Kinetics ❮ 209❯ Answer and Explanation
(a) i. This part of the problem begins with a generic rate equation: Rate = k[ClO 2 ]m [OH–]n. The
values of the exponents, the orders, must be determined. It does not matter which
exponent is done first. If you want to begin with ClO 2 , you must pick two experi-
ments from the table in which the concentration of ClO 2 changes but the OH– con-
centration does not change. These are experiments 1 and 3. Experiment 3 has twice
the concentration of ClO 2 as experiment 1. This doubling of the ClO 2 concentration
has quadrupled the rate. The relationship between the concentration (× 2) and the
rate (× 4 = × 22 ) indicates that the order for ClO 2 is 2 (= m). Using experiments 1
and 2 (in which only the OH– concentration changes), we see that doubling the con-
centration simply doubles the rate. Thus, the order for OH– is 1 (= n).
Give yourself 1 point for each order you got correct, for a maximum of 2 points.ii. Inserting the orders into the generic rate law gives Rate = k[ClO 2 ]^2 [OH–]^1 , which
is usually simplified to Rate = k[ClO 2 ]^2 [OH–].
Give yourself 1 point if you got this equation correct.
(b) Any one of the three experiments may be used to calculate the rate constant. If the
problem asked for an average rate constant, you would need to calculate a value for
each of the experiments and then average the values.
The rate law should be rearranged to k =
Rate
[ClO 2 ] ^2 [OH]−. Then, the appropriate
values are entered into the equation. Using experiment 1 as an example,k =
Rate
[ClO 2 ] ^2 [OH]−=
0.166 mol/L min
[0.020 ][0.030^2 ]⋅
MM
= 1.3833 × 104 M/M^3 min = 1.4 × 104 /M^2 minThe answer could also be reported as 1.4 × 104 L^2 /mol^2 min. You should not forget
that M = mol/L.
Give yourself 1 point for the correct numerical value. Give yourself 1 point for the
correct units. If you had the wrong rate law in part a. ii, but use it correctly in part b,
you will still get the points.
(c) The coefficients from the equation say that for every mole of ClO 3 – that forms,
2 moles of ClO 2 react. Thus, the rate of ClO 2 is twice the rate of ClO 3 –. Do not forget
that since ClO 3 – is forming, it has a positive rate, and since ClO 2 is reacting, it has a
negative rate. Rearranging and inserting the rate from experiment 1 gives t
∆
∆[ClO 2 ]
=
–2(0.166 mol/L min)] = –8.332 mol/L min.
Give yourself 2 points if you got the entire answer correct. You get only 1 point if
the sign or units are missing or incorrect.
(d) The rate-determining step must match the rate law. One approach is to determine the
rate law for each step in the mechanism. This gives the following:
Step 1: Rate = k[ClO 2 ]^2
Step 2: Rate = k[Cl 2 O 4 ] [OH–] = k[Cl 2 O]^2 [OH–]
Step 3: Rate = k[HClO 2 ] [OH–] = k[ClO 2 ] [OH–]^2