The rate constant of a reaction is `1.5xx10^(-4)s^(-1)" at "27^(@)C` and `3xx10^(-4)s^(-1)" at "127^(@)C`. The `E_a` is

Options

A. `1.663xx10^(3)" Cal"`
B. `3.326xx10^(3)" cal"`
C. `8.314xx10^(3)" cal"`
D. `2.255xx10^(3)" cal"`

Answer

To find the activation energy (Ea) for the given reaction using the Arrhenius equation, we can follow these steps:

### Step 1: Write down the given data
- Rate constant at T1 (k1) = \(1.5 \times 10^{-4} \, s^{-1}\) at \(T1 = 27^\circ C\)
- Rate constant at T2 (k2) = \(3.0 \times 10^{-4} \, s^{-1}\) at \(T2 = 127^\circ C\)

### Step 2: Convert temperatures to Kelvin
- \(T1 = 27^\circ C = 273 + 27 = 300 \, K\)
- \(T2 = 127^\circ C = 273 + 127 = 400 \, K\)

### Step 3: Use the Arrhenius equation
The Arrhenius equation in logarithmic form is given by:
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{2.303R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
Where:
- \(R\) is the gas constant. Since the answer is required in calories, we will use \(R = 2 \, \text{cal} \, \text{mol}^{-1} \, \text{K}^{-1}\).

### Step 4: Substitute values into the equation
Substituting the known values:
\[ \log\left(\frac{3.0 \times 10^{-4}}{1.5 \times 10^{-4}}\right) = \frac{-E_a}{2.303 \times 2}\left(\frac{1}{400} - \frac{1}{300}\right) \]

### Step 5: Calculate the left side
Calculating the logarithm:
\[ \frac{3.0 \times 10^{-4}}{1.5 \times 10^{-4}} = 2 \quad \Rightarrow \quad \log(2) \approx 0.301 \]

### Step 6: Calculate the right side
Calculate the difference in reciprocals:
\[ \frac{1}{400} - \frac{1}{300} = \frac{3 - 4}{1200} = -\frac{1}{1200} \]

### Step 7: Substitute and solve for Ea
Now substituting back:
\[ 0.301 = \frac{-E_a}{4.606} \left(-\frac{1}{1200}\right) \]
Rearranging gives:
\[ E_a = 0.301 \times 4.606 \times 1200 \]

### Step 8: Calculate Ea
Calculating the values:
\[ E_a = 0.301 \times 4.606 \times 1200 \approx 1663 \, \text{cal} \, \text{mol}^{-1} \]

### Final Answer
The activation energy \(E_a\) is approximately \(1.663 \times 10^3 \, \text{cal} \, \text{mol}^{-1}\).

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