Question

2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

Answer 1

The rate law is rate = k[ClO2]^2[OH-], with k ≈ 22.2222 M^-2s^-1. The calculated rate with [ClO2] = 0.100 M and [OH-] = 0.050 M is approximately 0.111 M/s.

Step-by-step explanation:

To determine the rate law for the reaction given, we should look at the changes in concentration and the effect on the initial rate.

Comparing experiment 1 and 2:

• When [ClO2] is changed from 0.060 M to 0.020 M (decreased by a factor of 3), the initial rate goes from 0.0248 M/s to 0.00276 M/s (decreased by a factor of ~9).
• This suggests that the rate is proportional to [ClO2] squared ([ClO2]^2).

Comparing experiment 2 and 3:

• When [OH-] is increased from 0.030 M to 0.090 M (tripled), the rate increases from 0.00276 M/s to 0.00828 M/s (tripled).
• This indicates the rate is first order with respect to [OH-].

Therefore, the rate law is: rate = k[ClO2]^2[OH-].

Using experiment 1 data to calculate the rate constant (k):

• 0.0248 M/s = k(0.060^2)(0.030)
• k = 0.0248 / (0.060^2 * 0.030)
• k ≈ 22.2222 M-2s-1

To calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M:

• rate = 22.2222 M-2s-1 * (0.100)^2 * 0.050
• rate = 0.111 M/s

Answer 2

Step-by-step explanation:

2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l)

The data is given as;

Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s)

1 0.060 0.030 0.0248

2 0.020 0.030 0.00276

3 0.020 0.090 0.00828

a) Rate law is given as;

Rate = k [ClO2]^x [OH-]^y

From experiments 2 and 3, tripling the concentration of [OH-] also triples the rate of the reaction. This means the reaction is first order with respect to [OH-]

From experiments 1 and 2, when the [ClO2] decreases by a factor of 3, the rate decreases by a factor of 9. This means the reaction is second order with respect to [ClO2]

Rate = k [ClO2]² [OH-]

b. Calculate the value of the rate constant with the proper units.

Taking experiment 1,

0.0248 = k (0.060)²(0.030)

k = 0.0248 / 0.000108

k = 229.63 M-2 s-1

c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.

Rate = 229.63 [ClO2]² [OH-]

Rate = 229.63 (0.100)²(0.050)

Rate = 0.1148 M/s