The Arrhenius law

Exercise 3

    

What is the activation energy of a reaction whose rate is multiplied by $10$ when the temperature increases from $300\;K$ to $310\;K$ ?

(1) At $300\;K$: $ln\;k$ $=$ $ln\;A$ $-$ $\frac{E_a}{R\cdot 300}$ (2) At $310\;K$: $ln(10\cdot k)$ $=$ $ln\;A$ $-$ $\frac{E_a}{R\cdot 310}$ (2)-(1) $ln\;10$ $=$ $E_a(\frac{1}{8.3\cdot 300}$ $-$ $\frac{1}{8.3\cdot 310})$ so: $E_a$ $=$ $1.78\cdot 10^5\frac{J}{mol}$