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Date May 2013 Marks available 1 Reference code 13M.2.hl.TZ1.3
Level HL Paper 2 Time zone TZ1
Command term Sketch Question number 3 Adapted from N/A

Question

To determine the activation energy of a reaction, the rate of reaction was measured at different temperatures. The rate constant, \(k\), was determined and \(\ln k\) was plotted against the inverse of the temperature in Kelvin, \({T^{ - 1}}\). The following graph was obtained.

M13/4/CHEMI/HP2/ENG/TZ1/03

Define the term activation energy, \({E_{\text{a}}}\).

[1]
a.

Use the graph on page 8 to determine the value of the activation energy, \({E_{\text{a}}}\), in \({\text{kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}\).

[2]
b.

On the graph on page 8, sketch the line you would expect if a catalyst is added to the reactants.

[1]
c.

Markscheme

minimum energy needed to react/start a reaction / energy difference between reactants and transition state;

a.

gradient of the line: –63;

Accept –60 to –65.

\({E_{\text{a}}}{\text{ }}( =  - R \times {\text{gradient}}) = 0.52{\text{ (kJ}}\,{\text{mo}}{{\text{l}}^{ - 1}}{\text{)}}\);

Accept 0.50 to 0.54.

b.

gradient of the line less steep (less negative);

Accept any position as long as gradient less steep.

c.

Examiners report

The idea of activation energy being a minimum was seldom communicated. Few were able to follow through all the mathematics to find \({E_{\text{a}}}\) by a graphical method and those that did had often omitted \({\text{1}}{{\text{0}}^{ - 2}}\) in their calculations. The answers were often poorly set out so it was difficult to assess the award of part marks; indeed, many candidates seemed to hope that a correct answer would somehow emerge from a mass of incomprehensible figures.

a.

The idea of activation energy being a minimum was seldom communicated. Few were able to follow through all the mathematics to find \({E_{\text{a}}}\) by a graphical method and those that did had often omitted \({\text{1}}{{\text{0}}^{ - 2}}\) in their calculations. The answers were often poorly set out so it was difficult to assess the award of part marks; indeed, many candidates seemed to hope that a correct answer would somehow emerge from a mass of incomprehensible figures.

b.

The idea of activation energy being a minimum was seldom communicated. Few were able to follow through all the mathematics to find \({E_{\text{a}}}\) by a graphical method and those that did had often omitted \({\text{1}}{{\text{0}}^{ - 2}}\) in their calculations. The answers were often poorly set out so it was difficult to assess the award of part marks; indeed, many candidates seemed to hope that a correct answer would somehow emerge from a mass of incomprehensible figures. The gradient of the graph for (c) was generously marked; all candidates had to do was to realize that the catalyst would lower the activation energy and thus the gradient would be less negative. As long as a line with less negative gradient was drawn, the mark was awarded.

c.

Syllabus sections

Core » Topic 11: Measurement and data processing » 11.2 Graphical techniques
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