Using the same procedure that you will use in this experiment, a student investi
ID: 1053258 • Letter: U
Question
Using the same procedure that you will use in this experiment, a student investigated the effect of changing reactant concentrations on the rate of the reaction: H_2 O_2_(aq) + 2 H^+_(aq) + 2 I^-_(aq) rightarrow I_2_(aq) + 2 H_2 O_(aq). The student made up four reaction mixtures with different initial concentrations of H_2 O_2, H^+, and I^-, which are listed in the table below. The student measured the time it took for each reaction to produce the first 8.33 times 10^-4 M of I_2. (That is, delta [I_2] = 8.33 times 10^-4 M.) For each of the four trials shown in the table, calculate the reaction rate, delta [I_2]/delta time. Determine the order of the reaction with respect to each of the three reactants. That is, determine the integer values of x, y, and z in the rate law:Explanation / Answer
a) For mixture 1 :-
rate = 8.33 x 10-4 M / 185 s = 4.50 x 10-6 M s-1
For mixture 2 :-
rate = 8.33 x 10-4 M / 389 s = 2.14 x 10-6 M s-1
For mixture 3 :-
rate = 8.33 x 10-4 M / 99 s = 8.41 x 10-6 M s-1
For mixture 4 :-
rate = 8.33 x 10-4 M / 205 s = 4.06 x 10-6 M s-1
b) The rate can be given as :-
Rate = [H2O2]a[H+]b[I-]c
For mixture 1 , we have :-
4.50 x 10-6Ms-1 = [5.60 x 10-2]a[1.8 x 10-5]b[8.3 x 10-3]c
for mixture 2 ,w have :-
2.14 x 10-6Ms-1 = [2.8 x 10-2]a[1.8 x 10-5]b[8.3 x 10-3]c
Dividing for mixture 1 by mixture 2 ,we have :-
4.50 / 2.14 = [2]a
2.10= [2]a
a= 1.07
Taking reaction mixture 3 ,We have :-
8.41 x 10-6Ms-1 = [5.60 x 10-2]a[1.8 x 10-5]b[1.67 x 10-2]c
for reaction mixture 4 we have :-
4.06 x 10-6Ms-1 = [5.60 x 10-2]a[1.8 x 10-5]b[8.3 x 10-3]c
dividing mixture 3 by mixture 2 we have ,
8.41 / 2.14 = [16.7/8.3]c[2]1.07
1.88 = [2.01]c
c = 0.90
Divide mixture 3 by mixture 4 we have
8.41/4.06 = [0.1]b[2.01]0.9
1.11 = [0.1]b
b = -0.04
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