AP Chemistry 5.2: Rate Laws

Practice AP Chemistry questions on extracting rate-law exponents from initial-rate data, calculating rate constants with units, and designing controlled rate experiments.

[Maximum number: 3]

Answer the following questions about ascorbic acid (vitamin C).

(a)

Dehydroascorbic acid (DHAsc) can be produced by reacting ascorbic acid with the triiodide

ion, $\mathrm{I}_{3}{ }^{-}$ , as represented by the following equation.

\mathrm{HAsc}+\mathrm{I}_{3}^{-} \rightarrow \mathrm{DHAsc}+3 \mathrm{I}^{-}+2 \mathrm{H}^{+}

The student runs three trials of the reaction with different initial concentrations of HAsc and

$\mathrm{I}_{3}{ }^{-}$ , producing the following data.

[ 3 ]

(i)

The rate law for the reaction is rate $=k[\mathrm{HAsc}]\left[\mathrm{I}_{3}{ }^{-}\right]$ . Explain how the data in the table

support the conclusion that the reaction is first order with respect to [HAsc].

[ 1 ]

C (i) For a correct explanation.

Examples of acceptable responses may include the following:

- $\frac{4.914 \times 10^{-4}}{2.457 \times 10^{-4}}=\left(\frac{0.900}{0.450}\right)^{a}$ , thus a=1

- Comparing trials 1 and 3, the rate doubles when the concentration of ascorbic acid is

doubled and the triiodide ion concentration is constant, indicating that the process is

first order with respect to ascorbic acid.

(ii)

Calculate the value of the rate constant, k, for the reaction. Include units with your

answer.

[ 2 ]

(ii) For the correct calculated value:

rate $=k[\mathrm{HAsc}]\left[\mathrm{I}_{3}{ }^{-}\right]$

Using trial 1 data:

$k=\frac{\mathrm{rate}}{[\mathrm{HAsc}]\left[\mathrm{I}_{3}^{-}\right]}=\frac{2.457 \times 10^{-4} \mathrm{M} / \mathrm{s}}{(0.450 \mathrm{M})(1.200 \mathrm{M})}=4.55 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}$

For the correct units:

$M^{-1} \mathrm{~s}^{-1}$

Point 07

Point 08

Point 09

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5.2.A—Represent experimental data with a consistent rate law expression question 1

Question (a)

Question (i)

Question (ii)