The half life for the viral inactivation if in the beginning 1.5% of the virus is inactivated per minute is________. (Given: The reaction is of first order) (in min)
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Solution
For the first order reaction for small finite change k1=1[A]Δ[Δ]Δt⇒Δ[A]/[A]Δt=1.5%min−1 = 0.015 min−1 t1/2=0.6930.015min−1 = 46.2 min ≈ 46 min