Reaction Kinetics: Question 5

Worked solution

Part (a): Showing the reaction is first order

Check the time taken for the concentration of X to halve at each stage:

• $0.800 \rightarrow 0.400\text{ mol dm}^{-3}$: from $t=0$ to $t=20$, taking $20$ minutes.
• $0.400 \rightarrow 0.200\text{ mol dm}^{-3}$: from $t=20$ to $t=40$, taking $40-20=20$ minutes.
• $0.200 \rightarrow 0.100\text{ mol dm}^{-3}$: from $t=40$ to $t=60$, taking $60-40=20$ minutes.

In each case, the half-life is the same, $20$ minutes, regardless of the concentration at the start of that interval. A constant half-life, independent of concentration, is the defining characteristic of a first-order reaction (for comparison, a zero-order reaction’s half-life would decrease as the reaction proceeds, and a second-order reaction’s half-life would increase). Since the half-life here is constant at $20$ minutes throughout, the reaction is first order with respect to X.

Part (b): Calculating the rate constant, k

For a first-order reaction, the rate constant is related to the half-life, $t_{\frac{1}{2}}$, by:

$k = \frac{0.693}{t_{\frac{1}{2}}}$

First convert the half-life into seconds, since $k$ is required in $\text{s}^{-1}$:

$t_{\frac{1}{2}} = 20\text{ min} = 20 \times 60 = 1200\text{ s}$

$k = \frac{0.693}{1200} = 5.775\times10^{-4}\text{ s}^{-1}$

Check by re-doing the division a different way: $1200 \times 5\times10^{-4} = 0.60$, and the remaining $0.693-0.600=0.093$ needs $0.093/1200 = 0.0000775$, so $k = 5\times10^{-4}+0.775\times10^{-4} = 5.775\times10^{-4}\text{ s}^{-1}$, the same value, confirming the result.

$k \approx 5.78\times10^{-4}\text{ s}^{-1} \text{ (3 s.f.)}$

Part (c): Identifying the rate-determining step

The mechanism is:

Step 1: $\text{X} \rightarrow \text{Y} + \text{Z}$ (slow)

Step 2: $\text{Y} + \text{X} \rightarrow 2\text{W}$ (fast)

Step 1 is explicitly labelled slow, so it is the rate-determining step. The slow step always controls (determines) the overall rate of a multi-step mechanism, however fast any later step is.

Step 1 involves only one molecule of X reacting, with no other species present in that step. The rate equation derived from a rate-determining step reflects the number and identity of species reacting in that step, so:

$\text{rate} = k[\text{X}]$

This is first order in X (and does not depend on any other species), which exactly matches the experimentally observed first-order kinetics found in parts (a) and (b). Step 2, although it also consumes X, is fast and therefore does not limit the overall rate. Its species do not appear in the overall rate equation, and Y is simply a reaction intermediate that is produced in Step 1 and consumed as fast as it forms in Step 2.

Final answers

• (a) Constant half-life of $20$ minutes at every stage $\Rightarrow$ first order with respect to X.
• (b) $k = 5.78\times10^{-4}\text{ s}^{-1}$ (using $t_{\frac{1}{2}}=1200\text{ s}$).
• (c) Step 1 is rate-determining (stated to be slow, and unimolecular in X), giving $\text{rate}=k[\text{X}]$, consistent with the observed first-order kinetics.