`70%` of a first order reaction was completed in `70 min`. What is the half life of the reaction?

Options

A. `4.2 min`
B. `42 min`
C. `4.2 hr`
D. `4.2 s`

Explanation

To solve the problem of finding the half-life of a first-order reaction given that 70% of the reaction was completed in 70 minutes, we can follow these steps:

### Step 1: Understand the given data
We know that 70% of the reaction is completed in 70 minutes. This means that if we start with 100 units of concentration, 30 units remain unreacted after 70 minutes.

### Step 2: Use the first-order reaction formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:

\[ k = \frac{2.303}{T} \log \left( \frac{A}{A - x} \right) \]

Where:
- \( T \) is the time (in minutes)
- \( A \) is the initial concentration
- \( x \) is the amount reacted

### Step 3: Substitute the values into the formula
Here, we have:
- \( A = 100 \) (initial concentration)
- \( x = 70 \) (amount reacted)
- \( A - x = 30 \) (amount remaining)
- \( T = 70 \) minutes

Substituting these values into the formula gives:

\[ k = \frac{2.303}{70} \log \left( \frac{100}{30} \right) \]

### Step 4: Calculate the logarithm
Calculate the logarithm:

\[ \log \left( \frac{100}{30} \right) = \log(3.33) \approx 0.5229 \]

### Step 5: Calculate \( k \)
Now substitute this value back into the equation for \( k \):

\[ k = \frac{2.303}{70} \times 0.5229 \approx 0.017 \]

### Step 6: Use the half-life formula
The half-life \( t_{1/2} \) for a first-order reaction is given by:

\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of \( k \):

\[ t_{1/2} = \frac{0.693}{0.017} \approx 40.38 \text{ minutes} \]

### Step 7: Conclusion
Thus, the half-life of the reaction is approximately 40.38 minutes.

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