The rate constants of a reaction at 700 K and 760 K are `0.011" M"^(-1)s^(-1)` and `0.105" M"^(-1)s^(-1)` respectively.
Calculate the values of Arrhenius parameters.

Answer

To solve the problem of finding the Arrhenius parameters (A and Ea) from the given rate constants at two different temperatures, we can follow these steps:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A \cdot e^{-\frac{E_a}{RT}} \]
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.

### Step 2: Set Up the Equations for Two Temperatures
For the two temperatures \( T_1 = 700 \, K \) and \( T_2 = 760 \, K \), we can write:
1. \( k_1 = A \cdot e^{-\frac{E_a}{RT_1}} \) (for \( T_1 \))
2. \( k_2 = A \cdot e^{-\frac{E_a}{RT_2}} \) (for \( T_2 \))

Given:
- \( k_1 = 0.011 \, M^{-1}s^{-1} \)
- \( k_2 = 0.105 \, M^{-1}s^{-1} \)

### Step 3: Divide the Two Equations
Dividing the two equations gives:
\[ \frac{k_2}{k_1} = \frac{A \cdot e^{-\frac{E_a}{RT_2}}}{A \cdot e^{-\frac{E_a}{RT_1}}} \]
This simplifies to:
\[ \frac{k_2}{k_1} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]

### Step 4: Substitute Known Values
Substituting the known values:
- \( k_1 = 0.011 \)
- \( k_2 = 0.105 \)
- \( R = 8.314 \, J/mol·K \)
- \( T_1 = 700 \, K \)
- \( T_2 = 760 \, K \)

Calculate:
\[ \ln\left(\frac{0.105}{0.011}\right) = \ln(9.545) \approx 2.259 \]
Now substituting into the equation:
\[ 2.259 = -\frac{E_a}{8.314}\left(\frac{1}{760} - \frac{1}{700}\right) \]

### Step 5: Calculate the Temperature Difference
Calculate the difference:
\[ \frac{1}{760} - \frac{1}{700} = \frac{700 - 760}{760 \cdot 700} = \frac{-60}{532000} \approx -0.0001125 \]

### Step 6: Solve for Activation Energy \( E_a \)
Now substituting this back into the equation:
\[ 2.259 = -\frac{E_a}{8.314} \cdot (-0.0001125) \]
Rearranging gives:
\[ E_a = \frac{2.259 \cdot 8.314}{0.0001125} \approx 166.3 \, kJ/mol \]

### Step 7: Calculate the Pre-exponential Factor \( A \)
Now we can use one of the original Arrhenius equations to solve for \( A \). Using \( k_1 \):
\[ 0.011 = A \cdot e^{-\frac{166300}{8.314 \cdot 700}} \]
Calculating the exponent:
\[ -\frac{166300}{8.314 \cdot 700} \approx -28.8 \]
Thus:
\[ 0.011 = A \cdot e^{-28.8} \]
Calculating \( e^{-28.8} \):
\[ e^{-28.8} \approx 0.00000005 \]
Now substituting back:
\[ A = \frac{0.011}{0.00000005} \approx 2.2 \times 10^6 \, M^{-1}s^{-1} \]

### Final Values
- Activation Energy \( E_a \approx 166.3 \, kJ/mol \)
- Pre-exponential Factor \( A \approx 2.2 \times 10^6 \, M^{-1}s^{-1} \)