CAIE A Level Chemistry 26 Reaction Kinetics

Practise rate equations, reaction orders, half-lives, rate constants and mechanisms with rate-determining steps, interpreting concentration-time and rate-concentration data.

Covered syllabus points

26.1 Rate equations, orders and rate constants 26.2 Homogeneous and heterogeneous catalysts

[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 \quad \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: 11]

(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 ]

(b)

Catalysts may be homogeneous or heterogeneous.

[ 6 ]

(i)

Identify two metals that act as heterogeneous catalysts in the removal of $\mathrm{NO}_{2}$ from the exhaust gases of car engines. and

[ 1 ]

(ii)

Iron acts as a heterogeneous catalyst in the Haber process.

Describe the mode of action of this iron catalyst.

[ 2 ]

(iii)

$\mathrm{Fe}^{2+}$ ions act as a homogeneous catalyst in the reaction between $\mathrm{I}^{-}(\mathrm{aq})$ and $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}(\mathrm{aq})$ .

Write equations for the two reactions that occur when $\mathrm{Fe}^{2+}(\mathrm{aq})$ is added to a mixture of $\mathrm{I}^{-}(\mathrm{aq})$ and $\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}(\mathrm{aq})$ .
equation $1 \quad \mathrm{~S}_{2} \mathrm{O}_{8}{ }^{2-}+$
equation 2

[ 2 ]

$\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}+2 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{SO}_{4}{ }^{2-}+2 \mathrm{Fe}^{3+}$ $2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_{2}$

[1] any multiple

[1] any multiple

(iv)

Explain the difference between a homogeneous catalyst and a heterogeneous catalyst.

[ 1 ]

[Maximum number: 7]

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 $\left[\mathrm{OH}^{-}\right]$ )

Marking guidance:

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)

(b)

A student suggests that the reaction between $\mathrm{H}_{2} \mathrm{PO}_{2}^{-}(\mathrm{aq})$ and $\mathrm{OH}^{-}(\mathrm{aq})$ might happen more quickly in the presence of a heterogeneous catalyst.

Describe the mode of action of a heterogeneous catalyst.

[ 2 ]

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Covered syllabus points

26. Reaction kinetics question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

26. Reaction kinetics question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question (iv)

26. Reaction kinetics question 3

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question (b)