The Arrhenius law

Exercise 4

For the reaction $2NO_2(g)$ $\longrightarrow$ $ 2NO(g)$ $+$ $O_2(g)$ rate constants (in $\frac{L}{mol\;s}$) have been determined for the following temperatures (in $^oC$):

kt
10410
14420
18430
24440

Determine the activation energy $E_a$ on a graph!

$ln\;k$ $=$ $ln\;A$ $-$ $\frac{E_a}{R\cdot T}$ is a linear equation $y=b+mx$ with $y=ln\;k$ $b=ln\;A$ $m=\frac{-E_a}{R}$ $x=\frac{1}{T}$ So let's just represent this on a graph in order to determine its slope $\frac{-E_a}{R}$
$ln\;k$$\frac{1}{T}$
2.30$1.464\cdot 10{-3}$
2.64$1.443\cdot 10{-3}$
2.89$1.422\cdot 10{-3}$
3.18$1.403\cdot 10{-3}$
We find: $-\frac{E_a}{R}$ $=$ $\frac{\Delta y}{\Delta y}$ $=$ $-1.4\cdot 10^4\;K$ $E_a$ $=$ $1.2\cdot 10^5\frac{J}{mol}$