4. Consider the reaction: (NH 4 ) 2 S 2 O 8 + 3KI (NH 4 ) 2 SO 4 + K 2 SO 4 + KI
ID: 490608 • Letter: 4
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?
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
Explanation / Answer
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)
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