Exercise practice: Chemical Kinetics

5.1 How the exercise set is arranged

The chapter has 30 exercises, numbered 3.1 to 3.30. They fall into four groups. Rate expressions and average rate cover 3.1 to 3.3. Order and rate law from data cover 3.4 to 3.12. Integrated rate equations, half-life and radioactive dating cover 3.13 to 3.21. Temperature dependence, Arrhenius and activation energy cover 3.22 to 3.30. The worked items below are the patterns that repeat.

5.2 Rates and rate laws

Exercises 3.1 and 3.2 ask for average and instantaneous rates from concentration data, so divide the change in concentration by the time interval and use the coefficients to make one rate for the reaction. Exercises 3.4 to 3.12 give initial rate tables. Compare two experiments in which only one concentration changes; the ratio of the rates gives the order in that reactant. Once the orders are known, substitute one experiment into the rate law to find k, watching the units: order two gives L mol^-1 s^-1, order three gives L^2 mol^-2 s^-1.

5.3 Integrated equations and half-life

Exercise 3.14 is carbon-14 dating: 80 percent of the original carbon-14 remains, so t equals 2.303 divided by k times log of 100 over 80, with k equal to 0.693 divided by 5730 per year. This gives about 1840 years.

Exercise 3.18 asks for a proof that the time for 99 percent completion is twice that for 90 percent completion in a first order reaction. The time is 2.303 divided by k times log of 100 over the amount left, so the ratio of the two times is log 100 over log 10, which is exactly two.

Exercise 3.19 gives a first order reaction that is 30 percent decomposed in 40 minutes. The amount left is 70 percent, so k is 2.303 divided by 40 times log of 100 over 70, which is 8.92 times 10^-3 per minute. The half-life is 0.693 divided by k, which is about 77.7 minutes.

Exercise 3.24 asks what fraction remains after 100 s when k is 2.0 times 10^-2 s^-1. The amount left is the initial amount times e to the power minus kt, and kt is 2.0, so the fraction left is e^-2, which is 0.135, or 13.5 percent.

5.4 Temperature and activation energy

Exercise 3.30 gives a rate that quadruples when the temperature rises from 293 K to 313 K. Using the two temperature form, the activation energy is 2.303R times T1T2 divided by (T2 - T1) times log 4. The temperature factor is 293 times 313 divided by 20, which is 4586 K. With 2.303R equal to 19.15 J K^-1 mol^-1, the activation energy is 19.15 times 4586 times 0.602, which is about 52.9 kJ mol^-1.

Exercise 3.28 uses the same equation in the opposite direction: with k at one temperature and Ea known, find the temperature at which k has a stated larger value. Keep the temperatures in kelvin and solve for the ratio of the rate constants first when both temperatures are unknown.

5.5 Two more worked templates

For an initial rate table, the multiplier in the rate divided by the multiplier in the concentration gives the order, provided only one concentration changes between the two experiments. When both change, use two pairs so that one reactant can be eliminated by division. Once the exponents are known, the rate constant comes from any one row, and the unit of k follows from the overall order. For a first order decay, the time to fall to a fraction f of the original amount is t equal to 2.303 divided by k times log of 1 over f, so the time to reach one sixteenth, for example, is four half-lives. For the pressure method, the total pressure at time t is related to the fall in the reactant pressure through the stoichiometry, and the difference gives the amount reacted. Finally, for an Arrhenius calculation with two rate constants, substitute into the two temperature form directly, keep the temperatures in kelvin and report the activation energy in kilojoules per mole.

5.6 Combining kinetics with stoichiometry

Some exercises give pressure data for a gas phase reaction and ask for the rate constant. Convert the pressure of the reactant into concentration using the ideal gas relation, or use the stoichiometry to relate the total pressure to the amount reacted. For a decomposition that produces more moles of gas, the rise in total pressure measures the extent of reaction directly. In radioactive dating questions, the fraction remaining is the activity ratio, so the activity can be used in place of concentration in the first order equation. These two translations turn an unfamiliar question into a standard calculation.

For a first order reaction, the time to reach a given fraction remaining does not depend on the starting concentration, so a single half-life measurement fixes the rate constant for every concentration. This is why the half-life method is so common: measure one clean half-life and the whole kinetic profile follows.

A final habit for the numericals: quote the order and the unit of k together, because the unit is itself evidence for the order. If a calculation gives a unit of s^-1 but the order claimed is zero, one of the two is wrong before the final answer is reached.