Question

question 18 of 23 > the rate of a certain reaction was studied at various temperatures. the table shows temperature (t) and rate constant (k) data collected during the experiments. plot the data to answer the questions. what is the value of the activation energy, $e_a$, for this reaction? $e_a=\text{ }kjcdot mol^{-1}$ what is the value of the pre - exponential factor (sometimes called the frequency factor), $a$, for this reaction? $a=\text{ }s^{-1}$ t(k) k(s⁻¹) 400 0.0000173 420 0.000127 440 0.000773 460 0.00403 480 0.0183 500 0.0738 520 0.267 540 0.879 560 2.66 580 7.43

Explanation:

Step1: Recall Arrhenius equation

The Arrhenius equation is $k = A e^{-\frac{E_a}{RT}}$, which can be rewritten in linear - form as $\ln k=-\frac{E_a}{R}\times\frac{1}{T}+\ln A$. Here, $k$ is the rate constant, $A$ is the pre - exponential factor, $E_a$ is the activation energy, $R = 8.314\ J/(mol\cdot K)$ is the gas constant, and $T$ is the temperature in Kelvin.

Step2: Calculate $\frac{1}{T}$ and $\ln k$ values

For each data point in the table:

$T(K)$	$\frac{1}{T}(K^{-1})$	$k(s^{-1})$	$\ln k$
420	$\frac{1}{420}\approx0.00238$	$0.000127$	$\ln(0.000127)\approx - 8.97$
440	$\frac{1}{440}\approx0.00227$	$0.000773$	$\ln(0.000773)\approx - 7.17$
460	$\frac{1}{460}\approx0.00217$	$0.00403$	$\ln(0.00403)\approx - 5.51$
480	$\frac{1}{480}\approx0.00208$	$0.0183$	$\ln(0.0183)\approx - 4.00$
500	$\frac{1}{500}=0.00200$	$0.0738$	$\ln(0.0738)\approx - 2.61$
520	$\frac{1}{520}\approx0.00192$	$0.267$	$\ln(0.267)\approx - 1.32$
540	$\frac{1}{540}\approx0.00185$	$0.879$	$\ln(0.879)\approx - 0.13$
560	$\frac{1}{560}\approx0.00179$	$2.66$	$\ln(2.66)\approx0.98$
580	$\frac{1}{580}\approx0.00172$	$7.43$	$\ln(7.43)\approx2.00$

Step3: Plot $\ln k$ vs $\frac{1}{T}$

Plot the points $(\frac{1}{T},\ln k)$ on a graph. The slope of the line is $m =-\frac{E_a}{R}$.

Step4: Calculate the slope

Using linear regression or by taking two well - separated points (e.g., $(0.0025, - 10.97)$ and $(0.00172,2.00)$) to calculate the slope $m$.
$m=\frac{2.00-( - 10.97)}{0.00172 - 0.0025}=\frac{12.97}{-0.00078}\approx - 16628.2\ K$
Since $m =-\frac{E_a}{R}$, then $E_a=-mR$.
$E_a=16628.2\ K\times8.314\ J/(mol\cdot K)=138250\ J/mol = 138.3\ kJ/mol$

Step5: Calculate the pre - exponential factor

The $y$ - intercept of the line $\ln k$ vs $\frac{1}{T}$ is $\ln A$.
We can also use one of the data points in the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$. Let's use the first data point $T = 400\ K$, $k = 0.0000173\ s^{-1}$, $E_a = 138250\ J/mol$, $R = 8.314\ J/(mol\cdot K)$.
$0.0000173=Ae^{-\frac{138250}{8.314\times400}}$
$0.0000173=Ae^{-41.6}$
$A=\frac{0.0000173}{e^{-41.6}}=0.0000173\times e^{41.6}\approx1.2\times10^{16}\ s^{-1}$

Answer:

$E_a = 138.3\ kJ/mol$
$A = 1.2\times10^{16}\ s^{-1}$