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 ....... x 10^–3 min^–1 . (Nearest integer) 

[Use : ln 10 = 2.303 ; log₁₀ 3 = 0.477; 

property of logarithm : log x^y = y log x]

Answer 1

Answer is (106)

As the unit of rate constant is min^–1 so it must be a first order reaction 

K × t = 2.303 log A0/At 

in 1 min 10% is in activated so tabing 

A₀ = 100 A_t = 90 in 1 min 

So K x 1 = 2.303 × log \(\frac{100} {90} \)

= 2.303 x (log 10 – 2log3) 

= 2.303 x (1 – 2 × 0.477) 

= 0.10593 

= 105.93 x 10^–3 

= 106

Answer 2

Correct answer: 106

Unit of rate constant is min⁻¹, so it must be a first order reaction. For first order reaction,

k × t = 2.303 log\(\frac{A_0}{A_t}\)

k is the rate constant

t is the time

A₀ is the initial conc.

A_t is the conc. at time, t

Using formula,

A₀ = 100, A_t = 90 min 1 min

So, K × 1 = 2.303 × log\(\frac{100}{90}\)

= 2.303 × (log 10 − 2 log 3)

= 2.303 × (1 − 2 × 0.477)

= 0.10593

= 105.93 × 10⁻³

≈ 106

Hence, the rate constant for viral inactivation is 106.