Iodine clock experiment

A typical iodine clock experiment seeks to determine the rate law of a reaction.

For the peroxide-iodide reaction $H_2O_2+2H^++2I^-\rightarrow 2H_2O+I_2$ , a proposed rate law is:

$rate=k[H_2O_2]^x[H^+]^y[I^-]^z\; \; \; \; \; \; \; \; 48$

Since the rate of the peroxide-iodide reaction is relatively slow, the method of initial rates can be used to determine the rate law. Let’s suppose a series of peroxide-iodide experiments is conducted at room temperature with the following parameters:

Experiment	Flask 1			Flask 2
Experiment	0.040M H₂O₂/cm³	0.05M H₂SO₄/cm³	Starch/ drops	0.010M KI /cm³	0.001M Na₂S₂O₃/cm³	H₂O/cm³
1	10	10	3	10	10	10
2	10	10	3	20	10	–
3	20	10	3	10	10	–
4	10	20	3	10	10	–

Water is added to experiment 1 so that all four experiments are conducted under the same constant volume. For every experiment, a stopwatch is turned on once the contents of the two flasks are mixed together and turned off when the mixture turns blue-black.

$H_2O_2+2H^++2I^-\rightarrow 2H_2O+I_2\; \; \; \; \; \; \; \; 49$

$2S_2O_3^{\; \: 2-}+I_2\rightarrow S_4O_6^{\; \: 2-}+2I^-\; \; \; \; \; \; \; \; 50$

When the mixture turns blue-black, the initial amount of 1.0×10^-5 moles of thiosulphate in each experiment has reacted with 5.0×10^-6 moles of iodine (eq50), which in turn required 1.0×10^-5 moles of iodide to produce (eq49). This means that we are measuring the time for a constant amount of 5.0×10^-6 moles of iodine to form in each experiment. Hence, we can rewrite eq48 as:

$\frac{d[I_2]}{dt}=\frac{\frac{n_{I_{2,f}}}{V}-\frac{n_{I_2,i}}{V}}{t_f-t_i}=k[H_2O_2]^x[H^+]^y[I^-]^z$

Since $n_{I_2,f}=5.0\times10^{-6}\: moles$ , $n_{I_2,i}=0$ , $V\approx0.05\, dm^3$ , and $t_i=0$

$\frac{\frac{5.0\times10^{-6}}{0.05}}{t_f}=k[H_2O_2]^x[H^+]^y[I^-]^z\; \; \; \; \; \; \; \; 51$

Comparing eq51 and eq48, $rate\propto\frac{1}{t_f}$ . Let’s assume the experiment is completed with the different times $t_f$ recorded in the table below:

Experiment	Flask 1			Flask 2			Time (t_f)/s	(1/t_f)/10^-2 s^-1
Experiment	0.040M H₂O₂/cm³	0.05M H₂SO₄/cm³	Starch/ drops	0.010M KI /cm³	0.001M Na₂S₂O₃/cm³	H₂O/cm³	Time (t_f)/s	(1/t_f)/10^-2 s^-1
1	10	10	3	10	10	10	21.5	4.65
2	10	10	3	20	10	–	11.2	8.93
3	20	10	3	10	10	–	10.4	9.62
4	10	20	3	10	10	–	22.3	9.18

Comparing experiments 1 and 2, doubling the concentration of KI increases the reaction’s initial rate by $\frac{8.93}{4.65}=1.92$ or approximately two fold. From experiments 1 and 3, doubling the concentration of H₂O₂ increases the reaction’s initial rate by $\frac{9.62}{4.65}=2.07$ or approximately two fold. With reference to experiments 1 and 4, doubling the concentration of the acid increases the initial rate of the reaction by $\frac{9.17}{4.65}=1.97$ or approximately two fold. Notice that the factor, $\frac{5.0\times10^{-6}}{0.05}$ , disappears when we divide $\frac{\frac{5.0\times10^{-6}}{0.05}}{t_f}$ from different experiments to compare the initial reaction rates. Therefore, we use $\frac{1}{t_f}$ instead of $\frac{\frac{5.0\times10^{-6}}{0.05}}{t_f}$ to represent the initial reaction rates and conclude that the rate law is first order with respect to each reactant: