Chemical Kinetics

University of Arizona, Chemistry 103b, Walter Miller

Among other things this semester we will approach the two basic questions concerning chemical reactions:

• Where is the position of equilibrium? (Does the reaction want to "go"?) Thermodynamics answers this question with G (& K). (later)
• How fast does it go? (Can it go if it wants to?) (Can it find a "reaction path"?)

examples
C(diamond) C(graphite) G⁰ = -2.3 kJ
H₂(g) + 1/2 O₂(g) H₂O(l) G⁰ = -237 kJ

This is related to the general problem of

"statics" versus "dynamics"

"equilibrium" versus "kinetics"

and involved with the problem of

"macroscopic" versus "microscopic"

Rates of chemical reactions

aA + bB cC + dD

Discuss "synthetic" experimental data for: F₂ + 2 NO₂ 2 NO₂F

This suggests:

Rate = -(1/a)([A]/t) = -(1/b)([B]/t) = (1/c)([C]/t) = (1/d)([D]/t)

This assumes that the stoichiometry is known and that no stoichiometrically significant intermediates are formed. Also, not everyone reports their data in this manner.

The data also suggests:

Rate = k[A]^[B]^[X]^[Y]^

This is an example of a rate law, with k being the (specific) rate constant and is the order with respect to A, is the order with respect to B, etc and the overall order is the sum of the individual orders. If your analysis is complete, k should depend only on T.

Questions:

• What is the rate law? (find orders)
• What is the rate constant? (find k)
• What is the activation energy? (find E_a)
• What is the mechanism?

Methods of finding the rate law:

• Curve fitting (the best)
• Method of half lives (the worst)
• Method of initial rates (fair, if you can use it)

Curve fitting:

Develop a vocabulary of resultlets: A products

In the following, if you don't like calculus, interpret "d" as .

0th order:

-d[A]/dt = k[A]⁰ = k

d[A] = -kdt

[A] - [A]₀ = -k(t - t₀)

let t₀ = 0

[A] = [A]₀ - kt

1st order:

-d[A]/dt = k[A]¹ = k[A]

ln ([A]/[A]₀) = -kt

[A] = [A]₀e^-kt

2nd order:

-d[A]/dt = k[A]²

d[A]/[A]² = -kdt

-1/[A] + 1/[A]₀ = -kt

More complicated cases (thousands) have been dealt with in the literature.

Examples:

Method of half lives:

More resultlets: A products

0th order:

[A]₀/2 = [A]₀ - kt_1/2

t_1/2 = [A]₀/2k

1st order:

ln ([A]₀/2[A]₀) = -kt_1/2

t_1/2 = (ln 2)/k (note independent of [A]₀)

2nd order:

-2/[A]₀ + 1/[A]₀ = -kt_1/2

t_1/2 = 1/k[A]₀

Revisit 1st order:

ln ([A]/[A]₀) = -kt = -(ln 2)(t/t_1/2)

[A]/[A]₀ = (e^{-ln 2})^(t/t1/2) = (1/2)^t/t1/2

Method of initial rates:

Work example.

Finding the rate constant:

Usually falls out during the analysis of the rate law.

Finding the activation energy:

When a rate law is fully analyzed, the specific rate constant should only depend on the temperature. The tradition is to analyze the temperature dependence using the Arrhenius equation:

k = Ae^-Ea/RT

If you have only two temperature measurements, you can also do this algebraically.

Although the activation energy is usually measured on the macroscopic scale it is often interpreted on the microscopic scale. In particular, it is popular to sketch "energy" versus "reaction coordinate".

Mechanisms:

This is the microscopic description of the chemical reaction. In general, one postulates a set of elementary steps which, when analyzed, gives you the observed rate law. There may be several mechanisms that superficially appear adequate to describe a single reaction. It is very important to separate in your mind the macroscopic analysis of rate data that we have been doing from the microscopic description that we are about to start. Confusing these issues is probably the most common mistake made by chemistry professionals.

When discussing elementary steps the term "molecularity" often replaces the term "order" in the language. This helps to remind people that this analysis is taking place at the microscopic scale and not the macroscopic scale.

1st order becomes unimolecular
2nd order becomes bimolecular (common)
3rd order becomes termolecular (rare)

Rules:

If there are no parallel steps, summing the individual elementary steps should give you the balanced equation.

Properly analyzed, the mechanism should give you the observed rate law.

Procedure for a single step mechanism:

Easy.

Procedure for a multistep mechanism:

Find the slow, "rate determining" step.

Use whatever tools are necessary (maybe none) to eliminate the concentration of any species that cannot be controlled (for instance, an intermediate) from the rate law.

The primary tool that we have available is equilibrium. This severely limits the number of mechanisms that we can analyze.

Examples:

For a good discussion of the SN2 reaction including a detailed graph of energy versus reaction coordinate and a movie go to SN2

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