A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is again 50% complete in 5 minutes at 350 K, Calculate the activation energy if it is a first order reaction. [`R = 8.314 JK^(-1) mol^(-1) , log 4 = 0.602`]

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the rate constants (K1 and K2)
Since the reaction is first order, we can use the half-life formula for first-order reactions:
\[ K = \frac{0.693}{t_{1/2}} \]
Where \(t_{1/2}\) is the half-life of the reaction.

For the first temperature (300 K):
\[ t_{1/2} = 20 \text{ minutes} = 20 \times 60 = 1200 \text{ seconds} \]
Thus,
\[ K_1 = \frac{0.693}{1200} = 0.0005775 \text{ s}^{-1} \]

For the second temperature (350 K):
\[ t_{1/2} = 5 \text{ minutes} = 5 \times 60 = 300 \text{ seconds} \]
Thus,
\[ K_2 = \frac{0.693}{300} = 0.00231 \text{ s}^{-1} \]

### Step 2: Use the Arrhenius equation
The relationship between the rate constants and temperature can be expressed using the Arrhenius equation:
\[ \ln\left(\frac{K_2}{K_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

### Step 3: Substitute known values
We know:
- \(K_1 = 0.0005775 \text{ s}^{-1}\)
- \(K_2 = 0.00231 \text{ s}^{-1}\)
- \(R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}\)
- \(T_1 = 300 \text{ K}\)
- \(T_2 = 350 \text{ K}\)

Now, we can substitute these values into the equation:
\[ \ln\left(\frac{0.00231}{0.0005775}\right) = -\frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{350}\right) \]

### Step 4: Calculate the left side
Calculating the logarithm:
\[ \frac{0.00231}{0.0005775} \approx 4 \]
Thus,
\[ \ln(4) \approx 1.386 \]

### Step 5: Calculate the right side
Calculating the temperature difference:
\[ \frac{1}{300} - \frac{1}{350} = \frac{350 - 300}{300 \times 350} = \frac{50}{105000} = \frac{1}{2100} \]

Now substituting back into the equation:
\[ 1.386 = -\frac{E_a}{8.314} \times \frac{1}{2100} \]

### Step 6: Solve for \(E_a\)
Rearranging gives:
\[ E_a = -1.386 \times 8.314 \times 2100 \]

Calculating:
\[ E_a \approx 1.386 \times 8.314 \times 2100 \approx 24205.8 \text{ J/mol} \]

### Final Answer
The activation energy \(E_a\) is approximately \(24205.8 \text{ J/mol}\).

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