A first order reaction is 20% complete in the 10 minutes. Calculate the time period for 75% completion of the reaction.

Explanation

For first order reaction, `K = (0.693)/(t_(1//2))`
`K_(653) = 0.693/360 = 1.925 xx 10^(-2) " min-"^(-1)`
`K_(723) = ?`
log `K_(723) - log K_(653) = (E_a)/(2.303R) [(T_2 - T_1)/(T_1T_2)]`
`log K_(723) - log 1.925 xx 10^(-3) "min"^(-1) = (200 xx 10^(1/3) Jmol^(-1))/(2.303 xx 8.314 JK mol^(-1)) xx [(723 - 653)/(723 xx 653 K)]`
`log K _(723) - (0.2844 – 3) = 1.55 `
`log K_(723) = 2.8344 `
`log K_(273) = 6.82 xx 10^(-2) "min"^(-1)`
For First order reaction,
`t = (2.303)/K "log " a/(a - x) = (2.303)/(6.82 xx 10^(-2)"min"^(-1)) "log" (100)/(100 - 75 )`
`= 33.768 "min log "4 = 33.768 xx 0.602 = 20.33 "min"`