Edexcel IAL Chemistry A2 11.10 Arrhenius Analysis

[Maximum number: 7]

A group of students investigated the kinetics of a 'clock' reaction.
The reaction investigated was that between hydrogen peroxide and iodide ions in the presence of acid.

Reaction $1 \quad \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{I}_{2}(\mathrm{aq})$

In this 'clock' reaction, a fixed volume of aqueous sodium thiosulfate, $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ , and a small amount of starch were added to the reaction mixture.

The added thiosulfate ions react with the iodine produced in Reaction 1.

Reaction $22 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$

When all the thiosulfate ions have reacted, the presence of iodine is detected by the formation of a starch-iodine complex. The students recorded the time taken for this complex to form.

(a)

The activation energy for Reaction 1 may be found by repeating the experiment at different temperatures.
Each student carried out an experiment at a different temperature.
One of the students misread the thermometer in their experiment.

Table for Question (a) — Edexcel A-Level Chemistry A2

The activation energy, $E_{\mathrm{a}}$ , for a reaction may be found by plotting a graph of In rate against 1 / T.
The gradient of the resulting line of best fit can be used in the Arrhenius equation to determine a value for $E_{\mathrm{a}}$ , in $\mathrm{kJ} \mathrm{mol}^{-1}$ .

[ 7 ]

(i)

Determine the value for $E_{\mathrm{a}}$ for Reaction 1 by plotting a graph using the axes provided.
You should take into account the error made by one of the students.

\text { In rate }=-\frac{E_{\mathrm{a}}}{R} \times \frac{1}{T}+\text { constant } \quad R=8.31 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

Figure for Question (i) — Edexcel A-Level Chemistry A2

$E_{a}$ $\mathrm{kJ} \mathrm{mol}^{-1}$

[ 5 ]

axes correct with scales allowing plot to cover at least half the grid and $\ln$ rate values becoming more negative down the axis

(1)

- both axes labelled with units $\mathrm{K}^{-1}$ on x axis

(1)

- points correctly plotted and best fit line drawn

(1)

- gradient in the range -7200 to -6800

(1)

- $\mathrm{E}_{\mathrm{a}}=-(-7000 \times 8.31) \div 1000$ $=+58.2\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

range 56.5 to 59.9 TE on incorrect gradient value but Ea

(1)

Example of calculation

Allow horizontal axis at bottom of graph Ignore extrapolation

Figure for Answer to Question (i) — Edexcel A-Level Chemistry A2

Allow plotting of all points except the incorrect one

TE on incorrect gradient

Ignore SF

(ii)

The students all used thermometers capable of reading to the same precision. Use your graph to deduce the temperature that the student who made the error should have read on their thermometer.

[ 2 ]

Edexcel A-Level Chemistry A2 11.10 Calculations and Graphical Methods to Find the Activation Energy for a Reaction From Experimental Data the Arrhenius Questions

Exam points

Edexcel A-Level Chemistry A2 11.10 Calculations and Graphical Methods to Find the Activation Energy for a Reaction From Experimental Data the Arrhenius Questions question 1

Question (a)

Question (i)

Question (ii)