A first order reaction has half life of `14.5` hrs. What percentage of the reactant will remain after 24 hrs ?

Options

A. `18.3%`
B. `31.8%`
C. `45.5%`
D. `68.2%`

Explanation

To solve the problem step by step, we will follow the principles of first-order kinetics and the relationship between half-life and rate constant.

### Step 1: Calculate the rate constant (k)
The half-life (t₁/₂) of a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

Given that the half-life is 14.5 hours, we can rearrange the formula to find k:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{14.5} \]
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