Animated Solution for Chemistry - Chemical Kinetics: The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ...... $\times 10^{-3}{\text{ min}}^{-1}$. (Nearest integer) [Use: $\ln 10=2.303$; ${\log }_{10}3=0.477$; property of logarithm : $\log x^y=y\log x$]

Have you ever wondered how scientists determine the shelf-life of a vaccine or the time required to completely disinfect a surface? The answer lies in the elegant mathematics of Chemical Kinetics. In this problem, we are tasked with finding the rate constant for the inactivation of a viral preparation. This isn't just an abstract textbook question; it's a real-world scenario that plays out in laboratories every single day.

Let's embark on this journey and decode the mathematics step-by-step.

The very first sentence of the problem holds the key to unlocking the entire solution: "The inactivation rate of a viral preparation is proportional to the amount of virus."

In the language of chemical kinetics, when the rate of a process is directly proportional to the first power of the concentration of the reactant, it is a first-order reaction.

Mathematically, we can express this as:

This simple deduction is our visual anchor. It tells us exactly which set of mathematical tools we need to pull from our kinetics toolbox.

For any first-order reaction, the relationship between the rate constant $k$, the time $t$, the initial concentration $[A{]}_0$, and the concentration at time $t$, $[A{]}_t$, is given by the integrated rate law:

This equation is the beating heart of our solution. Our goal is to find $k$, so we need to carefully extract $t$, $[A{]}_0$, and $[A{]}_t$ from the problem statement.

The problem states that in the first minute ($t=1\text{ min}$), 10% of the virus is inactivated.

Here lies a classic trap where many students stumble. The term $[A{]}_t$ in our master equation does not represent the amount that has reacted or been inactivated. It represents the amount that is left over or still active.

If we assume our initial amount $[A{]}_0$ is 100 (representing 100%), and 10% is inactivated, then the remaining active virus is:

Substituting these values into our rate law, we get:

Now we face a mathematical hurdle. We need to evaluate $\log \left(\frac{10}{9}\right)$ using only the given values: $\ln 10=2.303$ (which we already used in the formula) and ${\log }_{10}3=0.477$.

We must rely on the fundamental properties of logarithms. First, the logarithm of a quotient is the difference of the logarithms:

We know that ${\log }_{10}10=1$. But what about $\log 9$? We can express 9 as $3^2$, and use the power rule of logarithms ($\log x^y=y\log x$):

Substituting this back, our expression becomes:

We are now ready for the final execution. The problem generously provides the value of $\log 3$ as $0.477$. Let's plug it in:

Now, we multiply this result by 2.303 to find the rate constant $k$:

The question asks for the answer in the specific format of an integer multiplied by $10^{-3}$. To achieve this, we shift the decimal point three places to the right:

Rounding off to the nearest integer, we arrive at our final destination:

And there we have it! By carefully translating the physical scenario into a mathematical model, avoiding the trap of the reacted amount, and executing precise logarithmic manipulations, we have successfully determined the viral inactivation rate.