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