Iodine-clock runs with a fixed thiosulphate delay
In a kinetics practical, 10.0 mL of 0.20 M potassium persulphate, 10.0 mL of 0.20 M potassium iodide, 5.0 mL of 0.010 M sodium thiosulphate and 25.0 mL of water are mixed at 25 °C. The time until the starch turns blue is recorded as 45 s. In run 2, 20.0 mL of persulphate and 10.0 mL of iodide are used; in run 3, 10.0 mL of persulphate and 20.0 mL of iodide are used. The total volume is made up to 50.0 mL in every case and the thiosulphate volume is unchanged. Times are 22 s and 23 s respectively. The teacher says the blue colour appears only after all thiosulphate has been used up, so the amount of reaction occurring before the colour change is fixed. The group must decide the order with respect to each reactant, calculate a rate constant, and predict the time for a run using 15.0 mL of each 0.20 M reagent.
1. Explain why 1/t can be used as a relative rate here, calculate the concentrations of persulphate and iodide after mixing in each run, and determine the order with respect to each reactant.
Read the worked answer
The thiosulphate removes iodine as it is formed:
[ \mathrm{I_2+2S_2O_3^{2-}\rightarrow2I^-+S_4O_6^{2-}} ]
Moles of thiosulphate used = ((0.0050\ \mathrm{L})(0.010\ \mathrm{M})=5.0\times10^{-5}) mol. This consumes (2.5\times10^{-5}) mol of I2, which corresponds to (2.5\times10^{-5}) mol of persulphate consumed. In 50.0 mL, the fixed concentration change is:
[ \Delta[\mathrm{S_2O_8^{2-}}]=\frac{2.5\times10^{-5}}{0.0500}=5.0\times10^{-4}\ \mathrm{M} ]
Because the same amount of reaction occurs before the colour change, average rate = (\Delta[\mathrm{S_2O_8^{2-}}]/t), so (1/t) is proportional to rate.
After mixing:
- Run 1: ([\mathrm{S_2O_8^{2-}}]=0.040\ \mathrm{M}), ([\mathrm{I^-}]=0.040\ \mathrm{M}).
- Run 2: ([\mathrm{S_2O_8^{2-}}]=0.080\ \mathrm{M}), ([\mathrm{I^-}]=0.040\ \mathrm{M}).
- Run 3: ([\mathrm{S_2O_8^{2-}}]=0.040\ \mathrm{M}), ([\mathrm{I^-}]=0.080\ \mathrm{M}).
Doubling persulphate changes the time from 45 s to 22 s, so rate ratio = 45/22 = 2.05 ≈ 2; order in persulphate is 1. Doubling iodide changes the time to 23 s, ratio = 45/23 = 1.96 ≈ 2; order in iodide is 1. Overall order = 2.
2. Calculate the rate constant for each run and the mean value. Then predict the time for a run using 15.0 mL of each 0.20 M reagent, with the same total volume and thiosulphate amount.
Read the worked answer
Use rate = (k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]) and rate = ((5.0\times10^{-4})/t).
Run 1:
[ \text{rate}=\frac{5.0\times10^{-4}}{45}=1.11\times10^{-5}\ \mathrm{M\ s^{-1}} ]
[ k=\frac{1.11\times10^{-5}}{(0.040)(0.040)}=6.9\times10^{-3}\ \mathrm{M^{-1}s^{-1}} ]
Run 2:
[ \text{rate}=\frac{5.0\times10^{-4}}{22}=2.27\times10^{-5} ]
[ k=\frac{2.27\times10^{-5}}{(0.080)(0.040)}=7.1\times10^{-3}\ \mathrm{M^{-1}s^{-1}} ]
Run 3:
[ \text{rate}=\frac{5.0\times10^{-4}}{23}=2.17\times10^{-5} ]
[ k=\frac{2.17\times10^{-5}}{(0.040)(0.080)}=6.8\times10^{-3}\ \mathrm{M^{-1}s^{-1}} ]
Mean (k\approx6.9\times10^{-3}\ \mathrm{M^{-1}s^{-1}}).
For 15.0 mL of each reagent, total volume still 50.0 mL:
[ [\mathrm{S_2O_8^{2-}}]=[\mathrm{I^-}] =\frac{(15.0)(0.20)}{50.0}=0.060\ \mathrm{M} ]
[ \text{rate}=(6.9\times10^{-3})(0.060)(0.060) =2.5\times10^{-5}\ \mathrm{M\ s^{-1}} ]
[ t=\frac{5.0\times10^{-4}}{2.5\times10^{-5}}=20\ \mathrm{s} ]
The prediction is about 20 s, close to the trend of the earlier runs.
3. A repeat of run 1 at 35 °C gives 22 s. Estimate the temperature coefficient and explain the increase using collision theory. Why is it unsafe to compare runs if the starch was added at different times?
Read the worked answer
Rate ratio for a 10 K rise:
[ \frac{\text{rate at 35 °C}}{\text{rate at 25 °C}} =\frac{45}{22}\approx2.0 ]
The temperature coefficient is about 2 per 10 K. According to collision theory, raising the temperature increases the average kinetic energy of the molecules, so a larger fraction of collisions has energy equal to or greater than the activation energy. Collisions also become more frequent, but the increase in the fraction of effective collisions is usually the more important factor. The Arrhenius equation (k=Ae^{-E_a/RT}) expresses this quantitatively.
Starch is only an indicator; it does not change the reaction rate, but its timing affects when the blue colour is seen. If starch is added late, iodine may already have accumulated and the colour can appear immediately or after an irreproducible delay. If starch is added at the start in one run and near the end in another, the measured times are not comparable even though the chemistry is the same. The starch should be added in the same way in every run, and the tubes should be kept in the same thermostat bath.