The reaction C,H;(g) → 2 C,H, (g) has an activation energy of 262 kJ/mol. At 600.0 K, the rate constant, k, is 6.1 × 10-8 s-1. What is the value of the rate constant at 750.0 K? k = 1.34 х103 s-1 Incorrect

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### General Chemistry 4th Edition 
### McQuarrie • Rock • Gallogly 
### University Science Books presented by Macmillan Learning

---

### Reaction Analysis

The reaction:

\[ \text{C}_4\text{H}_8 (g) \longrightarrow 2\text{C}_2\text{H}_4 (g) \]

is characterized by an activation energy of 262 kJ/mol. At a temperature of 600.0 K, the rate constant, \( k \), is \( 6.1 \times 10^{-8} \, \text{s}^{-1} \). The question posed is to determine the value of the rate constant at 750.0 K.

A student attempted to solve the problem and submitted the following answer for the rate constant (\( k \)):

\[ k = 1.34 \times 10^3 \, \text{s}^{-1} \]

However, this answer is marked as incorrect.

---

To solve this problem correctly, one could use the Arrhenius equation:

\[ k = A e^{-\frac{E_a}{RT}} \]

where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor or frequency factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin.

By substituting the given values of activation energy (\( E_a = 262 \, \text{kJ/mol} \)), and the temperatures (initial \( T_1 = 600.0 \, \text{K} \) and final \( T_2 = 750.0 \, \text{K} \)), we can apply the two-point form of Arrhenius equation:

\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Given:
- \( k_1 = 6.1 \times 10^{-8} \, \text{s}^{-1} \)
- \( E_a = 262 \, \text{kJ/mol} = 262000 \, \text{J/mol} \)

After solving for
Transcribed Image Text:### General Chemistry 4th Edition ### McQuarrie • Rock • Gallogly ### University Science Books presented by Macmillan Learning --- ### Reaction Analysis The reaction: \[ \text{C}_4\text{H}_8 (g) \longrightarrow 2\text{C}_2\text{H}_4 (g) \] is characterized by an activation energy of 262 kJ/mol. At a temperature of 600.0 K, the rate constant, \( k \), is \( 6.1 \times 10^{-8} \, \text{s}^{-1} \). The question posed is to determine the value of the rate constant at 750.0 K. A student attempted to solve the problem and submitted the following answer for the rate constant (\( k \)): \[ k = 1.34 \times 10^3 \, \text{s}^{-1} \] However, this answer is marked as incorrect. --- To solve this problem correctly, one could use the Arrhenius equation: \[ k = A e^{-\frac{E_a}{RT}} \] where: - \( k \) is the rate constant, - \( A \) is the pre-exponential factor or frequency factor, - \( E_a \) is the activation energy, - \( R \) is the universal gas constant (8.314 J/mol·K), - \( T \) is the temperature in Kelvin. By substituting the given values of activation energy (\( E_a = 262 \, \text{kJ/mol} \)), and the temperatures (initial \( T_1 = 600.0 \, \text{K} \) and final \( T_2 = 750.0 \, \text{K} \)), we can apply the two-point form of Arrhenius equation: \[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] Given: - \( k_1 = 6.1 \times 10^{-8} \, \text{s}^{-1} \) - \( E_a = 262 \, \text{kJ/mol} = 262000 \, \text{J/mol} \) After solving for
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