A graph of ln(k)\ln(k)ln(k) is plotted against 1/T 1 / T\,1/T for a reaction, where k k\,k is the rate constant and T T\,T is the temperature in K\text{K}K.
The gradient of this line has a numerical value of −13 500-13\,500−13500.
What is the activation energy, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1, for this reaction? (Use R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}R=8.314 J K−1 mol−1)
+1.62+1.62+1.62
+112+112+112
+1620+1620+1620
+112 000+112\,000+112000
29 exam-style questions on OCR A A Level Chemistry How fast?. Each one has a worked solution and a mark scheme showing where the marks go.