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Question: A viral preparation was inactivated in a chemical bath. The inactivation process was found to be fir...

A viral preparation was inactivated in a chemical bath. The inactivation process was found to be first order in virus concentration. At the beginning of the experiment 2.0 % of the virus was found to be inactivated per minute . Evaluate k (in s–1) for inactivation process.

Answer

3.333e-4

Explanation

Solution

The inactivation process is first order in virus concentration. The rate of inactivation is given by:

Rate =−d[V]dt=k[V]t= - \frac{d[V]}{dt} = k[V]_t

At the beginning of the experiment (t=0t=0), the concentration of the virus is [V]0[V]_0. The initial rate of inactivation is given by:

Initial Rate =k[V]0= k[V]_0

The problem states that at the beginning of the experiment, 2.0% of the virus was found to be inactivated per minute. This means that the rate of decrease of virus concentration at t=0t=0 is 2.0% of the initial concentration per minute.

Initial Rate =(2.0% of [V]0) per minute= (2.0 \% \text{ of } [V]_0) \text{ per minute}

Initial Rate =2.0100[V]0 min−1= \frac{2.0}{100} [V]_0 \text{ min}^{-1}

Initial Rate =0.02[V]0 min−1= 0.02 [V]_0 \text{ min}^{-1}

Equating the two expressions for the initial rate:

k[V]0=0.02[V]0 min−1k[V]_0 = 0.02 [V]_0 \text{ min}^{-1}

We can cancel [V]0[V]_0 from both sides (assuming [V]0≠0[V]_0 \neq 0, which must be true for inactivation to occur):

k=0.02 min−1k = 0.02 \text{ min}^{-1}

The question asks for the value of kk in units of s−1s^{-1}. We need to convert the unit from per minute to per second.

1 minute = 60 seconds.

So, 1 min−1=11 min=160 s=160 s−1^{-1} = \frac{1}{1 \text{ min}} = \frac{1}{60 \text{ s}} = \frac{1}{60} \text{ s}^{-1}.

Substitute this conversion into the expression for kk:

k=0.02×160 s−1k = 0.02 \times \frac{1}{60} \text{ s}^{-1}

k=0.0260 s−1k = \frac{0.02}{60} \text{ s}^{-1}

k=2×10−260 s−1k = \frac{2 \times 10^{-2}}{60} \text{ s}^{-1}

k=1×10−230 s−1k = \frac{1 \times 10^{-2}}{30} \text{ s}^{-1}

k=13000 s−1k = \frac{1}{3000} \text{ s}^{-1}

To express this as a decimal:

k=13000=13×1000=13×10−3 s−1k = \frac{1}{3000} = \frac{1}{3 \times 1000} = \frac{1}{3} \times 10^{-3} \text{ s}^{-1}

k≈0.3333...×10−3 s−1k \approx 0.3333... \times 10^{-3} \text{ s}^{-1}

k≈3.333...×10−4 s−1k \approx 3.333... \times 10^{-4} \text{ s}^{-1}

Depending on the required precision, the answer can be given as a fraction or a decimal. The fraction 13000\frac{1}{3000} is the exact value.