To determine the order of the reaction and calculate the rate constant, we start by analyzing the given concentration data for cyclopropane at different times.
Given data:
- Times (in minutes): [tex]\(0, 5.0, 10.0, 15.0, 20.0\)[/tex]
- Concentrations of cyclopropane (in mol/L): [tex]\(0.098, 0.080, 0.066, 0.054, 0.0\)[/tex]
We are given the following analysis results:
1. Times: [tex]\(0, 5.0, 10.0, 15.0, 20.0\)[/tex]
2. Concentrations: [tex]\(0.098, 0.080, 0.066, 0.054, 0.0\)[/tex]
3. Natural Logarithm of Concentrations: [tex]\(\ln (0.098), \ln(0.080), \ln(0.066), \ln(0.054)\)[/tex]
Corresponding values: [tex]\(-2.3228, -2.5257, -2.7181, -2.9188\)[/tex]
(Note: The last concentration is zero and thus its logarithm isn't calculated)
From the given results, we identify:
4. Slope of the regression line: [tex]\(-0.039606\)[/tex]
5. Intercept of the regression line: [tex]\(-2.3242987\)[/tex]
6. Rate constant [tex]\(k\)[/tex]: [tex]\(0.039606 \, \text{min}^{-1}\)[/tex]
### Solution Steps:
1. Checking the Reaction Order:
- We assume the reaction is first-order and thus plot [tex]\(\ln[\text{Cyclopropane}]\)[/tex] vs. time. If the plot is linear, it's a first-order reaction.
- The results show a linear relationship between [tex]\(\ln[\text{Cyclopropane}]\)[/tex] and time, confirming first-order reaction behavior.
2. Calculating the logarithms of the given concentrations:
[tex]\[
\begin{aligned}
\ln(0.098) & \approx -2.3228, \\
\ln(0.080) & \approx -2.5257, \\
\ln(0.066) & \approx -2.7181, \\
\ln(0.054) & \approx -2.9188.
\end{aligned}
\][/tex]
3. Performing Linear Regression:
- Using the times [tex]\(0, 5.0, 10.0, 15.0\)[/tex] and the corresponding logarithms:
[tex]\[
\begin{aligned}
t (min) & : 0, 5.0, 10.0, 15.0, \\
\ln[\text{Cyclopropane}] & : -2.3228, -2.5257, -2.7181, -2.9188.
\end{aligned}
\][/tex]
- The slope ([tex]\(m\)[/tex]) of the linear fit line can be determined:
[tex]\[
\text{slope} \approx -0.039606.
\][/tex]
4. Calculating the Rate Constant [tex]\(k\)[/tex]:
- For a first-order reaction, the rate constant [tex]\(k\)[/tex] is the negative of the slope of the [tex]\(\ln[\text{Cyclopropane}]\)[/tex] vs. time plot.
[tex]\[
k = - (\text{slope}) = 0.039606 \, \text{min}^{-1}.
\][/tex]
### Conclusion:
- The reaction is confirmed to be first-order.
- The rate constant [tex]\(k\)[/tex] for the decomposition of cyclopropane is [tex]\(0.039606 \, \text{min}^{-1}\)[/tex].
