For a first order reaction with a rate constant k = 3.13 x 102 s-1, how long does it take for the concentration of the only reactant to become 40% of the original amount? please explain your answer...
[A]initial= 100 mol/L [A]= 40% of 100= 40 mol/L the equation used is ln[A]= -kt + ln[A initial] plug it the numbers into the equationln[40]= -(3.13 x 10^2/s)t + ln[100] and finally t= 2.93 x 10^-3 s
For a first order rxn: ln[A]=-kt+ln[A]0 [A]=40%[A]0=.4[A]0 ln[A]- ln[A]0=-kt ln([A]/ [A]0) =-kt ln([A]/ [A]0)/-k =t Plug in [A]=.4[A]0 and k=3.31x102 s-1 ln(.4[A]0/[A]0)/(- 3.31x102 s-1)=t ln(.4)/(-3.31x102 s-1)=t (-.916)/(-3.31x102 s-1)=t 2.93x10-3s=t
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let [A](initial)= 100 mol/L
ReplyDeleteso [A]= 40% of 100= 40 mol/L
ln[A]= -kt + ln[A initial]
ln[40]= -(3.13 x 10^2/s)t + ln[100]
t= 2.927 x 10^-3 s
ln[A]= -kt + ln[A]initial
ReplyDeletek = 3.13 x 102 s-1
t= ?
let [A] = 100 mol/L
then [A]initial= 40 mol/L
ln[40] = -3.13 x 102 s-1 (t) +ln[100]
t= .002927 = 2.927 x 10^-3 s
Since the reaction is 1st order, we use the equation:
ReplyDeleteln [A]t - ln [A]o = -kt
where
[A]t = concentration of the reactant
[A]o = initial concentration.
Since we are using percentage, we take a 100 mol/L sample to make the calculation simpler. Therefore, [A]t = 40 and [A]o = 100.
So...
ln [40] - ln [100] = -(3.13 x 10^2) x t
t = .002927
t = .00293
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ReplyDeleteln[A]=-kt+ln[A]initial
ReplyDeleteln(A/Ainitial)=-kt
ln(40/100)=-313t
-.91629/-313=t
.002927=t
t=2.927x10^-3 s
[A]initial= 100 mol/L
ReplyDelete[A]= 40% of 100= 40 mol/L
the equation used is ln[A]= -kt + ln[A initial]
plug it the numbers into the equationln[40]= -(3.13 x 10^2/s)t + ln[100]
and finally t= 2.93 x 10^-3 s
ln [A]= -kt+ln[A]initial
ReplyDelete[A] initial= 100 mol/L
[A]= 100 * .40 (40%)= 40 mol/L
ln[40]= -(3.13x10^2 s)t+ln[100]
t=2.927 x 10^-3 s
ln[A]=-kt+ln[A0]
ReplyDeletet=(ln[A]-ln[A0])/(-k)
t=(ln[40]-ln[100]/(-3.13*10^2s)
t=2.93*10^-3s
For a first order rxn: ln[A]=-kt+ln[A]0
ReplyDelete[A]=40%[A]0=.4[A]0
ln[A]- ln[A]0=-kt
ln([A]/ [A]0) =-kt
ln([A]/ [A]0)/-k =t
Plug in [A]=.4[A]0 and k=3.31x102 s-1
ln(.4[A]0/[A]0)/(- 3.31x102 s-1)=t
ln(.4)/(-3.31x102 s-1)=t
(-.916)/(-3.31x102 s-1)=t
2.93x10-3s=t
ln[A]= -kt + ln[A]initial
ReplyDeletek = 3.13 x 102 s-1
let[A] = 100 mol/L
then[A]initial= 40 mol/L
so ln[40] = -3.13 x 102 s-1 (t) +ln[100]
t=2.927 x 10^-3 s
Ln[A] = kt + ln[A] initial
ReplyDeleteLn[A] = 100 mol/L
Ln[A] Initial = 40 mol/L
Ln[40] = -(3.13*10^2)t + ln[100]
t = 0.002927
t = 2.927*10^-3 s
ln [A]= -kt+ln[A]initial
ReplyDelete[A] initial= 100 mol/L
[A]= 100 * .40 (40%)= 40 mol/L
ln[40]= -(3.13x10^2 s)t+ln[100]
t=2.927 x 10^-3 s
ln [A]= -kt+ln[A]initial
ReplyDelete[A] initial= 100 mol/L
[A]= 100 * .40 (40%)= 40 mol/L
ln[40]= -(3.13x10^2 s)t+ln[100]
t=2.927 x 10^-3 s
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