Chapter 125 of 2363 min read

Arrhenius equation and activation energy

Why this is asked: A catalyst speeds up a reaction by lowering the activation energy through an alternate pathway — it doesn't change ΔH, and because it accelerates the forward and reverse reactions equally, it never shifts the equilibrium position or the value of K. That "catalysts don't touch equilibrium, only kinetics" point is a recurring assertion-reason question.

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Test yourself

•The rate of a reaction quadruples when the temperature changes from $2 7^{\circ}$ C to $5 7^{\circ}$ C. Calculate the energy of activation. (Given $R = 8.314 J K^{- 1} mo l^{- 1}$ , $lo g 4 = 0.6021$ .)
•Which plot of $ln k$ vs $\frac{1}{T}$ (from the four graphs shown) is consistent with the Arrhenius equation?
•The activation energy of any chemical reaction can be calculated if one knows the value of:

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