Therate cosntantof firstorderreactionis 6.8 xx 10^(-4)s^(-1) Ifthe initialconcentrationof the reactantis0.04 M, Whatis itsmolarityafter 20minutes? Howlongwillit takefor 25%of thereactanttoreact ?

Therate cosntantof firstorderreactionis 6.8 xx 10^(-4)s^(-1) Ifthe ini

1.	Therate cosntantof firstorderreactionis 6.8 xx 10^(-4)s^(-1) Ifthe initialconcentrationof the reactantis0.04 M, Whatis itsmolarityafter 20minutes? Howlongwillit takefor 25%of thereactanttoreact ?
Answer» Solution :Given :Rateconstant`=k = 6.8xx 10^(-4) s^(-1)` `[A]_(0)=0.04 M` (i)Aftert= 20min= 20x 60 s= 1200s`[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%reactantto reactt= ? REACTANT reacted=25%of0.04 M `=(25)/(100) xx 0.04` `=0.01 M` `:.` Reactantleft `=[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`