The rate constant for the decomposition of `N_2O_5` in `CCl_4` is 6.2 x 10`^(-4) s^(-1)` at 45°C. Calculate the rate constant at 100°C if the activation energy is 103 kJ `mol^(-1)` [Ant (2.49) = 309]

Options

A. `1.9 x 10^(-1) s^(-1)`
B. `3.5 x 10^(-2) s^(-1)`
C. `4.5 x 10^(-1) s^(-1)`
D. `1.75 x 10^(-2) s^(-1)`

Answer

To solve the problem of calculating the rate constant \( k_2 \) for the decomposition of \( N_2O_5 \) at 100°C given the rate constant \( k_1 \) at 45°C and the activation energy \( E_a \), we can use the Arrhenius equation in its logarithmic form. Here’s a step-by-step breakdown of the solution:

### Step 1: Write down the Arrhenius equation in logarithmic form

The Arrhenius equation relates the rate constant \( k \) to the activation energy \( E_a \) and temperature \( T \):

\[ k = A e^{-\frac{E_a}{RT}} \]

Taking the logarithm, we have:

\[ \log k = \log A - \frac{E_a}{2.303RT} \]

### Step 2: Set up the equation for two temperatures

For two different temperatures \( T_1 \) and \( T_2 \), we can express the equation as:

\[ \log k_2 - \log k_1 = -\frac{E_a}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

Rearranging gives:

\[ \log \frac{k_2}{k_1} = -\frac{E_a}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

### Step 3: Convert temperatures to Kelvin

Convert the temperatures from Celsius to Kelvin:

- \( T_1 = 45°C = 45 + 273 = 318 \, K \)
- \( T_2 = 100°C = 100 + 273 = 373 \, K \)

### Step 4: Substitute known values into the equation

Given:
- \( k_1 = 6.2 \times 10^{-4} \, s^{-1} \)
- \( E_a = 103 \, kJ/mol = 103 \times 10^3 \, J/mol \)
- \( R = 8.314 \, J/(mol \cdot K) \)

Now, substitute these values into the equation:

\[ \log \frac{k_2}{6.2 \times 10^{-4}} = -\frac{103 \times 10^3}{2.303 \times 8.314} \left( \frac{1}{373} - \frac{1}{318} \right) \]

### Step 5: Calculate the right-hand side

First, calculate the term \( \frac{1}{373} - \frac{1}{318} \):

\[ \frac{1}{373} \approx 0.002684 \quad \text{and} \quad \frac{1}{318} \approx 0.003145 \]

So,

\[ \frac{1}{373} - \frac{1}{318} \approx 0.002684 - 0.003145 = -0.000461 \]

Now calculate \( -\frac{103 \times 10^3}{2.303 \times 8.314} \):

\[ -\frac{103000}{2.303 \times 8.314} \approx -\frac{103000}{19.091} \approx -5383.5 \]

Now multiply:

\[ -5383.5 \times (-0.000461) \approx 2.48 \]

### Step 6: Solve for \( k_2 \)

Now we have:

\[ \log \frac{k_2}{6.2 \times 10^{-4}} \approx 2.48 \]

Thus,

\[ \frac{k_2}{6.2 \times 10^{-4}} \approx 10^{2.48} \approx 301.2 \]

Therefore,

\[ k_2 \approx 301.2 \times 6.2 \times 10^{-4} \approx 0.186 \, s^{-1} \]

### Final Answer

The rate constant \( k_2 \) at 100°C is approximately:

\[ k_2 \approx 1.86 \times 10^{-1} \, s^{-1} \]