The inactivation of a viral preparation in a chemical bath is found to be a first order reaction. The rate constant for the viral inactivation if in beginning `1.5%` of the virus is inactivated per minute is (Given : In `(100)/(98.5)=0.01511`)

Options

A. `1.25xx10^(-4)"sec"^(-1)`
B. `2.5xx10^(-4)"sec"^(-1)`
C. `5xx10^(-4)"sec"^(-1)`
D. `2.5xx10^(-4)"min"^(-1)`

Answer

To solve the problem of finding the rate constant for the viral inactivation reaction, we can follow these steps:

### Step 1: Understand the first-order reaction
In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The mathematical representation of a first-order reaction is given by:

\[ k = \frac{1}{t} \ln \left( \frac{A_0}{A_t} \right) \]

where:
- \( k \) is the rate constant,
- \( A_0 \) is the initial concentration,
- \( A_t \) is the concentration at time \( t \),
- \( t \) is the time.

### Step 2: Define initial and final concentrations
Given that 1.5% of the virus is inactivated per minute, we can define:
- \( A_0 = 100 \) (initial concentration),
- \( A_t = 100 - 1.5 = 98.5 \) (concentration after 1 minute).

### Step 3: Substitute values into the formula
Using the formula for the rate constant \( k \):

\[ k = \frac{1}{t} \ln \left( \frac{A_0}{A_t} \right) \]

Substituting \( A_0 = 100 \), \( A_t = 98.5 \), and \( t = 1 \) minute (which we will convert to seconds later):

\[ k = \ln \left( \frac{100}{98.5} \right) \]

### Step 4: Calculate the natural logarithm
We are given that:

\[ \ln \left( \frac{100}{98.5} \right) = 0.01511 \]

### Step 5: Convert time from minutes to seconds
Since \( t \) is given in minutes and we want the rate constant in seconds, we convert 1 minute to seconds:

\[ t = 1 \text{ minute} = 60 \text{ seconds} \]

### Step 6: Calculate the rate constant
Now substituting back into the equation for \( k \):

\[ k = \frac{1}{60} \times 0.01511 \]

Calculating this gives:

\[ k = \frac{0.01511}{60} = 0.0002518333 \text{ s}^{-1} \]

### Step 7: Express in scientific notation
To express this in scientific notation:

\[ k \approx 2.5 \times 10^{-4} \text{ s}^{-1} \]

Thus, the rate constant for the viral inactivation is:

\[ \boxed{2.5 \times 10^{-4} \text{ s}^{-1}} \]

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