Question

Lab Reaction Rate and order of a Chemical Reaction

H₃AsO₃ (aq) + I₃^- (aq) + H₂O (l) → HAsO₄^2- (aq) + 3 I^- (aq) + 4H⁺ (aq)

The data has been collected and pre-lab completed. Need help with the results.

Results

            Laboratory temperature: 25°C

DATA TABLE 1:

Experiment	[IO3]0	[H+]0	[H3AsO3]0	[I]0
1	.005	.00001	.0015	.05
2	.01	.00001	.0015	.05
3	.005	.00001	.0015	.1
4	.005	.00002	.0015	.05

DATA TABLE 2:

Experiment	Δt (s)	Δ[IO3-]	Rate
1	74
2	37.3
3	18.7
4	18.0
5 Temp: 35°C	30
6 Temp: 45°C	13.7

Questions

- Calculate Δ[IO₃^-] = (1/3)[H₃AsO₃]₀ and record in Data Table 2. (Show calculation)

- Calculate the rates and record in Data Table 2 (Show calculations):

Rate=∆[IO3-]∆t

- Obtain the reaction orders with respect to IO₃^-, I^-, and H⁺ by comparing the rates as described in the lab discussion. Show the rates you are comparing.

- Write the rate law for this reaction with the correct values for a, b, and c.

- Calculate the rate constant, k, for Experiments 1, 2, 3, and 4, using the above rate law and the rates and initial concentrations calculated in Data Table 1. (Show calculations)

Experiment 1:

Experiment 2:

Experiment 3:

Experiment 4:

- How does the calculated value for k compare for the first four experiments, all at 25°C?

2) Calculate the rate constant, k, at the higher temperatures used in Experiment 5 and 6. Remember that you used the same initial concentrations as in Experiment 1. (Show calculations)

Experiment 5:_______

Experiment 6:_______

How does k change as the temperature increases?

Explain why the value for k at higher temperatures is different from the experiments at the lower temperature.

Answer 1

In all the experiment the concentration of H3AsO3 remain constant = 0.0015 M

{IO3-] = 1/3 [H3AsO3]

therefore [IO3] = 0.0015/3 = 0.0005 M in all the experiments.

From the given time taken for the experiment, the rate = 0.0005/t is calculated and are provided in the following table.

Let the rate law be [rate] = k [ IO3-]^a [H+]^b [I-]^c

From experiment (1)

                         6.7568 x 10^-6 = k [0.005]^a [0.00001]^b [0.05]^c ----------------- (1)

similarly from experiment (2), (3) and (4) we have

13.4048 x 10^-6 = k [0.01]^a [0.00001]^b [0.05]^c ----------------- (2)

26.7380 x 10^-6 = k [0.005]^a [0.00001]^b [0.1]^c ----------------- (3)

27.7778 x 10^-6 = k [0.005]^a [0.00002]^b [0.05]^c ----------------- (4)

Divide (1) / (2) we have

     6.7568 / 13.4048 = [1/2]^a

         0.5040 = 0.5^a

                        a = log 0.5040 / log 0.5 = -0.2976 / -0.3010 = 0.99 = 1

Thus a = 1

Divide (1) / (3) we have

   6.7568 / 26.7380 = [1/2]^c

        0.2527 = 0.5^c

          c = log 0.2527 / log 0.5 = -0.5974 / -0.3010 = 1.98 = 2

Thus c = 2

Divide (1) / (4) we have

6.7568 / 27.7778 = [1/2]^b

0.2432 = 0.5^b

b = log 0.2432 / log 0.5 = -0.614 / -0.3010 = 2.04 = 2

Thus b = 2

Thus the order of the reaction w.r.t IO3- is 1

   the order of the reaction w.r.t H+ is 2

         the order of the reaction w.r.t I- is 2

Hence the rate law is   [rate] = k [IO3-] [ H+]^2 [ I-]^2

Since we know the concentration and rate from the experiments, we can calculate the rate constant k.

k = [rate] / [IO3-] [H+]^2 [ I-]^2

For expt 1, k = 6.7568 x 10^-6 / [0.005] [ 0.00001}^2 [0.05]^2

                      = 5.4 x 10^9

similarly k is calculated for all other experiments and are given below

The rate constant is almost a constant in the first four experiments conducted at 25 deg C

The rate constant increases with increase in temperature as evidenced from experiment 5 and 6

Because increase in temperature increases the average kinetic motion of the molecules and thus increasing the collision frequency.

This increases the number of effective collisions and hence the rate increases with temperature.