A viral preparation was inactivated in a chemical bath. The activation process was found to be first order in virus concentration. The beginning of the experiment 2% of the virus was found to be inactivated per minute.
[i] Evaluate k for the inactivation process in units (1/s).
[ii] How much time would be required for the virus to become 50% inactivated?
[iii] How much time would be required for the virus to become 75% inactivated?
To evaluate the rate constant, k, for the inactivation process in units of 1/s, we can use the first-order rate equation:
Rate = k * [virus]
where [virus] represents the concentration of the virus.
In the given scenario, it is stated that at the beginning of the experiment, 2% of the virus was found to be inactivated per minute. This information can be used to determine the rate constant, k.
To do this, we can rearrange the rate equation as follows:
Rate = k * [virus]
2% per minute = k * [virus]
Since 2% per minute is equivalent to 0.02/min, we can substitute this value into the equation:
0.02/min = k * [virus]
At the beginning of the experiment, [virus] is the initial concentration of the virus, which we can assume to be 100% or 1.
Therefore, we have:
0.02/min = k * 1
Simplifying, we find that:
k = 0.02/min
To convert this to units of 1/s, we need to divide by 60 (since there are 60 seconds in a minute):
k = 0.02/min ÷ 60 = 0.0003333... 1/s
So, the rate constant, k, for the inactivation process is approximately 0.0003333... 1/s.
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To determine how much time would be required for the virus to become 50% inactivated, we can use the first-order rate equation again:
Rate = k * [virus]
Given that the rate constant, k, is 0.0003333... 1/s (as calculated in part [i]), and we want to find the time required for the virus to become 50% inactivated, we can substitute these values into the equation:
Rate = 0.0003333... 1/s * [virus]
Since we are looking for the time required for the virus to become 50% inactivated, the concentration of the virus at that point would be [virus] = 0.5.
Therefore, we have:
Rate = 0.0003333... 1/s * 0.5
Simplifying, we find that:
Rate = 0.000166666... 1/s
To find the time required, we can rearrange the rate equation:
Rate = k * [virus]
t = [virus] / (k * Rate)
Substituting the values we have:
t = 0.5 / (0.0003333... 1/s * 0.5)
Simplifying, we find that:
t = 1 / 0.0003333... s
t ≈ 3000 s
Therefore, it would take approximately 3000 seconds for the virus to become 50% inactivated.
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To determine how much time would be required for the virus to become 75% inactivated, we can once again use the first-order rate equation:
Rate = k * [virus]
Given that the rate constant, k, is 0.0003333... 1/s (as calculated in part [i]), and we want to find the time required for the virus to become 75% inactivated, we can substitute these values into the equation:
Rate = 0.0003333... 1/s * [virus]
Since we are looking for the time required for the virus to become 75% inactivated, the concentration of the virus at that point would be [virus] = 0.25.
Therefore, we have:
Rate = 0.0003333... 1/s * 0.25
Simplifying, we find that:
Rate = 0.000083333... 1/s
To find the time required, we can once again rearrange the rate equation:
Rate = k * [virus]
t = [virus] / (k * Rate)
Substituting the values we have:
t = 0.25 / (0.0003333... 1/s * 0.25)
Simplifying, we find that:
t = 1 / 0.0003333... s
t ≈ 12000 s
Therefore, it would take approximately 12000 seconds for the virus to become 75% inactivated.
ln(No/N) = kt
No = 100
N = 98
k is to be found
time is 1 min or 60 seconds depending upon how the answer is to be displayed.
For #2, N is 50
For #3 N = 25.