Half life of a first order reaction is 2 hours , what time is required for 90% of the reactant to be consumed ?

Options

A. 199 minutes
B. 398 minutes
C. 598 minutes
D. 798 minutes

Explanation

To solve the problem of determining the time required for 90% of the reactant to be consumed in a first-order reaction with a half-life of 2 hours, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Half-Life**:
The half-life (T½) of a first-order reaction is the time required for half of the reactant to be consumed. Given T½ = 2 hours.

2. **Determine Remaining Reactant after 90% Consumption**:
If 90% of the reactant is consumed, then 10% of the reactant remains. Therefore, if the initial concentration is C₀, the final concentration (Cₜ) after 90% consumption will be:
\[ C_t = 0.10 \times C_0 \]

3. **Use the First-Order Reaction Equation**:
The time (T) for a first-order reaction can be calculated using the formula:
\[ T = \frac{2.303}{k} \log\left(\frac{C_0}{C_t}\right) \]
Substituting \(C_t\):
\[ T = \frac{2.303}{k} \log\left(\frac{C_0}{0.10 C_0}\right) = \frac{2.303}{k} \log(10) \]

4. **Calculate the Value of k (Rate Constant)**:
The relationship between half-life and the rate constant for a first-order reaction is given by:
\[ T_{1/2} = \frac{0.693}{k} \]
Rearranging gives:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{2 \text{ hours}} = \frac{0.693}{2 \times 60 \text{ minutes}} = \frac{0.693}{120} \text{ min}^{-1} \]

5. **Substitute k back into the Time Equation**:
Now, substituting the value of k into the time equation:
\[ T = \frac{2.303}{\frac{0.693}{120}} \log(10) \]
Since \(\log(10) = 1\):
\[ T = \frac{2.303 \times 120}{0.693} \]

6. **Calculate the Time**:
Performing the calculation:
\[ T \approx \frac{276.36}{0.693} \approx 398.5 \text{ minutes} \]

7. **Final Answer**:
The time required for 90% of the reactant to be consumed is approximately 398 minutes.