Half life and activation energy

In summary: It's not 0.1. It's 0.01. And 90% of 0.1 is 0.09. So, your calculation is incorrect. In summary, the first question asks for the ratio of the rate constant at two different temperatures, given the activation energy. To solve it, we can use the Arrhenius equation and take logarithm on both sides to relate the two rate constants. The result is 2.5, which means the rate constant at 170°C is 2.5 times greater than at 150°C.For the second question, we are given that the reaction is second order and 10% complete after 20 seconds. We need to find the time it takes
  • #1
Hemolymph
30
0

Homework Statement


I have two questions that I just don't 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
 
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  • #2
Have you heard about Arrhenius equation?
 
  • #3
Borek said:
Have you heard about Arrhenius equation?

I can't see how I can apply the Arrhenius equation if I don't have any activation energies
 
  • #4
Hemolymph said:
I can't see how I can apply the Arrhenius equation if I don't 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
Pranav-Arora said:
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
Hemolymph said:
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
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
Hemolymph said:
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
Do you know how i could tackle the second one ?
 
  • #10
Hemolymph said:
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
Would I just be able to just substitute the percentages in as if they were concentrations?
 
  • #12
Hemolymph said:
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
(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
Hemolymph said:
(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
Pranav-Arora said:
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
What's the 10% of 0.1?
 

FAQ: Half life and activation energy

1. What is half life?

Half life is the amount of time it takes for half of a substance to decay or become inactive. This concept is commonly used in chemistry and nuclear physics to describe the rate at which radioactive elements decay.

2. How is half life calculated?

The half life of a substance can be calculated using the equation t1/2 = ln(2)/k, where t1/2 is the half life, ln is the natural logarithm, and k is the rate constant of the substance.

3. What is activation energy?

Activation energy is the minimum amount of energy required for a chemical reaction to occur. It is often represented by the symbol Ea and is an important factor in determining the rate of a reaction.

4. How does activation energy affect the rate of a reaction?

The higher the activation energy, the slower the rate of a reaction. This is because a higher activation energy means that more energy is required for the reaction to occur, making it less likely to happen. Lower activation energy means that the reaction can occur more easily and at a faster rate.

5. How can the activation energy of a reaction be lowered?

The activation energy of a reaction can be lowered by adding a catalyst, which is a substance that increases the rate of a reaction without being consumed in the process. Catalysts work by providing an alternate pathway for the reaction to occur with a lower activation energy.

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