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Activation energy results
|
Temperature /°C |
Temperature T /K |
1/T /K-1 |
Time t/s |
1/t (ie rate) /s-1 |
ln(rate) |
|
24.3 |
297.5 |
0.00336 |
53 |
0.019 |
-4.0 |
|
37.6 |
310.8 |
0.00322 |
24 |
0.042 |
-3.2 |
|
44.2 |
317.4 |
0.00315 |
16 |
0.063 |
-2.8 |
|
50.9 |
324.1 |
0.00309 |
12 |
0.083 |
-2.5 |
|
62.6 |
335.8 |
0.00298 |
6 |
0.17 |
-1.8 |
We now plot a graph of ln(rate) against 1/T and measure the gradient of this graph.
Graphs are very popular on the exam paper. Make sure you draw the points and best line in pencil so that it can easily be corrected. Fully, and correctly, label both axes - this is fairly easy to do as the details can just be copied from the results table. Choose suitable scales for both axes so that your plotting fills up most of the graph paper (you do not need to start your axes at zero). Be sure to correctly plot all the points, if any look particularly far out double check them. Finally, draw the best fit straight line, or smoothest curve, as close to the points as possible (don't join the dots though!).
Measuring the gradient of the above graph gives a value of - 5700K. To obtain a value for the activation energy, Ea, we must multiply the gradient value by - R (the gas constant, 8.314 J mol-1 K-1).
Activation energy, Ea, = - 5700 × - 8.314 = + 47 kJ mol-1
The official value is + 46.6 kJ mol-1. Avoid putting too much accuracy into your final answer. I have quoted the above value to 2 significant figures. Students often quote about six, but bear in mind this means that your experiment would have to be accurate up to 1 part in a million - unlikely!
