A second order reaction requires `70 min` to change the concentration of reactants form `0.08 M` to `0.01 M`. How much time will require to become `0.04 M` ?

Options

A. `10 min`
B. `20 min`
C. `30 min`
D. `40 min`

Explanation

To solve the problem step by step, we will use the integrated rate equation for a second-order reaction. The integrated rate law 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: Determine the change in concentration and calculate \(k\)

1. **Initial concentration, \([A_0]\)** = 0.08 M
2. **Final concentration, \([A]\)** = 0.01 M
3. **Change in concentration, \(x\)** = \([A_0] - [A] = 0.08 - 0.01 = 0.07\) M

Now, we can use the integrated rate law to find the rate constant \(k\) using the time given (70 minutes):

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

Substituting the values:

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

Calculating the left-hand side:

\[ 100 - 12.5 = k \cdot 70 \]

\[ 87.5 = k \cdot 70 \]

Now, solving for \(k\):

\[ k = \frac{87.5}{70} = 1.25 \, \text{M}^{-1}\text{min}^{-1} \]

### Step 2: Calculate the time required to change the concentration from 0.08 M to 0.04 M

1. **New final concentration, \([A]\)** = 0.04 M
2. **Change in concentration, \(x\)** = \([A_0] - [A] = 0.08 - 0.04 = 0.04\) M

Using the integrated rate law again:

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

Substituting the new values:

\[ \frac{1}{0.04} - \frac{1}{0.08} = k \cdot t \]

Calculating the left-hand side:

\[ 25 - 12.5 = k \cdot t \]

\[ 12.5 = k \cdot t \]

Now substituting the value of \(k\):

\[ 12.5 = 1.25 \cdot t \]

Solving for \(t\):

\[ t = \frac{12.5}{1.25} = 10 \, \text{minutes} \]

### Final Answer

The time required for the concentration to change from 0.08 M to 0.04 M is **10 minutes**.

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