The rate constant for the reaction below was determined to be 3.241x10-5 s-1 at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 9.30×10² K? N₂O(g) → N₂(g) + 0(g)

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### Example Problem: Determining the Rate Constant for a Reaction

#### Problem Statement
The rate constant for the reaction below was determined to be \( 3.241 \times 10^{-5} \, \text{s}^{-1} \) at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 930 K?

\[ \text{N}_2\text{O}(g) \rightarrow \text{N}_2(g) + \text{O}(g) \]

#### Steps to Solve

To determine the new rate constant at a different temperature, we can use the Arrhenius equation shown below:

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

However, since we are comparing two temperatures, we can use the following form of the Arrhenius equation:

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

Where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.314 J/mol·K),
- \( T_1 \) and \( T_2 \) are the absolute temperatures (in Kelvin).

Using the given values:
- \( k_1 = 3.241 \times 10^{-5} \, \text{s}^{-1} \)
- \( T_1 = 800 \, \text{K} \)
- \( T_2 = 930 \, \text{K} \)
- \( E_a = 225 \, \text{kJ/mol} = 225000 \, \text{J/mol} \)

Calculate:

\[ \ln \left( \frac{k_2}{3.241 \times 10^{-5}} \right) = \frac{225000}{8.314} \left( \frac{1}{800} - \frac{1}{930} \right) \]

Calculate inside the parenthesis first:

\[ \frac{1}{800} - \frac{
Transcribed Image Text:### Example Problem: Determining the Rate Constant for a Reaction #### Problem Statement The rate constant for the reaction below was determined to be \( 3.241 \times 10^{-5} \, \text{s}^{-1} \) at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 930 K? \[ \text{N}_2\text{O}(g) \rightarrow \text{N}_2(g) + \text{O}(g) \] #### Steps to Solve To determine the new rate constant at a different temperature, we can use the Arrhenius equation shown below: \[ k = Ae^{-\frac{E_a}{RT}} \] However, since we are comparing two temperatures, we can use the following form of the Arrhenius equation: \[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] Where: - \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively, - \( E_a \) is the activation energy, - \( R \) is the gas constant (8.314 J/mol·K), - \( T_1 \) and \( T_2 \) are the absolute temperatures (in Kelvin). Using the given values: - \( k_1 = 3.241 \times 10^{-5} \, \text{s}^{-1} \) - \( T_1 = 800 \, \text{K} \) - \( T_2 = 930 \, \text{K} \) - \( E_a = 225 \, \text{kJ/mol} = 225000 \, \text{J/mol} \) Calculate: \[ \ln \left( \frac{k_2}{3.241 \times 10^{-5}} \right) = \frac{225000}{8.314} \left( \frac{1}{800} - \frac{1}{930} \right) \] Calculate inside the parenthesis first: \[ \frac{1}{800} - \frac{
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