The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.

(N/A) From the given data,we calculate the values for the Arrhenius plot:

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$T / K$	$273$	$293$	$313$	$333$	$353$
$1/T / K^{-1}$	$3.66 \times 10^{-3}$	$3.41 \times 10^{-3}$	$3.19 \times 10^{-3}$	$3.00 \times 10^{-3}$	$2.83 \times 10^{-3}$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$
$\ln k$	$-7.147$	$-4.075$	$-1.359$	$-0.577$	$3.063$

Slope of the line is calculated as $\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \approx -12301 \ K$.
According to the Arrhenius equation,$\text{Slope} = -\frac{E_{a}}{R}$.
$E_{a} = -\text{Slope} \times R = -(-12301 \ K) \times (8.314 \ J \ K^{-1} \ mol^{-1}) \approx 102.27 \ kJ \ mol^{-1}$.
Using $\ln k = \ln A - \frac{E_{a}}{RT}$,we find $\ln A = \ln k + \frac{E_{a}}{RT}$.
At $T = 273 \ K$,$\ln A = -7.147 + \frac{102270}{8.314 \times 273} \approx 37.91$,so $A \approx 2.91 \times 10^{16} \ s^{-1}$.
For $30^{\circ}C$ $(303 \ K)$,$1/T \approx 3.30 \times 10^{-3} \ K^{-1}$,$\ln k \approx -2.8$,$k \approx 6.08 \times 10^{-2} \ s^{-1}$.
For $50^{\circ}C$ $(323 \ K)$,$1/T \approx 3.10 \times 10^{-3} \ K^{-1}$,$\ln k \approx -0.5$,$k \approx 0.607 \ s^{-1}$.