Quantitative kinetics: sampling, clocks and rate equations

What you’ll learn

• Deduce a rate equation from controlled concentration changes.
• Explain what titrimetric sampling and an iodine clock actually measure.

Understand the idea

Core practical 9 uses iodine–propanone kinetics and an iodine clock to turn an observable change into rate evidence. In a sampling experiment, a fixed aliquot represents the reaction mixture at a stated time. The sample must be quenched by an appropriate validated method so that its composition does not keep changing during analysis. When the titration amount is proportional to the iodine remaining, a titre–time graph traces iodine consumption; a tangent gives a rate at a particular instant.

An iodine clock records the time required to form a fixed small amount of iodine before a visible endpoint appears. Keeping the clock threshold and total volume constant makes $1/t$ a relative initial-rate measure, provided only a small fraction reacts before the endpoint. If the threshold concentration is supplied, dividing it by time gives an average production rate over that short interval. Changing the thiosulfate amount changes the threshold and invalidates a simple comparison of reciprocal times.

To find orders, compare experiments in which one initial concentration changes and the other relevant concentrations are fixed. The rate equation is experimental: coefficients in an overall equation do not by themselves establish orders. Initial concentrations are the values after mixing, so dilution must be included. Use the resulting total order to derive the units of the rate constant, and compare rate constants only at the same temperature.

Key formulae

$$r=k[A]^m[B]^n$$

$$\frac{r_2}{r_1}=\left(\frac{[A]_2}{[A]_1}\right)^m\quad\text{with other factors fixed}$$

$$\overline r=\frac{\Delta[\mathrm{I_2}]}{t}$$

$$[A]_{\mathrm{mixed}}=\frac{c_A V_A}{V_{\mathrm{total}}}$$

Worked example

For iodine–propanone, initial concentrations in $\mathrm{mol\,dm^{-3}}$ and rates in $\mathrm{mol\,dm^{-3}\,s^{-1}}$ are: experiment 1, propanone $0.20$, $\mathrm{H^+}$ $0.10$, iodine $0.0010$, rate $2.0\times10^{-5}$; experiment 2, $0.40$, $0.10$, $0.0010$, rate $4.0\times10^{-5}$; experiment 3, $0.20$, $0.20$, $0.0010$, rate $4.0\times10^{-5}$; experiment 4, $0.20$, $0.10$, $0.0020$, rate $2.0\times10^{-5}$. All temperatures match. Find the rate equation and $k$.

1. Experiments 1 and 2 double propanone alone and double the rate, so its order is $1$.
2. Experiments 1 and 3 double $[\mathrm{H^+}]$ alone and double the rate, so the hydrogen-ion order is $1$. Experiments 1 and 4 show zero order in iodine.
3. Thus $r=k[\mathrm{propanone}][\mathrm{H^+}]$. Experiment 1 gives $k=(2.0\times10^{-5})/(0.20\times0.10)=1.0\times10^{-3}\ \mathrm{dm^3\,mol^{-1}\,s^{-1}}$.

Answer: $r=k[\mathrm{propanone}][\mathrm{H^+}]$, with $k=1.0\times10^{-3}\ \mathrm{dm^3\,mol^{-1}\,s^{-1}}$.