Question

After 61.0 mins, 37% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

Answer 1

For first order reaction,

half life, t(1/2) = 0.693/K

and, K = 2.303/t log([R_{​​​​​​o}​​​​​]/[R])

t = 61.0 min

Let, [R_o] = 100 then [R] = 63

So, K = 2.303/61 log(100/63)

= 2.303×0.2/61 min^-1

  = 0.8×10^-2 min^​-1

t(1/2) = 0.693/0.8×10^-2 min = 86.6 min

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