The rate constant for the reaction PCl3 (g) + Cl2 (g) → PCl5 (g) is 1.3 x 103 M/min at 515 K, and increases to 2.6 x 103 M/min at 550 K. The activation energy, Eact, in J/mol, is а. 1.3 х 105. b. 1.9 х 105. с. 9.1 x 104. d. 4.7 x 104. е. 4.9х 10°.
The rate constant for the reaction PCl3 (g) + Cl2 (g) → PCl5 (g) is 1.3 x 103 M/min at 515 K, and increases to 2.6 x 103 M/min at 550 K. The activation energy, Eact, in J/mol, is а. 1.3 х 105. b. 1.9 х 105. с. 9.1 x 104. d. 4.7 x 104. е. 4.9х 10°.
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![**Reaction Rate and Activation Energy Calculation**
**Problem Statement:**
The rate constant for the reaction \( \text{PCl}_3 (g) + \text{Cl}_2 (g) \rightarrow \text{PCl}_5 (g) \)
is \(1.3 \times 10^{-3} \, \text{M/min}\) at \(515 \, \text{K}\), and increases to \(2.6 \times 10^{-3} \, \text{M/min}\) at \(550 \, \text{K}\).
The activation energy, \( E_{\text{act}} \), in J/mol, is:
a. \(1.3 \times 10^5\)
b. \(1.9 \times 10^5\)
c. \(9.1 \times 10^4\)
d. \(4.7 \times 10^4\)
e. \(4.9 \times 10^5\)
**Explanation:**
To find the activation energy, \( E_{\text{act}} \), we can use the Arrhenius equation:
\[ k = A e^{\frac{-E_{\text{act}}}{RT}} \]
Where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor (frequency factor)
- \( E_{\text{act}} \) is the activation energy
- \( R \) is the gas constant (\(8.314 \, \text{J/mol·K}\))
- \( T \) is the temperature in Kelvin
We can rearrange the Arrhenius equation to relate the two rate constants at different temperatures:
\[ \ln{\left(\frac{k_2}{k_1}\right)} = \frac{E_{\text{act}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Given:
- \( k_1 = 1.3 \times 10^{-3} \, \text{M/min} \) at \( T_1 = 515 \, K \)
- \( k_2 = 2.6 \times 10^{-3} \, \text{M/min}\) at \( T_2 = 550 \, K \)
The detailed calculation can be completed as follows:
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48deb947-3d41-43de-ac43-bef03e01cc59%2F85eed937-7abc-427a-9353-7ed0ea269fb8%2F2yim5o9.png&w=3840&q=75)
Transcribed Image Text:**Reaction Rate and Activation Energy Calculation**
**Problem Statement:**
The rate constant for the reaction \( \text{PCl}_3 (g) + \text{Cl}_2 (g) \rightarrow \text{PCl}_5 (g) \)
is \(1.3 \times 10^{-3} \, \text{M/min}\) at \(515 \, \text{K}\), and increases to \(2.6 \times 10^{-3} \, \text{M/min}\) at \(550 \, \text{K}\).
The activation energy, \( E_{\text{act}} \), in J/mol, is:
a. \(1.3 \times 10^5\)
b. \(1.9 \times 10^5\)
c. \(9.1 \times 10^4\)
d. \(4.7 \times 10^4\)
e. \(4.9 \times 10^5\)
**Explanation:**
To find the activation energy, \( E_{\text{act}} \), we can use the Arrhenius equation:
\[ k = A e^{\frac{-E_{\text{act}}}{RT}} \]
Where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor (frequency factor)
- \( E_{\text{act}} \) is the activation energy
- \( R \) is the gas constant (\(8.314 \, \text{J/mol·K}\))
- \( T \) is the temperature in Kelvin
We can rearrange the Arrhenius equation to relate the two rate constants at different temperatures:
\[ \ln{\left(\frac{k_2}{k_1}\right)} = \frac{E_{\text{act}}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Given:
- \( k_1 = 1.3 \times 10^{-3} \, \text{M/min} \) at \( T_1 = 515 \, K \)
- \( k_2 = 2.6 \times 10^{-3} \, \text{M/min}\) at \( T_2 = 550 \, K \)
The detailed calculation can be completed as follows:
1
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