The rate cosntant of first order reaction is `6.8 xx 10^(-4) s^(-1)`
If the initial concentration of the reactant is 0.04 M , What is its molarity after 20 minutes ? How long will it take for 25% of the reactant to react ?

Answer

(i) Molarity of reactant after 20 min =0.0177 M
(ii) Time for 25% of the reaction = 7.05 min

Explanation

Given : Rate constant `=k = 6.8 xx 10^(-4) s^(-1)`
`[A]_(0) =0.04 M`
(i) After t= 20 min = 20 x 60 s = 1200 s `[A]_(t) = ?`
`k= (2.303)/(t) log_(10) .([A]_(0))/([A]_(t))`
`:. log_(10).[[A]_(0)]/[[A]_(t)] = (kxxt)/(2.303) - (6.8 xx 10^(-4) xx 1200)/(2.303 ) = 0.3543`
`:. [[A]_(0)]/[[A]_(t)]=AL 0.3543 =2.26`
`:. [A]_(t) =[[A]_(0))/(2.26) =(0.04)/(2.26) =0.0177 M`
(ii) For 25% reactant to react t= ?
Reactant reacted =25% of 0.04 M
`=(25)/(100) xx 0.04`
`=0.01 M`
`:.` Reactant left `=[A]_(t) = 0.04 - 0.01 =0.03 M`
`t= (2.303)/(k) log_(10) .([[A]_(0)])/([[A]_(t)])`
`=(2.303)/(6.8 xx 10^(-4)) log_(10) .(0.04)/(0.03)`
`=(2.303)/(6.8xx 10^(-4))`
`=423 s`
`=(423)/(60) min = 7.05 min`