Sure, let's solve the problem step-by-step.
### Step 1: Understand the Given Values
- Rate constant at 120°C ([tex]\( k_1 \)[/tex]) = 1.173 hr[tex]\(^{-1}\)[/tex]
- Rate constant at 140°C ([tex]\( k_2 \)[/tex]) = 4.860 hr[tex]\(^{-1}\)[/tex]
- Temperature at 120°C = 120 + 273.15 K = 393.15 K
- Temperature at 140°C = 140 + 273.15 K = 413.15 K
- Gas constant ([tex]\( R \)[/tex]) = 1.987 cal/(mol·K)
### Step 2: Use the Arrhenius Equation
The Arrhenius equation is given by:
[tex]\[ k = A \cdot e^{-\frac{E_a}{RT}} \][/tex]
Taking the natural logarithm of both sides, we get:
[tex]\[ \ln(k) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T} \][/tex]
### Step 3: Formulate the Equations
By substituting the values at 120°C and 140°C, we form two equations:
[tex]\[ \ln(k_1) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T_1} \][/tex]
[tex]\[ \ln(k_2) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T_2} \][/tex]
### Step 4: Subtract the Equations to Eliminate [tex]\( \ln(A) \)[/tex]
Subtract the first equation from the second:
[tex]\[ \ln(k_2) - \ln(k_1) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \][/tex]
### Step 5: Solve for the Activation Energy [tex]\( E_a \)[/tex]
Rearranging the equation to solve for [tex]\( E_a \)[/tex]:
[tex]\[ E_a = \left(\frac{\ln(k_2) - \ln(k_1)}{\frac{1}{T_2} - \frac{1}{T_1}}\right) \cdot R \][/tex]
Given values:
[tex]\[ \ln\left(\frac{k_2}{k_1}\right) = \ln\left(\frac{4.860}{1.173}\right) \approx 1.550 \][/tex]
[tex]\[ \left(\frac{1}{393.15} - \frac{1}{413.15}\right) \approx -1.244 \times 10^{-4} \][/tex]
Now,
[tex]\[ E_a = \frac{1.550}{-1.244 \times 10^{-4}} \cdot 1.987 \][/tex]
[tex]\[ E_a \approx 22938.91 \, \text{cal/mol} \][/tex]
Convert cal/mol to kcal/mol:
[tex]\[ E_a \approx 22.94 \, \text{kcal/mol} \][/tex]
### Step 6: Calculate the Frequency Factor [tex]\( A \)[/tex]
Using the original Arrhenius equation:
[tex]\[ \ln(k_1) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T_1} \][/tex]
Rearrange to solve for [tex]\( \ln(A) \)[/tex]:
[tex]\[ \ln(A) = \ln(k_1) + \frac{E_a}{R} \cdot \frac{1}{T_1} \][/tex]
Given values:
[tex]\[ \ln(1.173) \approx 0.160 \][/tex]
[tex]\[ \frac{E_a}{R} \cdot \frac{1}{393.15} \approx \frac{22938.91}{1.987} \cdot \frac{1}{393.15} \approx 29.103 \][/tex]
Now,
[tex]\[ \ln(A) \approx 0.160 + 29.103 \approx 29.263 \][/tex]
[tex]\[ A \approx e^{29.263} \approx 6636866151436.70 \, \text{hr}^{-1} \][/tex]
### Step 7: Convert Frequency Factor [tex]\( A \)[/tex] from hr[tex]\(^{-1}\)[/tex] to sec[tex]\(^{-1}\)[/tex]
[tex]\[ A = 6636866151436.70 \, \text{hr}^{-1} \][/tex]
[tex]\[ A = \frac{6636866151436.70}{3600} \approx 1843573930.95 \, \text{sec}^{-1} \][/tex]
### Final Answer:
The activation energy ([tex]\( E_a \)[/tex]) for the decomposition of 5-hydroxymethyle furfural is:
[tex]\[ 22.94 \, \text{kcal/mol} \][/tex]
The frequency factor ([tex]\( A \)[/tex]) is:
[tex]\[ 1843573930.95 \, \text{sec}^{-1} \][/tex]
