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 x^y = y log x]

Unit of rate constant is min^$-$1, so it must be a first order reaction. For first order reaction,<br/><br/>k $\times$ t = 2.303 log${{{A_0}} \over {{A_t}}}$<br/><br/>k is the rate constant<br/><br/> t is the time<br/><br/> A₀ is the initial conc. <br/><br/>A_t is the conc. at time, t<br/><br/>Using formula,<br/><br/>A₀ = 100, A_t = 90 min 1 min<br/><br/>So, K $\times$ 1 = 2.303 $\times$ log${{100} \over {90}}$<br/><br/>= 2.303 $\times$ (log 10 $-$ 2 log 3)<br/><br/>= 2.303 $\times$ (1 $-$ 2 $\times$ 0.477)<br/><br/>= 0.10593<br/><br/>= 105.93 $\times$ 10^$-$3<br/><br/>$\approx$ 106<br/><br/>Hence, the rate constant for viral inactivation is 106.