Question

4.         Consider the reaction:

(NH₄)₂S₂O₈ + 3KI → (NH₄)₂SO₄ + K₂SO₄ + KI₃

            Some data for this reaction follows:

Experiment	[(NH4)2S2O8] (M)	[KI] (M)	Initial Rate of Appearance of KI3 (M/s)
1	0.200	0.100	4.76 × 10-3
2	0.050	0.200	2.38 × 10-3
3	0.200	0.200	9.52 × 10-3

            a.         Write the rate law for this reaction.

            b.         Calculate the rate constant.

            c.         At what rate does KI disappear if the initial concentrations are [KI] = 0.0450 M and [(NH₄)₂S₂O₈] = 0.120 M?

            d.         At what rate does (NH₄)₂S₂O₈ disappear under the conditions given in part c?

Answer 1

writing rate law for given reaction

r = k [(NH4)2S2O8]^a [KI]^b

from the given values trial 1 & 3 we can see when the concentration of KI doubles all else remaining same rate increases by times 2¹

hence b = 1

from the given values trial 1 & 2 we can see when the concentration of KI doubles and (NH4)2S2O8 decreases by 4 times rate is decreased to (2)¹(1/4)² = 0.125 times rate in trial1

hence a = 2

r = k [(NH4)2S2O8]² [KI]¹ Answer (a)

sub values of r , [(NH4)2S2O8] , [KI]

k = 0.238 M^-1 s^-1 Answer (b)

(c) given [(NH4)2S2O8] = 0.120 M, [KI] = 0.0450 M sub these values in rate equation

r = k [(NH4)2S2O8]² [KI]¹   

r = 0.238 * ( 0.120)² * (0.0450) = 1.54 * 10^-4  M/s this is the rate of formation of KI3

rate of disappearence of KI = 3 * rate of formation of KI3 = 3 * 1.54 * 10^-4  M/s = 4.62 * 10^-4  M/s   Answer (c)

(d) rate of disappearence of [(NH4)2S2O8] = rate of formation of KI3 = 1.54 * 10^-4  M/s    Answer (d)