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

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

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

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

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.

gradient of the line less steep (less negative);

Accept any position as long as gradient less steep.

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 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 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.