a student investigated the effect of changing reactant concentrations on the rat
ID: 916322 • Letter: A
Question
a student investigated the effect of changing reactant concentrations on the rate of the reaction:
H2O2 (aq) + 2 H+ (aq) + 2 I– (aq) ¾® I2 (aq) + 2 H2O (aq).
The student made up four reaction mixtures with different initial concentrations of H2O2, 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 ´ 10–4 M of I2. (That is, [I2] = 8.33 ´ 10–4 M.)
Mixture
[H2O2]
[H+]
[I–]
time
1
5.6 ´ 10–2 M
1.8 ´ 10–5 M
8.3 ´ 10–3 M
185 s
2
2.8 ´ 10–2 M
1.8 ´ 10–5 M
8.3 ´ 10–3 M
389 s
3
5.6 ´ 10–2 M
1.8 ´ 10–5 M
1.67 ´ 10–2 M
99 s
4
5.6 ´ 10–2 M
1.8 ´ 10–4 M
8.3 ´ 10–3 M
205 s
(a) For each of the four trials shown in the table, calculate the reaction rate, delta [I2]/delta t
(b) 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:
Rate = delta [I2]/delta t = k [H2O2]x [H+]y [I–]z.
(c) For each of the four trials, calculate the rate constant, k.
Mixture
[H2O2]
[H+]
[I–]
time
1
5.6 ´ 10–2 M
1.8 ´ 10–5 M
8.3 ´ 10–3 M
185 s
2
2.8 ´ 10–2 M
1.8 ´ 10–5 M
8.3 ´ 10–3 M
389 s
3
5.6 ´ 10–2 M
1.8 ´ 10–5 M
1.67 ´ 10–2 M
99 s
4
5.6 ´ 10–2 M
1.8 ´ 10–4 M
8.3 ´ 10–3 M
205 s
Explanation / Answer
a) reaction rate = dI2/ dt
mixture 1 :
reaction rate = 8.3 x 10-4 / 185 = 4.486 x 10-6
mixture 2 :
reaction ratate = 8.3 x 10-4 / 389 = 2.133 x 10-6
mixture 3:
reaction rate = 8.3 x 10-4 / 99 = 8.38 x 10-6
mixture 4 :
reaction rate = 8.3 x 10-4 / 205 = 4.049 x 10-6
b)
now
rate = k [H202] ^x [H+]^y [ I-]^z
now
for mixture 1 and 2 , [H+] and [I-] are constant
so
rate2 / rate 1 = ( [ H202]2 / [H202] 1 ) ^x
2.133 x 10-6 / 4.486 x 10-6 = ( 2.8 x 10-2 / 5.6 x 10-2 )^ x
x =1
so
first order with respect to H202
now
for mixture 1 and 4 , [H202] and [ I-] are constant
so
rate 4 / rate 1 = ( [H+] 4 / [H+] 1 ) ^y
4.049 x 10-6 / 4.486 x 10-6 = ( 1.8 x 10-4 / 1.8 x 10-5 ) ^ y
y = 0
so
zero order with respect to [H+]
now
for mixture 1 and 3 , [H202] and [H+] are constant
so
rate 3 ./ rate 1 = ( [I-]3 / [I-] 1 ) ^ z
8.38 x 10-6 / 4.486 x 10-6 = ( 1.67 x 10-2 / 8.3 x 10-3 )^z
z = 1
so
first order with respect to [I-]
so
rate = k [ H202] [I-]
c)
now
for mixture 1
4.486 x 10-6 = k x 5.6 x 10-2 x 8.3 x 10-3
k = 9.65 x 10-3
so
the rate constant k is 9.65 x 10-3
mixture 2 :
2.133 x 10-6 = k x 2.8 x 10-2 x 8.3 x 10-3
k = 9.20 x 10-3
mixture 3 :
8.38 x 10-6 = k x 5.6 x 10-2 x 1.67 x 10-2
k= 8.96 x 10-3
mixture 4 :
4.049 x 10-6 = k x 5.6 x 10-2 x 8.3 x 10-3
k = 8.7 x 10-3
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