3. The following data for the first-order decomposition of penicillin are obtained from Swintosky et al. Temp (°C) Тemp (К) 1/Тemp (1/K) 1st-order k (hr') In k 54 327 3.06 х 10-3 0.119 - 2.129 43 316 3.16 x 10-3 0.0403 - 3.211 37 310 3.23 х 10-3 0.0216 - 3.835 a) Plot the results and calculate the activation energy (E,) and the Arrhenius factor (A). 1/T (K)

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How did my teacher get the 1st order row and the ln k row

The table provided contains data for the first-order decomposition of penicillin, as referenced from Swintosky et al.

| Temp (°C) | Temp (K) | 1/Temp (1/K) | 1st-order k (hr⁻¹) | ln k   |
|-----------|----------|--------------|---------------------|--------|
| 54        | 327      | 3.06 x 10⁻³  | 0.119               | -2.129 |
| 43        | 316      | 3.16 x 10⁻³  | 0.0403              | -3.211 |
| 37        | 310      | 3.23 x 10⁻³  | 0.0216              | -3.835 |

The task is to plot the results and calculate the activation energy (Eₐ) and the Arrhenius factor (A).

The graph is a plot of ln k versus 1/T (K), showing a straight line that represents the relationship described by the Arrhenius equation:

\[ \ln k = \frac{-E_a}{RT} + \ln A \]

**Graph Details:**

- The x-axis represents \( 1/T \) in units of \( \text{K}^{-1} \), ranging from approximately 0.003 to 0.00325.
- The y-axis represents \( \ln k \), ranging from -4 to 0.
- The plot shows a linear trend with a line fitted through the data points, suggesting that the relationship follows the expected linear form of the Arrhenius equation.

**Equation of the Line:**

\[ y = -10820x + 28.987 \]
\[ R^2 = 1 \]

The equation suggests a perfect linear fit, signified by \( R^2 = 1 \). The slope of the line (-10820) can be used to calculate the activation energy \( E_a \), and the intercept (28.987) corresponds to \( \ln A \).
Transcribed Image Text:The table provided contains data for the first-order decomposition of penicillin, as referenced from Swintosky et al. | Temp (°C) | Temp (K) | 1/Temp (1/K) | 1st-order k (hr⁻¹) | ln k | |-----------|----------|--------------|---------------------|--------| | 54 | 327 | 3.06 x 10⁻³ | 0.119 | -2.129 | | 43 | 316 | 3.16 x 10⁻³ | 0.0403 | -3.211 | | 37 | 310 | 3.23 x 10⁻³ | 0.0216 | -3.835 | The task is to plot the results and calculate the activation energy (Eₐ) and the Arrhenius factor (A). The graph is a plot of ln k versus 1/T (K), showing a straight line that represents the relationship described by the Arrhenius equation: \[ \ln k = \frac{-E_a}{RT} + \ln A \] **Graph Details:** - The x-axis represents \( 1/T \) in units of \( \text{K}^{-1} \), ranging from approximately 0.003 to 0.00325. - The y-axis represents \( \ln k \), ranging from -4 to 0. - The plot shows a linear trend with a line fitted through the data points, suggesting that the relationship follows the expected linear form of the Arrhenius equation. **Equation of the Line:** \[ y = -10820x + 28.987 \] \[ R^2 = 1 \] The equation suggests a perfect linear fit, signified by \( R^2 = 1 \). The slope of the line (-10820) can be used to calculate the activation energy \( E_a \), and the intercept (28.987) corresponds to \( \ln A \).
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