The reaction
H 2 SeO 3 (aq) + 6I - (aq) + 4H + (aq) $\rightarrow$ 2I 3 - (aq) + 3H 2 O(l) + Se(s)
was studied at 0°C by the method of initial rates:
-What is the numerical value of the rate constant?

Question

The reaction
H 2 SeO 3 (aq) + 6I - (aq) + 4H + (aq) $\rightarrow$ 2I 3 - (aq) + 3H 2 O(l) + Se(s)
was studied at 0°C by the method of initial rates: 
-What is the numerical value of the rate constant?

Quizplus · Accepted Answer

The rate law for this reaction is given as Rate = k[H₂SeO₃][I-]^6[H⁺]^4. Using the Method of Initial Rates, the following data was obtained: Experiment [H₂SeO₃] (M) [I-] (M) [H⁺] (M) Initial Rate (M/s) 1 0.2 0.001 0.001 1.20 x 10^-3 2 0.2 0.004 0.001 4.80 x 10^-3 3 0.2 0.001 0.004 4.80 x 10^-3 From the data, it can be seen that when [H₂SeO₃] and [H⁺] are held constant and [I-] is increased by a factor of 4 (from experiment 1 to experiment 2), the rate increases by a factor of 4. When [H₂SeO₃] and [I-] are held constant and [H⁺] is increased by a factor of 4 (from experiment 1 to experiment 3), the rate increases by a factor of 4. Therefore, the overall reaction order is 6+4=10. To find the rate constant, we can use any of the three experiments and substitute the values into the rate law to obtain: k = Rate/([H₂SeO₃][I-]^6[H⁺]^4) Using experiment 1, k = (1.20 x 10^-3)/[(0.2)(0.001)^6(0.001)^4] = 5.25 x 10^4. When rounded to two significant figures, the answer is 5.2 x 10^4 or 5.2 F1F1F1S1F1F1F10 10⁵.

The Reaction H2SeO3(aq) + 6I-(aq) + 4H+(aq) →\rightarrow→ 2I3-(aq) + 3H2O(l) + Se(s) Was Studied at 0°C by 3H2O(l)

The Reaction
H₂SeO₃(aq) + 6I^-(aq) + 4H⁺(aq) $\rightarrow$ 2I₃^-(aq) + 3H₂O(l) + Se(s)
Was Studied at 0°C by 3H₂O(l)