JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Chemical Kinetics

The inactivation rate of a viral preparation in 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 1 0^{- 3} min^{- 1}

. (Nearest integer)
[ Use:

ℓ n 10 = 2.303; lo g_{10} 3 = 0.477

; property of logarithm:

lo g x^{y} = y lo g x

]

Answer: 106

Solution

A rate proportional to the amount present is first order, for which

k = \frac{2.303}{t} lo g \frac{a _{0}}{a}

After one minute a tenth has been inactivated, so

a = 0.9 a_{0}

and

k = 2.303 lo g \frac{10}{9} = 2.303 (1 - 2 lo g 3) = 2.303 (1 - 0.954) = 2.303 \times 0.046

k = 0.1059 mi n^{- 1}

To the nearest integer the rate constant for viral inactivation is

106 \times 1 0^{- 3} mi n^{- 1}

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer