CAIE A-Level Chemistry 26.1.2 Rate Equations and Reaction Order

Practise using initial-rate tables and graphs to determine zero-, first- or second-order dependencies and construct rate laws.

[Maximum number: 3]

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

[ 3 ]

(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 ]

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

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26.1.2—Rate equations and reaction order question 1

Question (a)

Question (i)

Question (ii)