Question

In a first reaction, `10%` of the reactant is consumed in 25 minutes. Calcualte.
i) Half life period `(t_(1//2))` ii) Time taken to complete `87.5%` of the reaction.

Answer 1

For the first order reaction
`k= 2.303/tloga/(a-x)`
`=2.303/(25 mm) log 100/(100-10)= 0.0042145 mm^(-1)`
Half life period `(t_(1//2))` = `0.693/k = 0.693/(0.0042145 mm^(-1))=164.43 mm`
ii) Again for the reaction:
`t= 2.303/t loga/(a-x)` or `t=2.303/tlog a/(a-x)`
`t= 2.303/(0.004215min^(-1)) log 100/(100-87.5) = 2.303/(0.0042145 mm^(-1)) xx 0.9030 = 493.45 mm`