A graph of lnk \ln k\,lnk against 1T\displaystyle \frac{1}{T}T1 (T T\,T in K\text{K}K) for a reaction has a gradient with the numerical value of -6800.
What is the activation energy, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1, for this reaction? (Use gas constant 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)
−818-818−818
−56.5-56.5−56.5
+56.5+56.5+56.5
+5.65×104+5.65 \times 10^4+5.65×104
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.