The Arrhenius law

Exercise 2

    

The first order reaction $C_2H_5Br(g)\longrightarrow C_2H_4(g) +HBr(g)$ has a rate constant: $k=2,0\cdot 10^{-5}\cdot \frac{1}{s}$ at $650\; K$ Its activation energy is: $E_a=2,26\cdot 10^5\frac{J}{mol}$ Calculate at what temperature we have: $k=6,0\cdot 10^{-5}\cdot \frac{1}{s}$

1) At $650\;K$: $ln\;k$ $=$ $ln\;A$ $-$ $\frac{E_a}{R\cdot T}$ $ln\;A$ $=$ $ln\;2.0\cdot 10^{-5}$ $+$ $\frac{2.26\cdot 10^5}{8.3\cdot 650}$ 2) At $T$ Kelvin: $ln\;k$ $=$ $ln\;A$ $-$ $\frac{E_a}{R\cdot T}$ $ln\;6.0\cdot 10^{-5}$ $=$ $ln\;2.0\cdot 10^{-5}$ $+$ $\frac{2.26\cdot 10^5}{8.3\cdot 650}$ $-$ $\frac{2.26\cdot 10^5}{8.3\cdot T}$ $\frac{2.26\cdot 10^5}{8.3}(\frac{1}{650}$ $-$ $\frac{1}{T})=ln3$ so: $T=667\;K$