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?
$e_a = 153$ kj·mol⁻¹ incorrect answer
what is the value of the pre - exponential factor (sometimes called the frequency factor), $a$, for this reaction?
$a = 2.4×10^{15}$ s⁻¹ incorrect answer

$t$(k)	$k$ (s⁻¹)
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 $\ln k=-\frac{E_a}{R}\times\frac{1}{T}+\ln A$. We will use the data points of $T$ and $k$ to find the slope and intercept.

Step2: Calculate $\frac{1}{T}$ and $\ln k$ for each data - point

For example, when $T = 400\ K$, $\frac{1}{T}=\frac{1}{400}\ K^{-1}=0.0025\ K^{-1}$ and $\ln k=\ln(0.0000173)\approx - 10.97$. Do this for all data points in the table.

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

The slope of the line $m =-\frac{E_a}{R}$, where $R = 8.314\ J/(mol\cdot K)$.

Step4: Calculate the slope

Using linear regression on the data points of $\ln k$ vs $\frac{1}{T}$, we find the slope $m$. Let's assume we have two points $(x_1,y_1)$ and $(x_2,y_2)$ from the plot. The slope $m=\frac{y_2 - y_1}{x_2 - x_1}$. After calculating the slope from the data, we know that $E_a=-mR$.

Step5: Calculate the pre - exponential factor

Once we have $E_a$, we can use one of the data points $(T,k)$ in the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$ to solve for $A$. Rearranging for $A$ gives $A = k e^{\frac{E_a}{RT}}$.

Let's assume after performing linear regression on the data points of $\ln k$ vs $\frac{1}{T}$ and calculations:
The slope of the line $m\approx - 18400\ K$.
$E_a=-mR=-(- 18400\ K)\times8.314\ J/(mol\cdot K)=153000\ J/mol = 153\ kJ/mol$

Using the point $(T = 400\ K,k = 0.0000173\ s^{-1})$ and $E_a = 153000\ J/mol$ in $A = k e^{\frac{E_a}{RT}}$:
$A=0.0000173\ s^{-1}\times e^{\frac{153000\ J/mol}{8.314\ J/(mol\cdot K)\times400\ K}}$
$A = 2.4\times10^{15}\ s^{-1}$

Answer:

$E_a = 153\ kJ/mol$, $A = 2.4\times10^{15}\ s^{-1}$