### Calculating Activation Energy using the Arrhenius EquationThe rate constant, \(k\), for a certain reaction is measured at two different temperatures:| **Temperature** | \( k \) ||-----------------|------------------|| 330.0 °C | \( 6.1 \times 10^{10} \) s\(^{-1}\) || 442.0 °C | \( 3.7 \times 10^{11} \) s\(^{-1}\) |Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy \( E_a \) for this reaction. Round your answer to 2 significant digits.### Key Concepts:1. **Arrhenius Equation**: The Arrhenius equation is given by:\[ k = A e^{-\frac{E_a}{RT}} \]where:- \( k \) is the rate constant,- \( A \) is the pre-exponential factor (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.2. **Linear Form of Arrhenius Equation**: By taking the natural logarithm of both sides, the equation can be transformed into a linear form:\[ \ln(k) = \ln(A) - \frac{E_a}{R} \frac{1}{T} \]From here, the activation energy \( E_a \) can be determined by measuring the rate constants at different temperatures and creating a plot of \( \ln(k) \) versus \( \frac{1}{T} \). The slope of this line will be \( -\frac{E_a}{R} \).### Step-by-Step Solution:1. **Convert Temperatures to Kelvin**: \[ T_1 = 330.0 + 273.15 = 603.15 \text{ K} \] \[ T_2 = 442.0 + 273.15 = 715.15 \text{ K} \]2. **Calculate \(\ln(k)\)**: \[ \ln(k_1) = \ln(6.1 \times 10^{10}) = 24.83 \] \[ \ln(k

Name: ### Calculating Activation Energy using the Arrhenius EquationThe rat
Uploaded: 2022-11-30T19:17:17.000Z
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Description: ### Calculating Activation Energy using the Arrhenius EquationThe rate constant, \(k\), for a certain reaction is measured at two different temperatures:| **Temperature** | \( k \) ||-----------------|------------------|| 330.0 °C |