JEE MAIN - Chemistry (2021 - 20th July Morning Shift - No. 19)
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 min$$-$$1. (Nearest integer)
[Use : ln 10 = 2.303; log10 3 = 0.477; property
of logarithm : log xy = y log x]
[Use : ln 10 = 2.303; log10 3 = 0.477; property
of logarithm : log xy = y log x]
Answer
106
Explanation
Unit of rate constant is min$$-$$1, so it must be a first order reaction. For first order reaction,
k $$\times$$ t = 2.303 log$${{{A_0}} \over {{A_t}}}$$
k is the rate constant
t is the time
A0 is the initial conc.
At is the conc. at time, t
Using formula,
A0 = 100, At = 90 min 1 min
So, K $$\times$$ 1 = 2.303 $$\times$$ log$${{100} \over {90}}$$
= 2.303 $$\times$$ (log 10 $$-$$ 2 log 3)
= 2.303 $$\times$$ (1 $$-$$ 2 $$\times$$ 0.477)
= 0.10593
= 105.93 $$\times$$ 10$$-$$3
$$\approx$$ 106
Hence, the rate constant for viral inactivation is 106.
k $$\times$$ t = 2.303 log$${{{A_0}} \over {{A_t}}}$$
k is the rate constant
t is the time
A0 is the initial conc.
At is the conc. at time, t
Using formula,
A0 = 100, At = 90 min 1 min
So, K $$\times$$ 1 = 2.303 $$\times$$ log$${{100} \over {90}}$$
= 2.303 $$\times$$ (log 10 $$-$$ 2 log 3)
= 2.303 $$\times$$ (1 $$-$$ 2 $$\times$$ 0.477)
= 0.10593
= 105.93 $$\times$$ 10$$-$$3
$$\approx$$ 106
Hence, the rate constant for viral inactivation is 106.
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