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Half life and activation energy

  • Thread starter Hemolymph
  • Start date
  • #1
30
0

Homework Statement


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

Homework Equations



I know the second one is a second order reaction rate which has
1/[A]=1/[A_0]+kt
 

Answers and Replies

  • #2
Borek
Mentor
28,448
2,843
Have you heard about Arrhenius equation?
 
  • #3
30
0
Have you heard about Arrhenius equation?
I cant see how I can apply the Arrhenius equation if I dont have any activation energies
 
  • #4
3,812
92
I cant see how I can apply the Arrhenius equation if I dont have any activation energies
Activation energy is given in the problem statement.

You 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
30
0
Activation energy is given in the problem statement.

You 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.
ln(k1/k2)=Ea/R(1/t_2)-(1/t_1)?
 
  • #6
3,812
92
ln(k1/k2)=Ea/R(1/t_2)-(1/t_1)?
Yes, that's right if you mean ln(k1/k2)=Ea/R((1/t_2)-(1/t_1)). (Take care of parentheses. :smile:)
 
  • #7
30
0
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
3,812
92
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
I haven't checked the calculations but your result matches with one of the options so I guess it is correct.
 
  • #9
30
0
Do you know how i could tackle the second one ?
 
  • #10
3,812
92
Do you know how i could tackle the second one ?
You do have posted an equation in the main post. Did you try applying it?
 
  • #11
30
0
Would I just be able to just substitute the percentages in as if they were concentrations?
 
  • #12
3,812
92
Would I just be able to just substitute the percentages in as if they were concentrations?
You can do that but the problem is you are not given the percentage of concentration left.
The percentage left after 20 seconds is 90% of the initial concentration. So substitute [A]=0.9[A0], assuming [A0] to be the initial concentration.
 
  • #13
30
0
(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
3,812
92
(1/.9)=1/.1+k(20 seconds)
How you got 0.1? Did you assume the initial concentration [A0] to be 0.1 M? If so, [A] would be equal to 0.09 M, not 0.9 M.
 
  • #15
30
0
How you got 0.1? Did you assume the initial concentration [A0] to be 0.1 M? If so, [A] would be equal to 0.09 M, not 0.9 M.
oh i did it wrong then i just assumed it was .1 because 10%=.1
same for 90% being .9
 
  • #16
3,812
92
What's the 10% of 0.1?
 

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