CAIE A-Level Chemistry 26.1 Rate Equations, Orders and Constants

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

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)

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Covered syllabus points

26.1 Rate equations, orders and rate constants question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

26.1 Rate equations, orders and rate constants question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)