Topic 10 · page 14 of 27

Core Practical 5: results and conclusion

Turn the times into rates, draw the two conclusions, class and halogen, and explain each with bond enthalpy or structure. Then evaluate: the end point is a judgement by eye, and fast reactions carry the biggest percentage error.

12 min read10.1110.1010.1210.13

Read first: Core Practical 5: method

Processing the results 10.11

The results below are illustrative: invented to show what a typical set looks like, not measured data. Each time is the time for the precipitate to hide the cross, in seconds, with three repeats.

Halogenoalkane Class Repeats / s Mean time, t / s
1-chlorobutane primary 296, 305, 299 300
1-bromobutane primary 88, 92, 90 90
1-iodobutane primary 13, 15, 14 14
2-bromobutane secondary 34, 36, 35 35
2-bromo-2-methylpropane tertiary 5, 6, 4 5

Take the mean of the repeats for each compound. The mean of 88, 92 and 90 is (88 + 92 + 90) ÷ 3 = 90 s.

The same amount of silver halide must form to hide the cross each time, so

rate∝1t \text{rate} \propto \frac{1}{t}

To keep the numbers readable, calculate 1000 ÷ t.

Halogenoalkane Class Mean t / s Rate ∝ 1000 ÷ t
1-chlorobutane primary 300 3.3
1-bromobutane primary 90 11
1-iodobutane primary 14 71
2-bromobutane secondary 35 29
2-bromo-2-methylpropane tertiary 5 200
A horizontal bar chart of 1000 divided by time for five halogenoalkanes. 1-chlorobutane 3.3, 1-bromobutane 11, 1-iodobutane 71, 2-bromobutane 29, 2-bromo-2-methylpropane 200. The bars are grouped: the three bromo isomers compare class, and the three primary compounds compare the halogen. 1-bromobutane appears in both groups.
Illustrative results. Left group: class, tertiary fastest. Right group: halogen, iodo fastest. 1-bromobutane is in both.

Conclusions 10.1010.1210.13

Class (same halogen, bromo). Tertiary 200, secondary 29, primary 11. The tertiary isomer is hydrolysed 18 times faster than the primary (90 ÷ 5), and the secondary 2.6 times faster (90 ÷ 35).

Halogen (same class, primary). Iodo 71, bromo 11, chloro 3.3. The iodoalkane is 6.4 times faster than the bromoalkane (90 ÷ 14), and the bromoalkane is 3.3 times faster than the chloroalkane (300 ÷ 90).

Write the observation first, then the explanation. "The precipitate formed fastest with the iodoalkane. The C–I bond has the lowest bond enthalpy, so it breaks most easily, so the iodoalkane is hydrolysed fastest."

Evaluating the method 10.11

Weakness Effect Improvement
The end point is judged by eye different observers stop at different moments one observer; or a colorimeter that records the cloudiness
Fast reactions have a short time a reaction time of about 1 s is a large share of 5 s cool the water bath, or dilute the silver nitrate, so the tertiary sample takes longer
Silver chloride, bromide and iodide scatter light differently the same amount of precipitate may not hide the cross each time, so the comparison between halogens is not exact use a colorimeter, or say the comparison is qualitative
Drops are not an accurate measure the amount of halogenoalkane differs a little between tubes use a small syringe or pipette
Temperature falls when the tube is lifted out slows the later part of the reaction do the timing in the water bath

Fast reactions carry the biggest percentage uncertainty. Suppose the combined reaction time and judgement gives an uncertainty of ±1 s in each time (illustrative):

Mean time / s Percentage uncertainty
5 1 ÷ 5 × 100 = 20%
14 1 ÷ 14 × 100 = 7.1%
35 1 ÷ 35 × 100 = 2.9%
90 1 ÷ 90 × 100 = 1.1%
300 1 ÷ 300 × 100 = 0.33%

The order of the five rates is still clear, because the differences are far larger than the uncertainties, but the size of the tertiary rate is much less certain than the size of the chloro rate.

Check yourself

  1. Calculate the rate ∝ 1000 ÷ t for a halogenoalkane whose mean time is 25 s.
  2. The repeats for one halogenoalkane are 33, 36 and 35 s. Calculate the mean.
  3. Using the table, state the order of reactivity of the three isomers and how many times faster the tertiary isomer is than the secondary.
  4. State the order for the three primary halogenoalkanes and give the explanation.
  5. Calculate the percentage uncertainty for a time of 5 s if the uncertainty is ±1 s, and say why this matters.
  6. Suggest an improvement to the method and explain how it helps.

Answers cover these until you have tried the questions

  1. 1000 ÷ 25 = 40.
  2. (33 + 36 + 35) ÷ 3 = 104 ÷ 3 = 34.7 s.
  3. Tertiary > secondary > primary. Tertiary versus secondary: 35 ÷ 5 = 7 times faster.
  4. Iodo > bromo > chloro. The C–I bond has the lowest bond enthalpy (228 kJ mol⁻¹) so it is the weakest and breaks most easily; C–Cl is the strongest (346 kJ mol⁻¹).
  5. 1 ÷ 5 × 100 = 20%. A short time has a large percentage uncertainty, so the size of the fastest rates is the least reliable.
  6. For example, use a colorimeter to record the cloudiness instead of judging by eye. This removes the subjective end point and the observer's reaction time.