A first order reaction `A rarr B` requires activation energy of `70 kJ mol^(-1)`. When a `20%` solution of `A` was kept at `25^(@)C` for `20 min`, `25%` decompoistion took place. What will be the percentage decompoistion in the same time in a `30%` solution maintained at `40^(@)C` ? (Assume that activation energy remanis constant in this range of temperature)

Explanation

To solve the problem step by step, we will follow the outlined procedure to find the percentage decomposition of a 30% solution of A maintained at 40°C after 20 minutes.

### Step 1: Calculate the rate constant \( k_1 \) for the 20% solution at 25°C
Given:
- Activation energy \( E_a = 70 \, \text{kJ/mol} = 70000 \, \text{J/mol} \)
- Initial concentration \( [A]_0 = 20\% \)
- Decomposed percentage = 25%
- Remaining concentration \( [A] = 100\% - 25\% = 75\% \)
- Time \( t = 20 \, \text{min} = 20 \times 60 \, \text{s} = 1200 \, \text{s} \)

Using the first-order rate equation:
\[ k_1 = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) \]
Substituting the values:
\[ k_1 = \frac{2.303}{1200} \log\left(\frac{100}{75}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{100}{75}\right) = \log(1.3333) \approx 0.1249 \]
Now substituting this back:
\[ k_1 = \frac{2.303}{1200} \times 0.1249 \approx 0.0144 \, \text{min}^{-1} \]

### Step 2: Calculate the rate constant \( k_2 \) for the 30% solution at 40°C
Using the Arrhenius equation:
\[ \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T_1 = 25°C = 298 \, \text{K} \)
- \( T_2 = 40°C = 313 \, \text{K} \)

Substituting the values:
\[ \log\left(\frac{k_2}{0.0144}\right) = \frac{70000}{2.303 \times 8.314} \left(\frac{1}{298} - \frac{1}{313}\right) \]
Calculating the right side:
\[ \frac{70000}{2.303 \times 8.314} \approx 3.75 \]
Calculating \( \left(\frac{1}{298} - \frac{1}{313}\right) \):
\[ \frac{1}{298} - \frac{1}{313} \approx 0.000168 \]
Now substituting:
\[ \log\left(\frac{k_2}{0.0144}\right) \approx 3.75 \times 0.000168 \approx 0.000630 \]
Thus:
\[ \frac{k_2}{0.0144} = 10^{0.000630} \approx 1.00145 \]
So:
\[ k_2 \approx 0.0144 \times 1.00145 \approx 0.0144 \, \text{min}^{-1} \]

### Step 3: Calculate the percentage decomposition for the 30% solution
Using the first-order rate equation again:
\[ k_2 = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]}\right) \]
Where \( [A]_0 = 30\% \) and \( [A] = 30\% - x \):
\[ 0.0557 = \frac{2.303}{20} \log\left(\frac{100}{100 - x}\right) \]
Rearranging gives:
\[ \log\left(\frac{100}{100 - x}\right) = 0.0557 \times 20 / 2.303 \]
Calculating:
\[ \log\left(\frac{100}{100 - x}\right) \approx 0.482 \]
Thus:
\[ \frac{100}{100 - x} = 10^{0.482} \approx 3.02 \]
So:
\[ 100 - x = \frac{100}{3.02} \approx 33.11 \]
Thus:
\[ x \approx 100 - 33.11 = 66.89\% \]

### Final Answer
The percentage decomposition in the 30% solution maintained at 40°C after 20 minutes is approximately **66.89%**.