The Arrhenius law

Exercise 1

For the second order reaction $2NOCl(g)\longrightarrow 2NO(g) +Cl_2(g)$ following results have been determined ($T$ in $K$, $k$ in $\frac{L}{mol\; s}$):

	$T$	$k$
(1)	300	$2.6\cdot 10^{-8}$
(2)	400	$4.9\cdot 10^{-4}$

a) Determine the activation energy $E_a$! b) Determine the pre-exponential factor $A$! c) State the Arrhenius law ! d) Calculate the rate constant $k_3 $ at $500 \; K $

a) $k=Ae^{-\frac{E_a}{R\cdot T}}$ $ln\;k$ $=$ $ln\;A$ $-$ $\frac{E_a}{R\cdot T}$ (1) $ln\;2.6\cdot 10^{-8}$ $=$ $ln\;A$ $-$ $\frac{E_a}{8.3\cdot 300}$ (2) $ln\;4.9\cdot 10^{-4}$ $=$ $ln\;A$ $-$ $\frac{E_a}{8.3\cdot 400}$ (2)-(1): $ln\;4.9\cdot 10^{-4}$ $-$ $ln\;2.6\cdot 10^{-8}$ $=$ $\frac{E_a}{8.3}(\frac{1}{300}$ $-$ $\frac{1}{400})$ so: $E_a$ $=$ $98.2\cdot 10^3 J$ b) (1): $A$ $=$ $k\;cdot\;e^{\frac{E_a}{R\cdot T}}$ $A$ $=$ $2.6\cdot 10^{-8}e^{\frac{98.2\cdot 10^3}{8.3\cdot 300}}$ $\approx$ $3.5\cdot 10^9 \frac{L}{mol\cdot s}$ c) $k$ $=$ $3.5\cdot 10^9e^{-\frac{98.2\cdot 10^3}{8.3\cdot T}}$ d) $ln\frac{k_3}{2.6\cdot 10^{-8}}$ $=$ $\frac{98.2\cdot 10^3 }{8.3}(\frac{1}{300}$ $-$ $\frac{1}{500})$ so: $k_3$ $=$ $0.19\frac{L}{mol\cdot s}$