The decomposition of a hydrocarbon follows the equationk `= (4.5 xx 10 ^(11) s^(-1) )e^(-28000) K/T.` Calculate `E_a`.

Explanation

To solve the problem of calculating the activation energy \(E_a\) for the decomposition of a hydrocarbon given the rate constant equation, we will follow these steps:

### Step 1: Identify the given equation
The given equation for the rate constant \(k\) is:
\[ k = (4.5 \times 10^{11} \, s^{-1}) e^{-\frac{28000 \, K}{T}} \]

### Step 2: Compare with 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 3: Relate the two equations
From the given equation, we can see that:
\[ -\frac{E_a}{R} = -\frac{28000}{T} \]
This implies:
\[ \frac{E_a}{R} = 28000 \]

### Step 4: Solve for \(E_a\)
Now, we can express \(E_a\) in terms of \(R\):
\[ E_a = 28000 \times R \]

### Step 5: Substitute the value of \(R\)
The value of \(R\) can be taken as:
- \(R = 8.314 \, J/(mol \cdot K)\) for calculations in Joules,
- \(R = 2 \, cal/(mol \cdot K)\) for calculations in calories.

#### For Joules:
\[ E_a = 28000 \times 8.314 \, J/(mol \cdot K) = 233592 \, J/mol \]

#### For Calories:
\[ E_a = 28000 \times 2 \, cal/(mol \cdot K) = 56000 \, cal/mol \]

### Final Result
Thus, the activation energy \(E_a\) is:
- \(E_a = 233592 \, J/mol\) or \(E_a = 56000 \, cal/mol\).