Question

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 ______ × 10^-3 min^-1. (Nearest integer)

[Use : ln 10 = 2.303; log₁₀3 = 0.477; property of logarithm : log x^y = y log x].

Options

• 250

• 106

• 365

• 198

Answer 1

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 106 × 10^-3 min^-1.

Explanation:

The inactivation rate of a viral preparation is proportional to the amount of virus. 

As it is a first order reaction, the rate constant can be expressed as,

K = `2.303/"t" log  "A"_0/"A"_"t"`   ...(i)

K is rate constant and t is time.

A₀ and A_t denoted the initial and final concentration.

The equation 1 can be written as,

`"K" xx "t" = 2.303  log "A"_0//"A"_"t"`    ...(ii)

In the first minute after preparation, the amount of virus inactivated was 10%.

Therefore, initial concentration A₀ = 100 and after 1  minute, A_t = 90

Here, t = 1 min

Substituting respective values in equation (ii), we get,

`"K" xx "t" = 2.303 xx log  100/90`

K = 2.303 × (log 10 - log 9)

K = 2.303 × (log 10 - 2 log 3)

K = 2.303 × (1 - 2 × 0.477)

K = 2.303 × (1 - 0.954)

K = 2.303 × 0.046 = 0.1059

K = 105.9 × 10^-3 ≈ 106 × 10^-3 

The rate constant for viral inactivation is 106 × 10^-3 min^-1.