CAIE A-Level Chemistry A2 26.1 Rate Equations Orders and Rate Constants Questions

Practise deriving rate equations from data, calculating k and half-life, and testing proposed mechanisms.

Exam points

compare experiments to determine each concentration exponent and sum them for overall order
substitute rate and concentrations to calculate k with units derived from the overall order
match species in the rate-determining step to the rate equation and identify intermediates or catalysts

Covered syllabus points

26.1.1—And use the terms rate equation, order 26.1.2—Rate equations and reaction order 26.1.3—Show understanding that the half-life 26.1.4—The numerical value of a rate constant, e.g 26.1.5—Mechanisms and rate-determining step 26.1.6—Effect of temperature change on the rate

[Maximum number: 4]

(a)

The equation for the decomposition of hydrogen peroxide without a catalyst is shown.

2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})

Under certain conditions this reaction is found to be first order with respect to hydrogen peroxide, with a rate constant, k, of $2.0 \times 10^{-6} \mathrm{~s}^{-1}$ at 298 K .

Calculate the initial rate of decomposition of a $0.75 \mathrm{moldm}^{-3}$ hydrogen peroxide solution at 298 K .
initial rate = $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

[ 1 ]

rate $=2.0 \times 10^{-6} \times 0.75=1.5 \times 10^{-6}$

(b)

A four-step mechanism is suggested for the reaction between hydrogen peroxide and iodide ions in an acidic solution.

step 1 $\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{I}^{-} \rightarrow \mathrm{IO}^{-}+\mathrm{H}_{2} \mathrm{O}$

step 2 $\mathrm{H}^{+}+\mathrm{IO}^{-} \rightarrow \mathrm{HIO}$

step 3 $\mathrm{HIO}+\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+\mathrm{OH}^{-}$

step 4 $\mathrm{OH}^{-}+\mathrm{H}^{+} \rightarrow \mathrm{H}_{2} \mathrm{O}$

Step 1 is the rate-determining step.

[ 3 ]

(i)

State what is meant by the term rate-determining step.

[ 1 ]

(ii)

Use this mechanism to construct a balanced equation for this reaction.

[ 1 ]

$\begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O} \mathrm{OR} \mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{HI} \rightarrow \mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O} \end{aligned}$

(iii)

Deduce the order of reaction with respect to each of the following.
$\mathrm{H}_{2} \mathrm{O}_{2}=$ $\mathrm{I}^{-}=$

\mathrm{H}^{+}=

[ 1 ]

$\mathrm{H}_{2} \mathrm{O}_{2}=1$ AND $\mathrm{I}^{-}=1$ AND $\mathrm{H}^{+}=0$

[Maximum number: 5]

Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ .

(a)

$\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})$ reacts with $\mathrm{OH}^{-}(\mathrm{aq})$ .

\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})

Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction.

Table 2.1

[ 5 ]

(i)

The rate equation was found to be:

\text { rate }=k\left[\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})\right]\left[\mathrm{OH}^{-}(\mathrm{aq})\right]^{2}

Show that the data in Table 2.1 is consistent with the rate equation.

[ 2 ]

[ $\mathrm{H}_{2} \mathrm{PO}_{2}{ }^{-}$ ] doubles / × 2 from experiments 1 to 2 ORA
volume of $\mathrm{H}_{2}$ produced doubles $/ \times 2$ ( ∴ first order wrt $\left[\mathrm{H}_{2} \mathrm{PO}_{2}{ }^{-}\right]$ )

[ $\mathrm{H}_{2} \mathrm{PO}_{2}{ }^{-}$ ] $\times 3$ and $\left[\mathrm{OH}^{-}\right] \times 1 / 2$ from experiments 1 to 3 ORA volume of $\mathrm{H}_{2}$ produced falls to $\frac{3}{4}$ original (if first order wrt $\left[\mathrm{H}_{2} \mathrm{PO}_{2}{ }^{-}\right]$ then must be second order wrt $\left[\mathrm{OH}^{-}\right]$ )

ALLOW input data into rate equation and show k is the same $k=2.8 \times 10^{-6} / k=6.7 \times 10^{-2}(1 / 15) / k=4$ for all experiments [2]

(ii)

State the units of the rate constant, k, for the reaction.

[ 1 ]

$\mathrm{mol}^{-2} \mathrm{dm}^{6} \mathrm{~s}^{-1}$

(iii)

The experiment is repeated using a large excess of $\mathrm{OH}^{-}(\mathrm{aq})$ .

Under these conditions, the rate equation is:

\begin{array}{ll} & \text { rate }=k_{1}\left[\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})\right] \\ k_{1}=8.25 \times 10^{-5} \mathrm{~s}^{-1} & \end{array}

Calculate the value of the half-life, $t_{\frac{1}{2}}$ , of the reaction.

\begin{aligned} & t_{\frac{1}{2}}= \\ & \text { s [1] } \end{aligned}

[ 1 ]

$t_{1 / 2}=0.693 / 8.25 \times 10^{-5}=\mathbf{8 4 0 0}(\mathrm{s})$ OR $t_{1 / 2}=\ln 2 / 8.25 \times 10^{-5}=\mathbf{8 4 0 1 . 8}(\mathrm{s})$

(iv)

Describe how an increase in temperature affects the value of the rate constant, $k_{1}$ .

[ 1 ]

( $k_{1}$ ) increases (with temperature)

[Maximum number: 5]

(a)

A and B react together to give product AB .

A+B \rightarrow A B

When the concentrations of A and B are both $0.0100 \mathrm{moldm}^{-3}$ , the rate of formation of AB is $7.62 \times 10^{-4} \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$ . When the concentrations of A and B are both $0.0200 \mathrm{moldm}^{-3}$ , the rate of formation of AB is $3.05 \times 10^{-3} \mathrm{moldm}^{-3} \mathrm{~s}^{-1}$ .

[ 5 ]

(i)

Complete the three possible rate equations that are consistent with these data.
rate =
rate =
rate =

[ 2 ]

rate $=\mathrm{k}[\mathrm{A}][\mathrm{B}]$

rate $=\mathrm{k}[\mathrm{A}]^{2} / \mathrm{k}[\mathrm{A}]^{2}[\mathrm{~B}]^{0}$ AND rate $=\mathrm{k}[\mathrm{B}]^{2} / \mathrm{k}[\mathrm{A}]^{0}[\mathrm{~B}]^{2}$

[1]

[1] BOTH

(ii)

Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, k. Include the units of k.
k= units

[ 2 ]

k=7.62 / 7.625 / 7.63

$\mathrm{mol}^{-1} \mathrm{dm}^{3} \mathrm{~s}^{-1}$

[1] ecf from 3a(i) min 3sf

[1] ecf from 3a(i)

(iii)

Explain why it is not possible to calculate a value for the half-life, $t_{\frac{1}{2}}$ , of this reaction using the value of the rate constant k calculated in (ii) and the equation $k=0.693 / t_{\frac{1}{2}}$ .

[ 1 ]

[Maximum number: 1]

A four-step mechanism is suggested for the reaction between hydrogen peroxide and iodide ions in an acidic solution.

step 1 $\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{I}^{-} \rightarrow \mathrm{IO}^{-}+\mathrm{H}_{2} \mathrm{O}$

step 2 $\mathrm{H}^{+}+\mathrm{IO}^{-} \rightarrow \mathrm{HIO}$

step 3 $\mathrm{HIO}+\mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+\mathrm{OH}^{-}$

step 4 $\mathrm{OH}^{-}+\mathrm{H}^{+} \rightarrow \mathrm{H}_{2} \mathrm{O}$

Step 1 is the rate-determining step.

State what is meant by the term rate-determining step.

[Maximum number: 1]

Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ .

$\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})$ reacts with $\mathrm{OH}^{-}(\mathrm{aq})$ .

\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})

Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction.

Table 2.1

The experiment is repeated using a large excess of $\mathrm{OH}^{-}(\mathrm{aq})$ .

Under these conditions, the rate equation is:

\begin{array}{ll} & \text { rate }=k_{1}\left[\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})\right] \\ k_{1}=8.25 \times 10^{-5} \mathrm{~s}^{-1} & \end{array}

Calculate the value of the half-life, $t_{\frac{1}{2}}$ , of the reaction.

\begin{aligned} & t_{\frac{1}{2}}= \\ & \text { s [1] } \end{aligned}

$t_{1 / 2}=0.693 / 8.25 \times 10^{-5}=\mathbf{8 4 0 0}(\mathrm{s})$ OR $t_{1 / 2}=\ln 2 / 8.25 \times 10^{-5}=\mathbf{8 4 0 1 . 8}(\mathrm{s})$

[Maximum number: 2]

A and B react together to give product AB .

A+B \rightarrow A B

Choose one of the rate equations you have written in (i), and calculate the value of the rate constant, k. Include the units of k.
k= units

k=7.62 / 7.625 / 7.63

$\mathrm{mol}^{-1} \mathrm{dm}^{3} \mathrm{~s}^{-1}$

[1] ecf from 3a(i) min 3sf

[1] ecf from 3a(i)

[Maximum number: 1]

Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}$ .

$\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})$ reacts with $\mathrm{OH}^{-}(\mathrm{aq})$ .

\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{HPO}_{3}^{2-}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})

Table 2.1 shows the results of a series of experiments used to investigate the rate of this reaction.

Table 2.1

Describe how an increase in temperature affects the value of the rate constant, $k_{1}$ .

( $k_{1}$ ) increases (with temperature)

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question 4

Question 5

Question 6

Question 7