At `380^(@)C`, the half-life period for the first order decomposition of `H_(2)O_(2)` is `360 min`. The energy of activation of the reaction is `200 kJ mol^(-1)`. Calculate the time required for `75%` decomposition at `450^(@)C`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert given temperatures to Kelvin
- The temperature at which the half-life is given is \( 380^\circ C \).
- To convert to Kelvin:
\[ T_1 = 380 + 273 = 653 \, K \]
- The temperature at which we want to find the time for 75% decomposition is \( 450^\circ C \).
- To convert to Kelvin:
\[ T_2 = 450 + 273 = 723 \, K \]

### Step 2: Calculate the rate constant \( k_1 \) at \( 380^\circ C \)
- The half-life \( t_{1/2} \) for a first-order reaction is given as \( 360 \, \text{min} \).
- The relationship between half-life and the rate constant \( k \) for a first-order reaction is:
\[ k_1 = \frac{0.693}{t_{1/2}} = \frac{0.693}{360 \, \text{min}} = 0.001925 \, \text{min}^{-1} \]

### Step 3: Use the Arrhenius equation to find \( k_2 \) at \( 450^\circ C \)
- The Arrhenius equation in logarithmic form is:
\[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
- Given:
- Activation energy \( E_a = 200 \, \text{kJ/mol} = 200 \times 10^3 \, \text{J/mol} \)
- Gas constant \( R = 8.314 \, \text{J/(mol K)} \)

- Substitute the values:
\[ \log \left(\frac{k_2}{0.001925}\right) = \frac{200 \times 10^3}{2.303 \times 8.314} \left(\frac{1}{653} - \frac{1}{723}\right) \]

### Step 4: Calculate the right-hand side
- Calculate \( \frac{1}{653} - \frac{1}{723} \):
\[ \frac{1}{653} \approx 0.001532 \, \text{K}^{-1}, \quad \frac{1}{723} \approx 0.001384 \, \text{K}^{-1} \]
\[ \frac{1}{653} - \frac{1}{723} \approx 0.001532 - 0.001384 = 0.000148 \, \text{K}^{-1} \]

- Now calculate the entire right-hand side:
\[ \frac{200 \times 10^3}{2.303 \times 8.314} \approx 10.55 \]
\[ 10.55 \times 0.000148 \approx 0.001564 \]

### Step 5: Solve for \( k_2 \)
- Now we have:
\[ \log \left(\frac{k_2}{0.001925}\right) \approx 0.001564 \]
- Convert from logarithmic form:
\[ \frac{k_2}{0.001925} = 10^{0.001564} \approx 1.0036 \]
\[ k_2 \approx 0.001925 \times 1.0036 \approx 0.00193 \, \text{min}^{-1} \]

### Step 6: Calculate time for 75% decomposition
- For 75% decomposition, 25% remains. The formula for time in first-order kinetics is:
\[ t = \frac{2.303}{k} \log \left(\frac{[A_0]}{[A]}\right) \]
- Here, \( [A_0] = 100 \) and \( [A] = 25 \):
\[ t = \frac{2.303}{0.00193} \log \left(\frac{100}{25}\right) = \frac{2.303}{0.00193} \log(4) \]
- Since \( \log(4) \approx 0.602 \):
\[ t \approx \frac{2.303}{0.00193} \times 0.602 \approx 20.3 \, \text{min} \]

### Final Answer
The time required for 75% decomposition at \( 450^\circ C \) is approximately **20.3 minutes**.

---