Certainly! Let's go through the solution step-by-step.
### Part A: Transforming the Data and Creating a Linear Plot
To begin with, let's transform the given data and then create a linear plot.
#### Transforming the Data
We have the temperature data [tex]\( T \)[/tex] and the rate constants [tex]\( k \)[/tex] as follows:
[tex]\[
\begin{array}{|c|c|}
\hline
T (K) & k (M^{-1}s^{-1}) \\
\hline
288 & 0.0521 \\
298 & 0.101 \\
308 & 0.184 \\
318 & 0.332 \\
\hline
\end{array}
\][/tex]
1. Calculate [tex]\( \frac{1}{T} \)[/tex]:
[tex]\[
\frac{1}{T} \text {(in K}^{-1}) \text{ for each temperature is:}
\][/tex]
[tex]\[
\frac{1}{288} = 0.003472 \, \text{K}^{-1}
\][/tex]
[tex]\[
\frac{1}{298} = 0.003356 \, \text{K}^{-1}
\][/tex]
[tex]\[
\frac{1}{308} = 0.003247 \, \text{K}^{-1}
\][/tex]
[tex]\[
\frac{1}{318} = 0.003145 \, \text{K}^{-1}
\][/tex]
2. Calculate [tex]\( \ln(k) \)[/tex]:
[tex]\[
\ln(0.0521) = -2.955
\][/tex]
[tex]\[
\ln(0.101) = -2.293
\][/tex]
[tex]\[
\ln(0.184) = -1.693
\][/tex]
[tex]\[
\ln(0.332) = -1.103
\][/tex]
So, the transformed data is:
[tex]\[
\begin{array}{|c|c|}
\hline
\frac{1}{T} (K^{-1}) & \ln(k) \\
\hline
0.003472 & -2.955 \\
0.003356 & -2.293 \\
0.003247 & -1.693 \\
0.003145 & -1.103 \\
\hline
\end{array}
\][/tex]
#### Creating the Linear Plot
Using Excel or any graphing tool, plot [tex]\( \ln(k) \)[/tex] (y-axis) vs. [tex]\( \frac{1}{T} \)[/tex] (x-axis).
1. Labeling:
- Title: Arrhenius Plot
- x-axis: [tex]\( \frac{1}{T} \)[/tex] (K[tex]\(^{-1}\)[/tex])
- y-axis: [tex]\( \ln(k) \)[/tex]
2. Plot the Data Points and Add Linear Fit:
- Display the equation of the line on the graph.
The equation of the line should look something like this (based on linear regression trendline):
[tex]\[ \ln k = -5638.34 \left(\frac{1}{T}\right) + 16.6231 \][/tex]
### Part B: Calculating Activation Energy [tex]\( E_a \)[/tex]
From the equation of the line generated from plotting the data, we have:
[tex]\[ \ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A \][/tex]
Comparing this with the equation of our line:
[tex]\[ \ln k = -5638.34 \left(\frac{1}{T}\right) + 16.6231 \][/tex]
The slope of the line is [tex]\(-5638.34\)[/tex], and it is equal to [tex]\(-\frac{E_a}{R}\)[/tex].
1. Calculate [tex]\( E_a \)[/tex]:
[tex]\[
-5638.34 = -\frac{E_a}{R}
\][/tex]
[tex]\[
5638.34 = \frac{E_a}{8.314}
\][/tex]
[tex]\[
E_a = 5638.34 \times 8.314 = 46877.16 \, \text{J/mol}
\][/tex]
2. Convert [tex]\( E_a \)[/tex] to kJ/mol:
[tex]\[
E_a (kJ/mol) = \frac{46877.16}{1000} = 46.877 \, \text{kJ/mol}
\][/tex]
### Part C: Determining the Order of the Reaction
From the units of the rate constant [tex]\( k \)[/tex]:
- The units of [tex]\( k \)[/tex] are [tex]\( M^{-1} s^{-1} \)[/tex].
For a reaction of order [tex]\( n \)[/tex]:
[tex]\[ \text{Rate Constant (k)} \text{ units} = \left( \text{Concentration} \right)^{1-n} \left( \text{Time} \right)^{-1} = \left(M^{1-n}\right)s^{-1} \][/tex]
Given that [tex]\( k \)[/tex] has units [tex]\( M^{-1} s^{-1} \)[/tex]:
- Thus, [tex]\( 1-n = -1 \)[/tex]
- This implies that [tex]\( n = 2 \)[/tex].
The reaction is second order.
Rate Law Expression for a second-order reaction:
[tex]\[ \text{Rate} = k[B]^2 \][/tex]
### Summary
- Transformed data and created a linear plot with the equation of the line as [tex]\( \ln k = -5638.34 \left(\frac{1}{T}\right) + 16.6231 \)[/tex].
- Calculated the activation energy [tex]\( E_a \)[/tex] to be [tex]\( 46.877 \, \text{kJ/mol} \)[/tex].
- Determined that the reaction is second order with the rate law expression: [tex]\( \text{Rate} = k[B]^2 \)[/tex].
