3)can someone explain the question ### Chemical Kinetics Problem**Problem Statement:**"A reaction has a rate constant of 0.0117 M/s at 400.0 K and 0.689 M/s at 450.0 K. What is the activation energy of the reaction?"**Graphical Explanation (if applicable):**To calculate the activation energy, you can use the Arrhenius equation:\[ k = A e^{-\frac{E_a}{RT}} \]where:- \( k \) is the rate constant,- \( A \) is the pre-exponential factor,- \( E_a \) is the activation energy,- \( R \) is the universal gas constant (8.314 J/(mol·K)),- \( T \) is the temperature in Kelvin.Taking the natural logarithm of both sides gives:\[ \ln(k) = \ln(A) - \frac{E_a}{RT} \]For two different temperatures, the equation becomes:\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]Given:- \( k_{1} = 0.0117 \, \text{M/s} \) at \( T_{1} = 400.0 \, \text{K} \)- \( k_{2} = 0.689 \, \text{M/s} \) at \( T_{2} = 450.0 \, \text{K} \)We can solve for \( E_a \):\[ \ln\left(\frac{0.689}{0.0117}\right) = -\frac{E_a}{8.314} \left( \frac{1}{450.0} - \frac{1}{400.0} \right) \]Calculating the left side:\[ \ln\left(\frac{0.689}{0.0117}\right) = \ln(58.8034) \approx 4.0732 \]Calculating the right side:\[ \frac{1}{450.0} - \frac{1}{400.0} = 0.002222 - 0.0025 = -0.000278 \]Then:\[

Name: 3)can someone explain the question ### Chemical Kinetics Problem**Prob
Uploaded: 2022-11-30T19:17:17.000Z
Channel: Bartleby
Description: 3)can someone explain the question ### Chemical Kinetics Problem**Problem Statement:**"A reaction has a rate constant of 0.0117 M/s at 400.0 K and 0.689 M/s at 450.0 K. What is the activation energy of the reaction?"**Graphical Explanation (if applic