AP Chemistry – Chapter 11: Chemical Kinetics (Reaction Rates)
Meaning of a Reaction Rate______
- The rate of a reaction is the positive quantity that expresses how the concentration of a reactant or product changes with time.
- The rates of reactions span an enormous range. From those that are complete within seconds, such as explosions, to those that take thousands or even millions of years, such as the formation of diamonds or other minerals in the Earth’s crust.
- Factors that affect reaction rates:
- The physical state of the reactants: The ability of the reactants to collide lead to a reaction. The more readily molecules collide with each other, the more rapidly they react. A solid that is broken in up in to pieces will react faster than one that is not because of a greater surface area for reactions.
- The concentrations of reactants: Most chemical reactions occur faster if the concentration of the reactants is increased. A higher concentration calls for more collisions, and more collisions lead to a faster reaction.
- The temperature at which the reaction occurs: The rates of chemical reactions increase as temperature is increased. An increase in temperature leads to higher kinetic energy of the substance and more collisions for the reactions to take place faster.
- The presence of a catalyst: Catalysts are agents that increase reaction rates without being used up. They affect the kinds of collisions (mechanisms) that lead up to a reaction.
* On a molecular level: Rates depend on the frequency of the collisions between molecules, the greater the frequency of collisions, the greater the rate of reaction.
Reaction Rates______
- A reaction rate is the change in the concentration of reactants or products per unit time.
- The unit for a reaction rate is then molarity per seconds (M/s).
- The rate of a reaction can be based on the rate of disappearance of a reactant of the rate of appearance for a product.
- Example: A B
Time (s) / [A] / [B] / Rate (M/s)
0 / 1.00 / 0.00 / --
20 / 0.54 / 0.46
40 / 0.30 / 0.70
Rate of appearance of B = change in concentration of B ([B]) = [B] at t2 - [B] at t1
change in time (t) t2 – t1
Average rate = 0.46M – 0.00M = 2.3 x 10-2 M/s
20 s – 0.0 s
Rate of disappearance of A= - change in concentration of A ([A])= -[A] at t2 - [A] at t1
change in time (t) t2 – t1
Average rate = - 0.54M – 1.00M = 2.3 x 10-2 M/s
20 s – 0.0 s
Reaction Rates & Stoichiometry ______
- When stoichiometric relationships in a chemical reaction are not one to one, the molar ratio must be applied to determining reaction rates.
- For a general reaction,
aA + bB → cC + dD,
the reaction velocity (reaction rate) can be written in a number of different but equivalent ways:
Example: 2HI (g) H2 (g) + I2 (g)
Rate = - 1 ] = H2] = [I2]
2 t t t
The Rate Law______
In general the rate law expression for the general reaction aA + bB → cC + dD, is
Rate = k[A]m[B]n
-Rate = M/s
-k = units will vary depending on the rate order of the reactants
-[A] & [B] = M
-m & n = will be whole numbers that determine the rate order of the reactants, they can vary and are no way related to the coefficients.
Rate Order______
- One way of studying the effect of concentration on reaction rate is to determine the way in which the rate at the beginning of a reaction depends on the starting concentrations. Let us consider the following reaction to illustrate:
NH4+(aq) + NO2 (aq) N2 (g) + 2H2O (l)
Experiment # / Initial [NH4+] / Initial [NO2-] / Observed Initial Rate (M/s)1 / 0.0100 / 0.200 / 5.4 x 10-7
2 / 0.0200 / 0.200 / 10.8 x 10-7
3 / 0.0400 / 0.200 / 21.5 x 10-7
4 / 0.200 / 0.0202 / 10.8 x 10-7
5 / 0.200 / 0.0404 / 21.6 x 10-7
6 / 0.200 / 0.0808 / 43.3 x 10-7
The rate law expression for this reaction is:
Rate = k[NH4+]1[NO2-]1
- take note only reactants appear in the rate law expression
- k represents a proportionality constant called the rate law constant (sometimes simply the rate constant).
- overall rate order for the reaction is the sum of the rate orders for the reactants. So for the above reaction the rate order is 2.
0 order: The concentration of the reactants has no impact on the rate of the reaction.
1st order: The rate doubles when concentration is doubles, triples when concentration is tripled, quadruples when the concentration is quadrupled, and so forth (raised to the 1st power in the rate law).
2nd order: The rate quadruples when the concentration is doubled ([2]2 = 4), the rate increases ninefold when the concentration is triples ([3]2 = 9), and so forth (raised to the 2nd power in the rate law).
- Once the rate law constant is determined the rate could be calculated at any reactant concentration.
First Order Rate Law______
The integrated form of the first-order rate law equation is:
- Where X is the concentration of a reactant at any moment in time, (X)o is the initial concentration of this reactant, k is the constant for the reaction, and t is the time since the reaction started.
- This equation is useful in calculating how much of a substance remains after a certain amount of time has passed, or to calculate how long it takes until the concentration is at a certain point.
First-Order Reaction
/ If the rate law of a reaction is first order
with respect to [A], then the graph of ln[A]
versus time (t) creates a straight line with a
negative slope. The value of the slope of the
line is equal to the negative value of the
rate constant (k).
- The equation for the half-life of a substance is derived from this equation.
- Half-life - The length of time it takes for exactly half of the nuclei of a radioactive sample to decay.
Second Order Rate Law______
The integrated form of the second-order rate law equation is:
- Where X is the concentration of a reactant at any moment in time, (X)o is the initial concentration of this reactant, k is the constant for the reaction, and t is the time since the reaction started.
- This equation is useful in calculating how much of a substance remains after a certain amount of time has passed, or to calculate how long it takes until the concentration is at a certain point.
Second-Order Reaction
/ If the rate law for a reaction is second order
with respect to [A], a graph of 1/[A] versus
time (t) creates a straight line with a positive
slope. The value of the slope of the line is
equal to the value of the rate constant (k).
Reaction Mechanisms ______
- A reaction mechanism is the sequence of steps of a multi-step chemical reaction.
- Important points to remember:
- The slow step is the rate determining step.
- A catalyst will not appear as a reactant or a product.
- An intermediate is a material the is produced by a step and consumed later, it does not show as either a product or reactant in the overall equation.
Two-Step Mechanism:
Consider the following reaction:
2 NO + O22 NO2
This reaction actually is the sum of two steps:
Step 1: 2 NO N2O2(fast step)
Step 2: N2O2+ O22 NO2(slow step)
When the two equations are added together, the resulting equation is the original equation:
2 NO N2O2+ N2O2+ O22 NO2
2 NO + O22 NO2
Since the rate of the entire reaction is dependent upon the rate limiting step, the rate law can be written as:
rate = k[N2O2][O2]
However, there is one problem. The rate law of a reaction has to be in terms of the concentrations of the reactants.
- N2O2 is an intermediate, not an original reactant.
- Therefore, we must solve for [N2O2] in terms of [NO].
First, find the rate law of the other equilibrium reaction (fast step):
- NOTE: Since the reaction is an equilibrium reaction, it can be assumed that at equilibrium, the rates of the forward and reverse reactions are equal.
We end up with the rate law in terms of the original reactants.
- Even though the coefficients of the reactants in the balanced equation happened to be the exponents of the concentrations of the reactants in the rate law, this is not always true.
- The rate law for a reaction cannot be determined by the coefficients of the balanced reaction; it must be determined experimentally.
Three-Step Mechanism:
2 A + 2 B C + D
Step 1: A + A X (fast)
Step 2: X + B C + Y (slow)
Step 3: Y + B D (fast)
FRQ Practice Problems:
Experiment / Initial Concentration of NO (mol L-1) / Initial Concentration of O2 (mol L-1) / Initial Rate of Formation of NO2 (mol L-1 s-l)1 / 0.0200 / 0.0300 / 8.52 10-2
2 / 0.0200 / 0.0900 / 2.56 10-1
3 / 0.0600 / 0.0300 / 7.67 10-1
1. Determine the order of the reaction with respect to each of the following reactants (based on the experimental data above). Give details of your reasoning, clearly explaining or showing how you arrived at your answers.
(a)NO
(b)O2
(c)Write the expression for the rate law for the reaction as determined from the experimental data.
(d)Determine the value of the rate constant for the reaction, clearly indicating the units.
Answer:
(a) second order with respect to NO; using experiments 1 & 3,
; m = 2
(b) first order with respect to O2; using experiments 1 & 2,
; n = 1
(c)rate = k [NO]2[O2]
(d)0.0852 mol L-1s-1 = k (0.200 mol L-1)2(0.0300 mol L-1)
k = 7100 L2 mol-2s-1
2. An environmental concern is the depletion of O3 in Earth's upper atmosphere, where O3 is normally in equilibrium with O2 and O. A proposed mechanism for the depletion of O3 in the upper atmosphere is shown below.
Step IO3 + Cl O2 + ClO
Step IIClO + O Cl + O2
(a)Write a balanced equation for the overall reaction represented by Step I and Step II above.
(b)Clearly identify the catalyst in the mechanism above. Justify your answer.
(c)Clearly identify the intermediate in the mechanism above. Justify your answer.
(d)If the rate law for the overall reaction is found to be rate = k[O3] [Cl], determine the following.
(i)The overall order of the reaction
(ii)Appropriate units for the rate constant, k
(iii) The rate-determining step of the reaction, along with justification for your answer
Answer:
(a)O3 + O 2 O2
(b)Cl; used in step I and regenerated in step II, the amount at the end is the same as the beginning
(c)ClO; product of step I and used in step II, an intermediate is a material the is produced by a step and consumed later, it does not show as either a product or reactant in the overall equation.
(d)(i) second order overall
(ii) k unit is M-1time-1
(iii) step 1. the rate law applies to the concentration of the materials in the slowest step, the rate determining step.
3.
5 Br–(aq) + BrO3–(aq) + 6 H+(aq) 3 Br2(l) + 3 H2O(l)
In a study of the kinetics of the reaction represented above, the following data were obtained at 298 K.
Experiment / Initial [Br–] (mol L-1) / Initial [BrO3–] (mol L-1) / Initial [H+] (mol L-1) / Rate of Disappearance of BrO3– (mol L-1 s-1)1 / 0.00100 / 0.00500 / 0.100 / 2.5010-4
2 / 0.00200 / 0.00500 / 0.100 / 5.0010-4
3 / 0.00100 / 0.00750 / 0.100 / 3.7510-4
4 / 0.00100 / 0.01500 / 0.200 / 3.0010-3
(a)From the data given above, determine the order of the reaction for each reactant listed below. Show your reasoning.
(i)Br–
(ii)BrO3–
(iii)H+
(b)Write the rate law for the overall reaction.
(c)Determine the value of the specific rate constant for the reaction at 298 K. Include the correct units.
Answer:
(a)(i)1st order with respect to Br–; in experiments 1 and 2, a doubling of the [Br–] results in the doubling of the initial rate, and indication of 1st order
(ii)1st order with respect to BrO3–
using expt. 1 & 3
rate1 = k[Br–]1[BrO3–]m[H+]n
= k[Br–]1[H+]n
rate3 = k[Br–]1[BrO3–]m[H+]n
= k[Br–]1[H+]n
m = 1
(iii)2nd order with respect to H+
using expt. 3 & 4
rate3 = k[Br–]1[BrO3–]1[H+]n
= k[Br–]1
rate4 = k[Br–]1[BrO3–]1[H+]n
= k[Br–]1
n = 2
(b)rate = k[Br–]1 [BrO3–]1 [H+]2
(c)2.50x10–4 = k (0.00100) (0.00500) (0.100)2
k = 5000 mol–3L3s-1