Consider the reaction, S2O82- + 3 I- à 2 SO42- + I3-. Data obtained in measuring rate of formation of I3- are listed in the table.
Experiment |
[S2O82-], M |
[I‑], M |
Initial rate, Ms-1 |
1 |
0.035 |
0.055 |
1.5 x 10-5 |
2 |
0.070 |
0.055 |
3.0 x 10-5 |
3 |
0.070 |
0.110 |
6.0 x 10-5 |
Solution :-
Suppose, the order with respect to S2O82- is p and with respect to I- is q. The rate law can thus be written as: Rate = k [S2O8]p [I-]q
Substituting the values of rate, [S2O8] and [I-] for various sets of experiments, we have
1.5 × 10-5 = k [0.035]p [0.055]q ............….(i)
3.0 × 10-5 = k [0.070]p [0.055]q ................(ii)
6.0 × 10-5 = k [0.070]p [0.110]q ...............(iii)
Dividing eq. (ii) by eq. (i) , we get,
(3.0 × 10-5) / (1.5 × 10-5) = (0.070)p / (0.035)p
2 = 2p
p = 1
Dividing eq. (iii) by eq. (ii), we get,
(6.0 × 10-5) / (3.0 × 10-5) = (0.110)q / (0.055)q
2 = 2q
q = 1
1) Thus, the order of reaction with respect to S2O82- is = 1
2) The order of reaction with respect to l- is = 1
3) Overall rate of reaction = (p + q) = (1 + 1) = 2
4) Rate Law ,
Rate = k [S2O82-]1 [I-]1
5) Putting values of order of reactants in equation 1, we get,
1.5 × 10-5 = k [0.035]1 [0.055]1
or, k (Rate constant) = (1.5 × 10-5) / (0.035 × 0.055) = 7.8 × 10-3 M-1 s-1
6) If [S2O82-] = 0.030 M, and [I-] = 0.050 M
k = 7.8 × 10-3 M-1 s-1
Rate =?
Rate = (7.8 × 10-3) × (0.030)× (0.050)
Rate = 1.17 × 10-5 M s-1
7) Rate of formation of I3- = 1/3 × Rate of disappearance of I-
or , Rate of disappearance of I- = 3 × Rate of formation of I3-
or, Rate of disappearance of I- = 3 × (3.0 × 10-5) = 9.0 × 10-5 M/s
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Consider the reaction, S2O82- + 3 I- à 2 SO42- + I3-. Data obtained in measuring rate of formation of I3- are...
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