This is a 5 part question: To Get credit all 5 parts must be answered! Consider

Question

This is a 5 part question:

To Get credit all 5 parts must be answered!

Consider the following reaction:

S2O8^-2 (aq) + 2I-(aq)    2SO4^-2(aq) + I2 (aq)

The persulfate and iodide are mixed in the following quantities at 25 C.

                   S2O8^-4                   I-                           Initial Rate of formation of I2 (M/min)  

Ex. #1      1.23 x 10^-4            1.25x10^-2                          9.99 x 10^-7

Ex. #2       2.50x10^-4             1.25x10^-2                        2.03x10^-6

Ex. #3        2.50x10^-4            2.40x 10^-2                         3.60x10^-6

1. What is the order of reaction with respect to persulfate ion?

2. What is the order of reaction with respect to iodide ion?

3. Write the rate law for the reaction.

4. Calculate k for the reaction.

5 What do you expect the rate of reaction to be if the temperature was increased to 35 C?

Explanation / Answer

1. When persulfate changed from 1.23 to 2.5, rate changes in the same ratio

i.e. 2.5 / 1.23 = 2.032

2.03 * 10^-6 / 9.99 * 10^-7 = 2.032

So, directly proportional

So, Order = 1

2.ratio of I- = 2.4 / 1.25 = 1.92

ratio of rate = 3.6 * 10^-6 / 9.9 * 10^-7 = 3.636

Hence ratio of rate = (ratio of I-) ^2

So proportional to square

order = 2

3. R = k.[persulfate].[I-]^2

4. From case (1) put values to get k = 51.98

5.If temp is increased, the activation energy is reduced and hence k increases. So, Rate increases. The value of k doubles as there is an increase in 10 degrees

So, rate approximately doubles.

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