The rate constant for the decompoistion of a certain reaction is described by the equation:
`log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)`
Energy of activation (in `kcal`) is

Options

A. `57.6 kcal`
B. `1.25 xx 10^(4) kcal`
C. `14.0 kcal`
D. `14 xx 10^(4) kcal`

Explanation

To find the energy of activation (Ea) for the decomposition reaction described by the equation:

\[ \log k (s^{-1}) = 14 - \frac{1.25 \times 10^{4} K}{T} \]

we can follow these steps:

### Step 1: Identify the Arrhenius Equation
The given equation is in the form of the Arrhenius equation, which can be expressed as:

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

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: Compare the Given Equation with the Arrhenius Equation
From the given equation:

\[ \log k = 14 - \frac{1.25 \times 10^{4} K}{T} \]

We can compare this with the Arrhenius equation:

1. From the equation, we can see that \( \log A = 14 \).
2. Also, we can equate the coefficients of \( \frac{1}{T} \) to find \( \frac{E_a}{2.303 R} \):

\[ \frac{E_a}{2.303 R} = 1.25 \times 10^{4} \]

### Step 3: Solve for Activation Energy (Ea)
To find \( E_a \), we need to rearrange the equation:

\[ E_a = 1.25 \times 10^{4} \times 2.303 R \]

### Step 4: Substitute the Value of R
The value of the gas constant \( R \) in kilocalories is:

\[ R = 2 \times 10^{-3} \text{ kcal/mol K} \]

Substituting this value into the equation:

\[ E_a = 1.25 \times 10^{4} \times 2.303 \times (2 \times 10^{-3}) \]

### Step 5: Calculate the Activation Energy
Now, we perform the calculation:

1. Calculate \( 1.25 \times 10^{4} \times 2.303 \):
\[ = 28862.5 \]

2. Now multiply by \( 2 \times 10^{-3} \):
\[ E_a = 28862.5 \times 2 \times 10^{-3} = 57.725 \text{ kcal} \]

### Step 6: Round the Result
Rounding \( 57.725 \) to one decimal place gives us:

\[ E_a \approx 57.6 \text{ kcal} \]

### Final Answer
The energy of activation \( E_a \) is approximately \( 57.6 \text{ kcal} \).

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