦π¦β1
ππππ ππ = βππππ + ππππ ππ0
ππππ(0.096 mg) = β(1.21 Γ 10β4 π¦π¦πππππ¦π¦β1)(12530 π¦π¦πππππ¦π¦) + ππππ ππ0
π΅π΅ππ = ππ.ππππππ π¦π¦π¦π¦
14
Population Growth: Bacteria
- In biology labs, we study bacteria growth under various conditions. Population growth is mathematically equivalent to a first order reaction, but the slope is positive because we are modelling population growth. In these types of problems, the rate constant k is referred to as the growth constant, and N is the number of bacteria and some time, t.
Integrated Rate Law π₯π₯π₯π₯ π΅π΅ = πππ π + π₯π₯π₯π₯π΅π΅ππ
Raise both sides to the power of e ππln ππ = ππππππ ππln ππ0
Simplify to get exponential form π΅π΅ = π΅π΅ππ πππππ π
β’ Suppose the initial number of bacteria in your culture dish is 1.40 million. You check back in an hour and the number of bacteria has increased to 10.35 million. What is the growth rate of this bacterial species (per min)?
One can use the linear or exponential equation above where the initial number of bacteria is No and the number of bacteria after an hour is N.
ππ = ππ0 ππππππ
[10.35] = [1.40]ππππ(60 ππππππ)
7.39 = ππ(60 ππππππ)ππ
ln(7.39) = ππππ ππ60ππ
2.00 = 60ππ
ππ = ππ.ππππππ ππππππβππ
= ππ.ππ% ππππππβππ
15
Drug Concentration
- In biology labs, we study the effectiveness of the body to metabolize certain administered drugs. The body's metabolism of drugs can be defined as a first order reaction. In these types of problems, the rate constant k is characterized as the removal rate of the drug through metabolism or excretion, and C is the concentration of the substance in the body.
Integrated Rate Law π₯π₯π₯π₯ [πͺπͺ] = βπππ π + π₯π₯π₯π₯[ππ]ππ
Raise both sides to the power of e ππln [πΆπΆ] = ππβππππ ππln [πΆπΆ]0
Simplify to get exponential form [πͺπͺ] = [πͺπͺ]ππ ππβπππ π
β’ Patients diagnosed with depression have a decreased concentration of serotonin in their brain. To help increase this concentration, doctors administer Selective Serotonin Reuptake Inhibitors (SSRIs) to the patients. Normal levels of serotonin in a healthy individual range from 101-283 ng/ml. Serotonin is metabolized by the body at a rate of 9.63 x 10-6 s-1. Assume that when the SSRI is administered, the serotonin concentration increases to 175.0 ng/ml. How long will it take for the serotonin concentration to fall below 100.0 ng/ml?
[πΆπΆ] = [πΆπΆ]0ππβππππ
[100.0] = [175.0]ππβ(9.63 Γ 10β6 π π β1)ππ
0.571 = ππβ(9.63 Γ 10β6 π π β1)ππ
ln 0.571 = ππππ ππβ(9.63 Γ 10β6 π π β1)ππ
β0.560 = β9.63 Γ 10β6 ππ
π π = ππ.ππππ Γ ππππππ ππ