Question

An iodine-clock kinetics experiment was performed by varying the composition of reactant solutions and measuring the resulting rates of reaction.

	Initial Rate Experiment 1	Initial Rate Experiment 2	Initial Rate Experiment 3
Test tube contents	10.0 mL 0.10 M NaI(aq) 10.0 mL 0.10 M Na2S2O8(aq) 5.0 mL 0.0050 M Na2S2O3(aq) 1.0 mL starch solution	10.0 mL 0.10 M NaI(aq) 5.0 mL 0.10 M Na2S2O8(aq) 5.0 mL 0.10 M Na2SO4(aq) 5.0 mL 0.0050 M Na2S2O3(aq) 1.0 mL starch solution	5.0 mL 0.10 M NaI(aq) 10.0 mL 0.10 M Na2S2O8(aq) 5.0 mL 0.10 M NaCl(aq) 5.0 mL 0.0050 M Na2S2O3(aq) 1.0 mL starch solution
Time between last addition and first appearance of colour change	32 second	64 seconds	64 seconds

Based on these results, what is the rate law for the reaction of iodide anion and persulfate anion?

Question 4 options:

	Rate = k
	Rate = k[I-]
	Rate = k[I-]2
	Rate = k[I-][S2O82-]
	Rate = k[I-]2[S2O82-]
	Rate = k[I-][S2O82-]2
	Rate = k[I-]2[S2O82-]2
	Rate = k[S2O82-]2
	Rate = k[S2O82-]

Answer 1

In experiment 2, 10 mL 0.1 M NaI and 5 mL0.1 M Na₂S₂O₈ used and time is 64 seconds.

In experiment 1 10 mL 0.1 M NaI and 10 mL 0.1 M Na₂S₂O₈ used and time is 32 seconds.

So, when concentration of Na₂S₂O₈ is doubled rate is also doubled.

So,rate is 1 times that of [S₂O₈^2-]

Similarly in experiment 3, 5 mL 0.1 M NaI and 10 mL 0.1 M Na₂S₂O₈ used.

With respect to experiment 3,in experiment 1 only NaI amount is doubled and rate also doubled.

Rate is 1 times that of [I^-]

Rate=k[I^-][S₂O₈^2-]