Effect of concentration on the initial rate for the reduction of persulfate ion

ID: 833657 • Letter: E

Question

Effect of concentration on the initial rate for the reduction of persulfate ion with iodide ion.

[S2O8-2],M [I-], M Initial rate ,M/s

0.22 0.21 1.67
0.30 0.42 4.54
0.44 0.21 3.33

0.44 0.42 6.65

Use the rate data to answer the following questions.

The answers are in bold i just dont understand how to get them! PLEASE EXPLAIN!

What is the reaction order for [S2O8-2]? a. Zero b. Third c. First d. Fourth e. Second

What is the reaction order for [I-]?
a. Zero b. First c. Second d. Third e. Fourth

What is the value of the rate constant?
a. 86 M-2s-1 b. 196 M-3s-1 c. 120 M-2s-1 d. 43 M-1s-1 e. 36 M-1s-1

[S2O8-2],M [I-], M Initial rate ,M/s

0.22 0.21 1.67..............trial1
0.30 0.42 4.54....................trial2
0.44 0.21 3.33....................trial3

0.44 0.42 6.65..........................trial 4

if we observe trial 1& trial 3

trial1 ==> rate = K[S2O8-2]^m[I-]^n = 1.67 = K[0.22]^m[0.21]^n..............exp1

trial3 ==> 3.33 = K[0.44]^m[0.21]^n .............................exp2

if we do Exp 1/exp2 = 1.67/3.33 = (0.22/0.44)^m

= (1/2)^1 = (1/2)^m

m = 1 so order w.r.t [S2O8^-2] is 1

for increasing the conc. of [S2O8-2] rate of reaction increases 2 times

so order w.r.t [S2O8-2] is 1

if we observe trial 3& trial 4

similarly do above process to Know the order w.r.t [I-]

for increasing the conc. of [I-] rate of reaction increases 2 times

so order w.r.t [I-] is 1

rate = K2 [S2O8^-2][I-]

1.67 = K2 (0.21)*(0.22)

K2 = 36M^-1 s^-1

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