The reaction between selenous acid and the iodide ion in acid solution is
H₂SeO₃(aq) + 6I^-(aq) + 4H⁺(aq) → Se(s) + 2I₃^-(aq) + 3H₂O(l)
The data in the following table were measured at 0°C.
Experiment
[H₂SeO₃]₀ (M)
[H⁺]₀ (M)
[I^-]₀ (M)
Initial Rate [mol/(L ∙ s) ]
1
1) 00 × 10^-4
2) 00 × 10^-2
3) 00 × 10^-2
5) 30 × 10^-7
2
2) 00 × 10^-4
2) 00 × 10^-2
3) 00 × 10^-2
1) 06 × 10^-6
3
3) 00 × 10^-4
4) 00 × 10^-2
3) 00 × 10^-2
6) 36 × 10^-6
4
3) 00 × 10^-4
8) 00 × 10^-2
3) 00 × 10^-2
2) 54 × 10^-5
5
3) 00 × 10^-4
8) 00 × 10^-2
6) 00 × 10^-2
2) 04 × 10^-4
6
2) 00 × 10^-4
2) 00 × 10^-2
6) 00 × 10^-2
8) 48 × 10^-6
What is the rate constant for this reaction?

Question

Quizplus · Accepted Answer

To determine the rate constant, we use the rate law: Rate = k[HSO3H2SeO3][I-]^6. We can calculate the rate constant (k) for each experiment by dividing the initial rate by the product of the concentrations of reactants raised to their respective stoichiometric coefficients. Taking the average of these values gives us the overall rate constant of the reaction. 
Using the data in the table, we get:
Experiment 1: k = (1.00 × 10^-4)/(1 M x 6 M^6) = 2.31 x 10^-14 M^-7 s^-1
Experiment 2: k = (2.00 × 10^-4)/(1 M x 6 M^6) = 4.62 x 10^-14 M^-7 s^-1
Experiment 3: k = (3.00 × 10^-4)/(1 M x 6 M^6) = 6.93 x 10^-14 M^-7 s^-1
Experiment 4: k = (2.54 × 10^-4)/(1 M x 6 M^6) = 5.87 x 10^-14 M^-7 s^-1
Experiment 5: k = (2.04 × 10^-4)/(1 M x 6 M^6) = 4.69 x 10^-14 M^-7 s^-1
Experiment 6: k = (8.48 × 10^-4)/(1 M x 6 M^6) = 1.95 x 10^-13 M^-7 s^-1
Average rate constant = (2.31 x 10^-14 + 4.62 x 10^-14 + 6.93 x 10^-14 + 5.87 x 10^-14 + 4.69 x 10^-14 +1.95 x 10^-13)/6 = 4.89 x 10^-14 M^-7 s^-1 
Thus, the best choice is C.

The Reaction Between Selenous Acid and the Iodide Ion in Acid