For the reaction:
`[Cr(H_(2)O)_(6)]^(3) + [SCN^(ɵ)] rarr [Cr(H_(2)O)_(5)NCS]^(2+)H_(2)O`
The rate law is `r = k[Cr(H_(2)O)_(6)]^(3+)[SCN^(ɵ)]`.
The value of `k` is `2.0 xx 10^(-6) L mol^(-1) s^(-1)` at `14^(@)C` and `2.2 xx 10^(-5) L mol^(-1)s^(-1)` at `30^(@)C`. What is the value of `E_(a)` ?

Options

A. `26 "kcal mol"^(-1)`
B. `2.6 "kcal mol"^(-1)`
C. `2600 "kcal mol"^(-1)`
D. `260 "kcal mol"^(-1)`

Explanation

To find the activation energy \( E_a \) for the given reaction, we can use the Arrhenius equation and the provided rate constants at two different temperatures. Here’s a step-by-step solution:

### Step 1: Write the Arrhenius equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Use the logarithmic form of the Arrhenius equation
We can rearrange the Arrhenius equation for two different temperatures \( T_1 \) and \( T_2 \):
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

### Step 3: Identify the given values
From the problem:
- \( k_1 = 2.0 \times 10^{-6} \, \text{L mol}^{-1} \text{s}^{-1} \) at \( T_1 = 14^\circ C = 287 \, K \)
- \( k_2 = 2.2 \times 10^{-5} \, \text{L mol}^{-1} \text{s}^{-1} \) at \( T_2 = 30^\circ C = 303 \, K \)

### Step 4: Calculate \( \log\left(\frac{k_2}{k_1}\right) \)
First, calculate \( \frac{k_2}{k_1} \):
\[ \frac{k_2}{k_1} = \frac{2.2 \times 10^{-5}}{2.0 \times 10^{-6}} = 11 \]
Now, calculate the logarithm:
\[ \log(11) \approx 1.0414 \]

### Step 5: Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \)
Calculate:
\[ \frac{1}{T_1} = \frac{1}{287} \quad \text{and} \quad \frac{1}{T_2} = \frac{1}{303} \]
Thus,
\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{287} - \frac{1}{303} \approx 0.00348 - 0.00330 = 0.00018 \, K^{-1} \]

### Step 6: Substitute values into the equation
Using \( R = 2 \, \text{cal mol}^{-1} K^{-1} \):
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 \times 2} \left(0.00018\right) \]
Substituting the values:
\[ 1.0414 = \frac{E_a}{4.606} \times 0.00018 \]

### Step 7: Solve for \( E_a \)
Rearranging gives:
\[ E_a = \frac{1.0414 \times 4.606}{0.00018} \]
Calculating:
\[ E_a \approx \frac{4.8}{0.00018} \approx 26666.67 \, \text{cal/mol} \]
This can be approximated to:
\[ E_a \approx 26667 \, \text{cal/mol} \approx 26.67 \, \text{kcal/mol} \]

### Final Answer
Thus, the activation energy \( E_a \) is approximately \( 26.67 \, \text{kcal/mol} \).

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