Iodine clock experiment (chemical kinetics)

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
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 (tf)/s (1/tf)/10-2 s-1
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 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:

rate=k[H_2O_2][H^+][I^-]

 

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