Question

Exp No.	[S2O82-]	[I-]	Intial Rate of I2, M/min
1	1.23 x 10-4	1.25 x 10-2	9.99 x 10-7
2	2.50 x 10-4	1.25 x 10-2	2.03 x 10-6
3	2.50 x 10-4	2.40 x 10-2	3.60 x 10-6

The persulfate and iodide are mixed in the following quantities at 25^oC:

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

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

3.Write the rate law for the reaction.

4. Calculate K for the reaction

5. What do you expect the rate of the reaction to be if the temperature was increased to 35 degree Celcius.

*Can you explain the step for me I dont understand the questions.

Answer 1

rate= k [S₂O₈^2-]^x[I^-]^y

from data

9.99 x 10^-7=k [1.23 x 10^-4]^x [1.25 x 10^-2]^y ------------->(1)

2.03 x 10^-6=k [2.50 x 10^-4]^x [1.25 x 10^-2]^y ------------->(2)

3.60 x 10^-6=k [2.50 x 10^-4]^x [2.40 x 10^-2]^y ------------->(3)

by solving 1 and 2

x= 1

by solving 2 and 3

y = 1

so rate= k [S₂O₈^2-]¹[I^-]¹

1. order of reaction with respect to the persulfate ion   =1

2. order of the reaction with respect to the iodide ion =1

3.rate law for the reaction. = rate= k [S₂O₈^2-]¹[I^-]¹

4. K for the reaction

9.99 x 10^-7=k [1.23 x 10^-4]¹ [1.25 x 10^-2]¹

k = 0.6498 M-1 s-1

5. What do you expect the rate of the reaction to be if the temperature was increased to 35 degree Celcius.

rate becomes 2 times.

for every increase in 10 oC temperature rate becomes 2 times