18.3 Hydrogen peroxide and the iodide ion react in acidic solution as follows: H
ID: 1045110 • Letter: 1
Question
18.3 Hydrogen peroxide and the iodide ion react in acidic solution as follows: HO (aq) + 31(aq) + raq) I,(aq) + 2H,O(1) The kinetics of this reaction were studied by following the concentration of H,O and constructing plots of In[H,O,] vs. time. All plots were linear and all solutions had [H,O].-8.0x10 mole/L. The slopes of these straight lines depended on the initial concentrations of I and H' as follows: [l1. (mole/L) [H'1. (mole/L) Slope (min) 0.1000 0.3000 0.4000 0.0750 0.0750 0.0750 0.0400 0.0400 0.0400 0.0200 0.0800 0.1600 -0.120 -0.360 -0.480 -0.0760 0.118 -0.174 The rate law for this reaction has the form b. Calculate the values of the rate constants k, and k C. a. Specify the orders of the reaction with respect to [I'] and H,0,] c. What reason could there be for the two-term dependence of the rate on [H']?Explanation / Answer
a, b)
Rate = (k1 + k2[H+])[I-]m[H2O2]n
S No.
[H2O2]o
[I-]o
[H+]o
slope
1
8E-04
0.1
0.04
-0.12
2
8E-04
0.3
0.04
-0.36
3
8E-04
0.4
0.04
-0.48
4
8E-04
0.75
0.02
-0.076
5
8E-04
0.75
0.08
-0.118
6
8E-04
0.75
0.16
-0.174
Rate = -d[H2O2]/dt
Ln(Rate) = -d(ln([H2O2]))/dt = - slope
Since ln([H2O2]) vs time is linear, n = 1 (first order with respect to H2O2)
Initial Rate1 = (k1 + k2 * 0.04) * 0.1m [H2O2]n
Ln(Rate1) = m ln(0.1) + c = -slope = 0.12
Initial Rate2 = (k1 + k2 * 0.04) * 0.3m [H2O2]n
Ln(Rate2) = m ln(0.3) + c = -slope = 0.36
Subtracting we get,
0.36 – 0.12 = 0.24 = m *( ln(0.3) – ln(0.1))
m = 0.22
Rate4 = (k1 + k2 * 0.02) * 0.75m [H2O2]n
Rate5 = (k1 + k2 * 0.08) * 0.75m [H2O2]n
Ln(Rate4) = ln((k1 + k2 * 0.02)) + c1 = -slope = 0.076
Ln(Rate5) = ln((k1 + k2 * 0.08)) + c1 = -slope = 0.118
Subtracting we get,
0.118 – 0.076 = 0.042 = ln((k1 + k2 * 0.08) - ln((k1 + k2 * 0.02)
(k1 + k2 * 0.08)/(k1 + k2 * 0.02) = 1.043
Solving we get,
k1= 1.379 k2
Ln(Rate6) = ln((k1 + k2 * 0.08)) + m ln(0.75) + n ln(8 x 10-4) = -slope = 0.174
ln((1.379 k2 + k2 * 0.08)) + n ln(8 x 10-4) = 0.237
ln((1.459 k2) + n ln(8 x 10-4) = 0.237
ln(k2) -7.13 n = - 0.14
ln(k2) = 6.99
k2 = 1085.88
k1 = 1.379 k2 = 1497.19
c)
The decomposition of hydrogen peroxide catalyzed by iodide ion proceeds by a two-step mechanism.
Therefore the combined rate law can be written as
Rate = (k1 + k2[H+])[I-]m[H2O2]n
S No.
[H2O2]o
[I-]o
[H+]o
slope
1
8E-04
0.1
0.04
-0.12
2
8E-04
0.3
0.04
-0.36
3
8E-04
0.4
0.04
-0.48
4
8E-04
0.75
0.02
-0.076
5
8E-04
0.75
0.08
-0.118
6
8E-04
0.75
0.16
-0.174
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