5 Steps to a 5 AP Chemistry

 Answers and Explanations

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

is done first. If you want to begin with ClO 2 , you must pick two experiments from the

table where its concentration changes but the OH−concentration 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 (only the OH−concentra-

tion changes), we see that doubling the concentration simply doubles the rate. Thus,

the order for OH−is 1 (=n). Give yourself 1 point for each order you got correct.

ii. Inserting the orders into the generic rate law gives: Rate =k[ClO 2 ]^2 [OH−]^1 ,which is usu-

ally simplified to: Rate =k[ClO 2 ]^2 [OH−]. Give yourself 1 point if you got this 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 appro-

priate values are entered into the equation. Using experiment 1 as an example:

k=(0.166 mol/L min)/[(0.020 M)^2 (0.030 M)]

=1.3833 × 104 M/M^3 min =1.4 × 104 /M^2 min

The 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

correctunits.

c. The coefficients from the equation say that for every mole of ClO 3 −that forms, 2 mol

of ClO 2 reacted. 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 nega-

tive rate. This gives:

Rearranging and inserting the rate from experiment 1 gives:

Δ[ClO]/Δt=−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.

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:

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

For steps 2 and 3, the intermediates must be replaced with reactants. Step 2 gives a rate

law matching the one derived in part a. Give yourself 1 point if you picked step 2, or if you

picked a step with a rate law that matches a wrong answer for part a. Give yourself 1 more

point if you explained the substitution of reactants for intermediates.

To see if the stoichiometry is correct, simply add the three steps together and cancel the

intermediates (materials that appear on both sides of the reaction arrow).

−=−

1

2 23

ΔΔΔΔ[] []ClO /tt ClO /

Kinetics  209