C/P 07 · Reaction kinetics
Rate Laws and Temperature
Find reaction orders from a table, then use the rate law, temperature and catalysis.
Passage
The rate of a chemical reaction describes how quickly reactants are consumed or products are formed. For a reaction between two reactants, A and B, the rate law has the form rate = k[A]ᵐ[B]ⁿ, where k is the rate constant and the exponents m and n are the orders of the reaction with respect to A and B. The exponents cannot be found from the balanced equation and must be determined by experiment. The sum m + n is the overall order, and the units of k depend on the overall order.
The rate constant also depends on temperature. Reactions occur when molecules collide with enough energy to overcome an energy barrier called the activation energy, Ea. Raising the temperature increases the fraction of molecules with that much energy, and a catalyst provides a different pathway with a lower Ea. Neither of these changes the overall free energy change of the reaction.
Part 1. A chemist studies the reaction A + B → products in solution at 300 K. She measures the initial rate of the reaction for different starting concentrations, in moles per liter (M). The rate is given in M/s. Table 1 shows the results.
Table 1. Initial rates at 300 K.
| Experiment | [A] (M) | [B] (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 8.0 × 10⁻³ |
| 4 | 0.20 | 0.20 | 1.6 × 10⁻² |
Each rate is measured by following the disappearance of A with a spectrophotometer, because A absorbs light at a wavelength where B and the products do not. The cuvette is held at 300 K in a thermostated holder, and the reaction is started by adding B last.
Part 2. The chemist next measures the rate constant at different temperatures, using the same starting concentrations as in Experiment 1. She also repeats the 300 K measurement in the presence of a catalyst. Table 2 shows the results. The units of k are those of the rate law found in Part 1.
Table 2. Rate constant under different conditions.
| Condition | Temperature (K) | Rate constant k |
|---|---|---|
| No catalyst | 300 | 2.0 |
| No catalyst | 310 | 4.0 |
| No catalyst | 320 | 8.0 |
| With catalyst | 300 | 20 |
A typical rule of thumb holds that many reactions at room temperature proceed about twice as fast for each 10 K rise in temperature, which is close to what the chemist sees. The chemist also notes that the reaction mixture is stirred to keep the concentrations uniform, and that each rate in Table 1 is measured before more than a few percent of the reactants have been used, so that the concentrations are close to their starting values.
In a mechanism, a reaction takes place through a series of elementary steps. The slowest elementary step determines the overall rate, and the rate law of an elementary step can be read directly from its molecularity. The chemist wants to propose a mechanism that is consistent with the rate law she finds, but she must remember that a proposed mechanism can only be disproved by experiment, not proved.
Questions
Question 1 of 6
What is the order of the reaction with respect to A?
If the reaction were zero order in A, changing [A] would not change the rate. In Experiments 1 and 2 doubling [A] doubles the rate.
A third-order dependence would multiply the rate by 8.
A second-order dependence would quadruple the rate when [A] is doubled. The rate only doubles.
Why the answer is what it is
Comparing Experiments 1 and 2, [B] is constant and [A] doubles, and the rate doubles (2.0 × 10⁻³ to 4.0 × 10⁻³). The reaction is first order in A.
Review: Rate Laws & Reaction Order · Reaction Kinetics & Activation Energy · Equations: Reactions · Gibbs Free Energy & Spontaneity
Question 2 of 6
What is the order of the reaction with respect to B?
Doubling [B] would then double the rate. In Experiments 1 and 3 the rate increases fourfold.
Doubling [B] would then increase the rate eightfold. The observed increase is fourfold.
Experiments 1 and 3 show that the rate depends strongly on [B].
Why the answer is what it is
Comparing Experiments 1 and 3, [A] is constant and [B] doubles, and the rate rises from 2.0 × 10⁻³ to 8.0 × 10⁻³, a factor of 4 = 2². The reaction is second order in B. Experiment 4 agrees: doubling both gives 2 × 4 = 8 times the rate of Experiment 1.
Review: Rate Laws & Reaction Order · Reaction Kinetics & Activation Energy · Equations: Reactions · Gibbs Free Energy & Spontaneity
Question 3 of 6
What are the units of the rate constant k?
These are the units for a first-order reaction. This reaction is third order overall.
These are the units for an overall second-order reaction.
These are the units of a rate, not of a third-order rate constant.
Why the answer is what it is
The rate law is rate = k[A][B]², with units of M/s on the left. k = rate ÷ ([A][B]²) has units (M s⁻¹) ÷ M³ = M⁻² s⁻¹. From Experiment 1, k = 2.0 × 10⁻³ ÷ (0.10 × 0.010) = 2.0 M⁻² s⁻¹.
Review: Rate Laws & Reaction Order · Reaction Kinetics & Activation Energy · Equations: Reactions · Gibbs Free Energy & Spontaneity
Question 4 of 6
Using the rate law, what is the predicted initial rate when [A] = 0.30 M and [B] = 0.30 M at 300 K?
This is 2.0 × 0.30 × 0.30, which treats the reaction as first order in B. B is second order, so [B] must be squared.
This is the rate in Experiment 4, where both concentrations are 0.20 M.
This is smaller by a factor of 10 than the correct value. Check the arithmetic: [B]² = 0.090, and 2.0 × 0.30 × 0.090 = 0.054.
Why the answer is what it is
rate = k[A][B]² = 2.0 × 0.30 × (0.30)² = 2.0 × 0.30 × 0.090 = 0.054 M/s = 5.4 × 10⁻² M/s.
Review: Rate Laws & Reaction Order · Reaction Kinetics & Activation Energy · Equations: Reactions · Gibbs Free Energy & Spontaneity
Question 5 of 6
Why does the rate constant increase as the temperature is raised from 300 K to 320 K?
The activation energy is a property of the reaction pathway and does not change with temperature in this treatment. A catalyst would change it.
The concentrations are kept the same in every measurement.
The free energy change relates to the position of equilibrium, not to the rate.
Why the answer is what it is
At higher temperature a larger fraction of molecules have kinetic energy at or above Ea, so more collisions lead to reaction. The rate constant roughly doubles with each 10 K.
Review: Rate Laws & Reaction Order · Reaction Kinetics & Activation Energy · Equations: Reactions · Gibbs Free Energy & Spontaneity
Question 6 of 6
In the presence of the catalyst, k is ten times larger at 300 K. Which statement about the catalyst is correct?
A higher barrier would lower the rate constant. The rate constant here increased.
A catalyst changes the pathway and the speed, not the free energy change of the overall reaction.
Catalysts speed up the forward and reverse reactions equally, and the position of equilibrium is unchanged.
Why the answer is what it is
A catalyst provides an alternative pathway with a lower activation energy, so more collisions are successful and k is larger. The initial and final states are the same, so ΔG and the equilibrium constant do not change.
Review: Rate Laws & Reaction Order · Reaction Kinetics & Activation Energy · Equations: Reactions · Gibbs Free Energy & Spontaneity