
#1
Dec1612, 10:48 AM

P: 30

1. The problem statement, all variables and given/known data
I have two questions that I just dont even know where to start The activation energy for the reaction CH3CO CH3 + CO is 71 kJ/mol. How many times greater is the rate constant for this reaction at 170°C than at 150°C? A) 0.40 B) 1.1 C) 2.5 D) 4.0 E) 5.0 and A certain reaction A products is second order in A. If this reaction is 10.% complete after 20. s, how long would it take for the reaction to be 90.% complete? A) 180 s B) 1600 s C) 440 s D) 18,000 s E) 540 s 2. Relevant equations I know the second one is a second order reaction rate which has 1/[A]=1/[A_0]+kt 



#3
Dec1612, 11:59 AM

P: 30





#4
Dec1612, 01:55 PM

P: 3,545

Half life and activation energyYou will have to derive an equation using the Arrhenius equation to relate the rate constants. Take logarithm on both the sides of Arrhenius equation. Let the rate constant at temperature T1 be k1 and at temperature T2, let the rate constant be k2. You get the two equations: [tex]k_1=Ae^{E_a/RT_1}[/tex] [tex]k_2=Ae^{E_a/RT_2}[/tex] Take logarithm on both the sides of the equation and subtract the equations you get. 



#5
Dec1612, 02:01 PM

P: 30





#6
Dec1612, 02:05 PM

P: 3,545





#7
Dec1612, 02:08 PM

P: 30

ok so I did
ln(Rate 2/rate1)=71/.008314((1/423)(1/443)) got .911452 took e^.911452 = 2.5 so 2.5 times greater is rate 2 to rate 1 



#8
Dec1612, 02:10 PM

P: 3,545





#9
Dec1612, 02:14 PM

P: 30

Do you know how i could tackle the second one ?




#10
Dec1612, 02:18 PM

P: 3,545





#11
Dec1612, 02:22 PM

P: 30

Would I just be able to just substitute the percentages in as if they were concentrations?




#12
Dec1612, 02:26 PM

P: 3,545

The percentage left after 20 seconds is 90% of the initial concentration. So substitute [A]=0.9[A_{0}], assuming [A_{0}] to be the initial concentration. 



#13
Dec1612, 02:37 PM

P: 30

(1/.9)=1/.1+k(20 seconds)
I got k to be .005555 so If I find t i get (1/.1)=(1/.9)+.005555(t) 9/.0055555=t t=1600s that look like the right path to get to the answer? 



#14
Dec1612, 02:41 PM

P: 3,545





#15
Dec1612, 02:45 PM

P: 30

same for 90% being .9 


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