If the rate of a reaction increases by a factor of 2.5 when the
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Question

If the rate of a reaction increases by a factor of 2.5 when the 
temperature is raised from 20 °C to 30 °C, what is the value 
of the activation energy in kJ/mol? By what factor does the 
rate of this reaction increase when the temperature is raised 
from 120 °C to 130 °C?

Accepted Answer

Hello everyone today. We are being given the falling problem, calculate the activation energy in joules per mole over reaction. If the rate of the reaction increases by a factor of 3.5 when the temperature is increased from 40 degrees Celsius to 60 degrees Celsius. If the temperature is increased raised from 1 10 to 30 degrees Celsius by how much will the rate of the reaction increase? So first we need to write out our two point form of the Iranians equation which is the natural log of our two constants divided by each other. So we have the second constant divided by the first constant is equal to the negative activation energy over our gas constant Times one over our final temperature -1 over our initial temperature. So First we need to find our temperatures in Kelvin. So we're gonna do is we're gonna take our final temperature, our final temperature was 60 degrees And we're gonna add to 73.15 And that's gonna give us 333.15 Kelvin. And for initial temperature which is 40°C, we're gonna do the same thing. That's gonna give us 3 13.15 Kelvin. And so we can simply plug in these values. So we're gonna have the natural log of our 3. is equal to our activation energy over r gas constant which is 8. joules per mole kelvin times one over our final temperature which is 3 33.15 kelvin minus one over initial temperature which was 3 13.15 kelvin. When we saw for our activation energy, our activation energy will equal 554,329. jewels per mole. However, we need this in terms of units of killing jewels per mole. So we're going to use the conversion factor that one kill a jewel is equal to 10 to the third jewels. When we do that, we get an activation energy, that's equal to 54. kg joules per mole. And so this is going to be our activation energy for that increase. Secondly, we're going to do the same but we are going to find the change like the factor change with the rate of the reaction. So we're gonna do the same process, we're gonna need our final temperature and initial temperature and degrees of kelvin. So for our final temperature, that was 1 30 degrees Celsius plus our 73.15 kelvin. And that's gonna give us 403.15 Kelvin. Their initial temperature was 1 10 degrees Celsius plus 73.15. And that's gonna equal 3 83.15 Kelvin. And our activation energy As we have cells before was 54.33 kg per mole or 54,330 joules per mole. So in setting this up, we're going to have our natural log of our two constants. Our final over our initial is equal to negative activation energy which was 54,330 jules primal over our rate or gas constant, which is 8.3 14 jules per mole kelvin. And we're going to multiply that by our one over our final temperature which was 403.15 kelvin over our initial temperature 1/3 83.15 kelvin. In doing that, we're going to get our natural log Of our constants equal to 0.84611. Now, to get rid of this natural log, we simply need to raise each of these to the power of E. So when you raise these to the power of E, we're just going to get our two constants is equal to 2.3. So our final answers will be 2.3 for the rate of the reaction that's increased with a factor that it increases by And then we're gonna have our activation energy which is 54.33 kg per mole. And with that we have answered the question overall, I hope this helped. And until next time

If the rate of a reaction increases by a factor of 2.5 when the temperature is raised from 20 °C to 30 °C, what is the value of the activation energy in kJ/mol? By what factor does the rate of this reaction increase when the temperature is raised from 120 °C to 130 °C?

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