A second order reaction requires `70 min` to change the concentration of reactants form `0.08 M` to `0.01 M`. The time required to become `0.04 M = 2x min`. Find the value of `x`.

Explanation

To solve the problem, we need to use the integrated rate law for a second-order reaction. The integrated rate equation for a second-order reaction is given by:

\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]

Where:
- \([A]\) is the final concentration,
- \([A_0]\) is the initial concentration,
- \(k\) is the rate constant,
- \(t\) is the time.

### Step 1: Calculate the rate constant \(k\)

Given:
- Initial concentration \([A_0] = 0.08 \, M\)
- Final concentration \([A] = 0.01 \, M\)
- Time \(t = 70 \, min\)

Using the integrated rate equation:

\[ \frac{1}{0.01} - \frac{1}{0.08} = k \times 70 \]

Calculating the left side:

\[ \frac{1}{0.01} = 100 \quad \text{and} \quad \frac{1}{0.08} = 12.5 \]
\[ 100 - 12.5 = k \times 70 \]
\[ 87.5 = k \times 70 \]
\[ k = \frac{87.5}{70} = 1.25 \, M^{-1}min^{-1} \]

### Step 2: Calculate the time required to change the concentration to \(0.04 M\)

Now, we need to find the time required to change the concentration from \(0.08 M\) to \(0.04 M\).

Using the same integrated rate equation:

\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]

Where:
- \([A_0] = 0.08 \, M\)
- \([A] = 0.04 \, M\)

Substituting the values:

\[ \frac{1}{0.04} - \frac{1}{0.08} = 1.25 \times t \]

Calculating the left side:

\[ \frac{1}{0.04} = 25 \quad \text{and} \quad \frac{1}{0.08} = 12.5 \]
\[ 25 - 12.5 = 1.25 \times t \]
\[ 12.5 = 1.25 \times t \]
\[ t = \frac{12.5}{1.25} = 10 \, min \]

### Step 3: Relate the time to \(2x\)

We are given that the time required to become \(0.04 M\) is \(2x\):

\[ 2x = 10 \, min \]

### Step 4: Solve for \(x\)

To find \(x\):

\[ x = \frac{10}{2} = 5 \, min \]

Thus, the value of \(x\) is \(5\).

### Final Answer:
The value of \(x\) is \(5 \, min\).

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