The rate constant of a reaction with a virus is `3.3xx10^(-4)S^(-1)`. Time required for the virus to become `75%` inactivated is

Options

A. `35 mi n`
B. `70 mi n`
C. `105 mi n`
D. `17.5 mi n`

Explanation

To solve the problem of determining the time required for a virus to become 75% inactivated, we will use the formula for the half-life of a first-order reaction. Here are the steps:

### Step 1: Understand the Reaction Order
The problem states that the reaction is first-order. For a first-order reaction, the time required for a certain percentage of the reactant to be consumed can be calculated using the first-order rate equation.

### Step 2: Identify Given Values
- Rate constant (k) = \(3.3 \times 10^{-4} \, \text{s}^{-1}\)
- Percentage inactivated = 75%
- Remaining concentration = 100% - 75% = 25%

### Step 3: Set Up the Equation
The formula for the time \( t \) required for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right) \]
Where:
- \([A]_0\) = initial concentration (100%)
- \([A]\) = remaining concentration (25%)

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ t = \frac{2.303}{3.3 \times 10^{-4}} \log\left(\frac{100}{25}\right) \]

### Step 5: Calculate the Logarithm
Calculate the logarithm:
\[ \log\left(\frac{100}{25}\right) = \log(4) \]
Using the known value:
\[ \log(4) \approx 0.602 \]

### Step 6: Substitute the Logarithm Value
Now substitute the logarithm back into the equation:
\[ t = \frac{2.303}{3.3 \times 10^{-4}} \times 0.602 \]

### Step 7: Perform the Calculation
Calculate the time:
1. Calculate \( \frac{2.303}{3.3 \times 10^{-4}} \):
\[ \frac{2.303}{3.3 \times 10^{-4}} \approx 6970.3 \]
2. Multiply by \( 0.602 \):
\[ t \approx 6970.3 \times 0.602 \approx 4201 \, \text{seconds} \]

### Step 8: Convert Seconds to Minutes
To convert seconds into minutes:
\[ \text{Minutes} = \frac{4201}{60} \approx 70.02 \, \text{minutes} \]

### Final Answer
The time required for the virus to become 75% inactivated is approximately **70 minutes**.

---