# Half life and activation energy

by Hemolymph
Tags: activation, energy, life
 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
P: 30
 Quote by Borek Have you heard about Arrhenius equation?
I cant see how I can apply the Arrhenius equation if I dont have any activation energies

P: 3,807
Half life and activation energy

 Quote by Hemolymph 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:
$$k_1=Ae^{-E_a/RT_1}$$
$$k_2=Ae^{-E_a/RT_2}$$
Take logarithm on both the sides of the equation and subtract the equations you get.
P: 30
 Quote by Pranav-Arora 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: $$k_1=Ae^{-E_a/RT_1}$$ $$k_2=Ae^{-E_a/RT_2}$$ 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)?
P: 3,807
 Quote by Hemolymph 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. )
 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
P: 3,807
 Quote by Hemolymph 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.
 P: 30 Do you know how i could tackle the second one ?
P: 3,807
 Quote by Hemolymph 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?
 P: 30 Would I just be able to just substitute the percentages in as if they were concentrations?
P: 3,807
 Quote by Hemolymph 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.
 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?
P: 3,807
 Quote by Hemolymph (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.
P: 30
 Quote by Pranav-Arora 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
 P: 3,807 What's the 10% of 0.1?

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