Topic 11 · page 21 of 21

Core Practical 9b: results, order and activation energy

A graph of 1 ÷ t against the volume of iodide is a straight line through the origin, so the reaction is first order in iodide. Repeating at different temperatures and plotting ln(1/t) against 1/T gives the activation energy.

14 min read11.1211.4(i)11.5(iii)11.10

Read first: Core Practical 9b: the iodine clock, method, Activation energy from a graph

The results 11.1211.4(i)

Illustrative results for the method on page 20. They are not measured data. The rate is taken as 2.5 × 10⁻⁴ mol dm⁻³ divided by the time.

KI(aq) / cm³ [I⁻] / mol dm⁻³ t / s 1/t / 10⁻³ s⁻¹ rate / 10⁻⁶ mol dm⁻³ s⁻¹
5.0 0.010 119 8.40 2.10
10.0 0.020 61 16.4 4.10
15.0 0.030 41 24.4 6.10
20.0 0.040 30 33.3 8.33
25.0 0.050 24 41.7 10.4

The order with respect to iodide 11.5(iii)

Rate is proportional to 1/t, so plot 1/t on the y axis. Because the total volume is constant, the concentration of iodide is proportional to the volume of potassium iodide solution, so the volume can go on the x axis in place of concentration.

A graph of 1 over t in units of 10 to the minus 3 per second against the volume of potassium iodide solution in cubic centimetres. Five points lie close to a straight line that passes very close to the origin. The gradient of the line is 1.67 per cubic centimetre.
Illustrative results. 1/t is proportional to the volume of KI, so the rate is proportional to [I⁻]: first order.

Worked exampleOrder from the table

  1. Experiments 1 and 2: the volume (and so [I⁻]) doubles, and 1/t rises from 8.40 to 16.4, a factor of 1.95. The rate is about ×2.
  2. Experiments 1 and 4: [I⁻] is ×4, and 1/t rises from 8.40 to 33.3, a factor of 3.97. The rate is about ×4.
  3. The graph is a straight line through the origin, so the rate is directly proportional to [I⁻]: the reaction is first order with respect to iodide.
  4. Rate ÷ [I⁻] is almost constant: 2.1 × 10⁻⁴, 2.05 × 10⁻⁴, 2.03 × 10⁻⁴, 2.08 × 10⁻⁴ and 2.08 × 10⁻⁴ s⁻¹. This constant includes k and the fixed concentrations of the other reagents.

If you plotted t against volume you would get a curve (a time that halves when the volume doubles), which is hard to interpret. Plotting 1/t gives a straight line.

To find the orders for the other reagents, repeat the experiment changing the volume of that reagent only, with water making up the total volume to 50.0 cm³.

The activation energy 11.1211.10

Repeat the experiment with the same volumes (10.0 cm³ of KI) at four temperatures, with all the reagents warmed or cooled in a water bath before mixing. Because the concentrations are the same each time, 1/t is proportional to k.

Temperature / °C T / K 1/T / 10⁻³ K⁻¹ t / s ln(1/t)
15 288 3.472 118 −4.771
25 298 3.356 61 −4.111
35 308 3.247 33 −3.497
45 318 3.145 18 −2.890

The graph of ln(1/t) against 1/T is a straight line, as on page 15. The line of best fit passes through (3.15, −2.93) and (3.45, −4.65).

Worked exampleActivation energy from the clock

  1. Δy = −4.65 − (−2.93) = −1.72
  2. Δx = (3.45 − 3.15) × 10⁻³ = 0.30 × 10⁻³ K⁻¹
  3. gradient = −1.72 ÷ 0.30 × 10⁻³ = −5.73 × 10³ K
  4. EaE_\text{a} = 5.73 × 10³ × 8.31 = 4.76 × 10⁴ J mol⁻¹
  5. EaE_\text{a} = 47.6 kJ mol⁻¹

A quick check with only the end values, 288 K and 318 K, using the two-temperature equation from page 16: the ratio of the times is 118 ÷ 18 = 6.56, so ln 6.56 = 1.880. Then 1/288 − 1/318 = 3.276 × 10⁻⁴ K⁻¹ and EaE_\text{a} = 8.31 × 1.880 ÷ 3.276 × 10⁻⁴ = 4.77 × 10⁴ J mol⁻¹, which is 47.7 kJ mol⁻¹. The two methods agree.

Evaluating the experiment 11.12

Source of error Effect Improvement
Judging the first appearance of the blue-black colour an error of about 1 s: 5.6% of an 18 s run but 0.8% of a 118 s run use a colorimeter, or have the same person judge each time against white paper
Reagents not at the temperature of the bath when mixed the temperature of the run is not the one recorded warm or cool each solution in the bath first and check the temperature at the end
Hydrogen peroxide solution that has partly decomposed its concentration is less than stated, so every time is long use a fresh solution, and the same bottle throughout
One run per condition a random error is not averaged out repeat each and take the mean
The clock only approximates the initial rate small systematic error use a small amount of thiosulfate (1.25% of the peroxide is used)

Check yourself

  1. Calculate 1/t for the experiments with t = 61 s and t = 24 s, in s⁻¹.
  2. In an iodine clock the time falls from 61 s to 30 s when the volume of KI is doubled from 10.0 to 20.0 cm³ (total volume constant). Deduce the order with respect to iodide.
  3. Explain why a graph of t against the volume of KI is a curve and not a straight line.
  4. Use the data at 298 K (61 s) and 308 K (33 s) to calculate EaE_\text{a}.
  5. A stopwatch is started and stopped with an uncertainty of about 1 s. Calculate the percentage uncertainty for a time of 18 s and for a time of 118 s.
  6. A student plots ln t against 1/T and gets a gradient of +5.73 × 10³ K. Explain how this relates to the activation energy.

Answers cover these until you have tried the questions

  1. 1/61 = 0.0164 s⁻¹ (1.64 × 10⁻²) and 1/24 = 0.0417 s⁻¹ (4.17 × 10⁻²).
  2. 1/t rises from 1/61 = 0.0164 to 1/30 = 0.0333, a factor of 2.03, so the rate approximately doubles when [I⁻] doubles. The reaction is first order with respect to iodide.
  3. The rate is proportional to the volume (so to [I⁻]), and t is proportional to 1 ÷ rate, so t is inversely proportional to the volume. That is a curve, a hyperbola. A graph of 1/t is the straight line.
  4. ln(t₁ ÷ t₂) = ln(61 ÷ 33) = 0.6144 (the rate ratio is the inverse of the time ratio). 1/298 − 1/308 = 1.0895 × 10⁻⁴ K⁻¹. EaE_\text{a} = 8.31 × 0.6144 ÷ 1.0895 × 10⁻⁴ = 4.69 × 10⁴ J mol⁻¹ = 46.9 kJ mol⁻¹.
  5. 1 ÷ 18 × 100 = 5.6%. 1 ÷ 118 × 100 = 0.85%.
  6. ln t = −ln(1/t), so a plot of ln t against 1/T is the same line turned upside down and has the opposite gradient, +EaE_\text{a}/R. The activation energy is the gradient × R = 5.73 × 10³ × 8.31 = 4.76 × 10⁴ J mol⁻¹ = 47.6 kJ mol⁻¹.