The initial concentration of both the reactants of a second order reaction are equal and `60%` of the reaction gets completed in `30s`. How much time will be taken in `20%` completion of the reaction?

Explanation

To solve the problem step-by-step, we will use the formula for the rate constant of a second-order reaction and apply it to find the time taken for 20% completion of the reaction.

### Step 1: Understand the reaction and given data
We know that the reaction is second-order and the initial concentrations of both reactants are equal. We are given that 60% of the reaction gets completed in 30 seconds.

### Step 2: Define the variables
- Let the initial concentration of each reactant be \( A \).
- The amount of reaction completed, \( X \), when 60% is completed is \( 60\% \) of \( A \).
- Therefore, \( A = 100\% \) and \( A - X = 100\% - 60\% = 40\% \).
- Time, \( T = 30 \) seconds.

### Step 3: Use the second-order reaction formula
The formula for the rate constant \( k \) for a second-order reaction is given by:
\[ k = \frac{1}{T} \cdot \frac{X}{A \cdot (A - X)} \]

### Step 4: Substitute the known values into the formula
Substituting \( T = 30 \) seconds, \( X = 60 \), and \( A = 100 \):
\[ k = \frac{1}{30} \cdot \frac{60}{100 \cdot 40} \]

### Step 5: Calculate \( k \)
Now we simplify the expression:
\[ k = \frac{1}{30} \cdot \frac{60}{4000} = \frac{1}{30} \cdot \frac{3}{200} = \frac{1}{2000} \]

### Step 6: Find the time for 20% completion
Now we need to find the time taken for 20% completion of the reaction. Here:
- \( X = 20 \)
- \( A = 100 \)
- \( A - X = 100 - 20 = 80 \)

Using the same formula for time:
\[ T = \frac{1}{k} \cdot \frac{X}{A \cdot (A - X)} \]

### Step 7: Substitute the values into the formula
Substituting \( k = \frac{1}{2000} \), \( X = 20 \), \( A = 100 \), and \( A - X = 80 \):
\[ T = \frac{1}{\frac{1}{2000}} \cdot \frac{20}{100 \cdot 80} \]

### Step 8: Calculate \( T \)
Simplifying:
\[ T = 2000 \cdot \frac{20}{8000} = 2000 \cdot \frac{1}{400} = 5 \text{ seconds} \]

### Final Answer
The time taken for 20% completion of the reaction is **5 seconds**.

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