Calculate the temperature for this reaction to occur

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SUMMARY

The discussion centers on calculating the temperature required for a reaction involving deuterium and protons using the kinetic energy formula KE = 3/2 R/NA T. Participants clarify that the average kinetic energy (KE) for each particle type is derived from the total KE of the system, which is shared between the two particle types. The average KE for each particle is 1.2 x 10^-14 J, indicating that this energy level is necessary for the reaction to commence. The conversation emphasizes the importance of understanding how kinetic energy is distributed among particles in a thermal motion context.

PREREQUISITES
  • Understanding of kinetic energy equations, specifically KE = 3/2 R/NA T
  • Familiarity with the Boltzmann Constant and its application in thermodynamics
  • Basic knowledge of particle physics, particularly regarding deuterium and protons
  • Concept of root mean square velocity (v(rms)) in relation to temperature
NEXT STEPS
  • Study the derivation and applications of the kinetic energy formula KE = 3/2 R/NA T
  • Explore the concept of thermal motion and its impact on particle interactions
  • Learn about the distribution of kinetic energy among different particle types in a mixture
  • Investigate the role of collision dynamics in reaction rates and temperature calculations
USEFUL FOR

This discussion is beneficial for students in physics or chemistry, particularly those studying thermodynamics, reaction kinetics, or particle physics. It is also useful for educators seeking to clarify concepts related to kinetic energy and temperature in chemical reactions.

Janiceleong26
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1. Homework Statement
image.jpg

Homework Equations


KE = 3/2 R/NA T

The Attempt at a Solution


By using the formula above, I used 2.4 x 10^-14 J as the mean translational KE , but it should be 1.2x10^-14 , why? I thought the mixture contains both deuterium nucleus and the proton? Why half KE of mixture?
 
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Janiceleong26 said:
View attachment 100665 1. Homework Statement
View attachment 100666

Homework Equations


KE = 3/2 R/NA T

The Attempt at a Solution


By using the formula above, I used 2.4 x 10^-14 J as the mean translational KE , but it should be 1.2x10^-14 , why? I thought the mixture contains both deuterium nucleus and the proton? Why half KE of mixture?
what happens to the particles kept at some temperature-they get energy and start thermal motion - the average K.E. of particles is related to the temp. and Boltzmann Constant.
3/2 is a factor depending on degree of freedom.
so, what amount of energy a particle say proton will get if you are given with Total K.E. of proton and deuterium? that one should find and then apply the energy-temp relation.
 
Janiceleong26 said:
By using the formula above, I used 2.4 x 10^-14 J as the mean translational KE , but it should be 1.2x10^-14 , why? I thought the mixture contains both deuterium nucleus and the proton? Why half KE of mixture?
At a given temperature, how much KE will each particle have? (You have to assume some collisions will be head on.)
 
drvrm said:
what happens to the particles kept at some temperature-they get energy and start thermal motion - the average K.E. of particles is related to the temp. and Boltzmann Constant.
3/2 is a factor depending on degree of freedom.
so, what amount of energy a particle say proton will get if you are given with Total K.E. of proton and deuterium? that one should find and then apply the energy-temp relation.
haruspex said:
At a given temperature, how much KE will each particle have? (You have to assume some collisions will be head on.)
Half of the total KE, but why? I thought they are asking for the sample? Why do we need to consider KE of just one of them?
 
Janiceleong26 said:
Half of the total KE, but why? I thought they are asking for the sample? Why do we need to consider KE of just one of them?
They are asking for the temperature at which you would expect the reaction to occur. It doesn't have to occur for all particle pairs immediately. Leaving aside that some will have greater than the average KE, they will be moving in various directions. From the temperature you can find the average KE of any given particle, but the reaction involves two particles. Their relative directions of travel matter. Which case makes them most likely to reach the required KE level?
 
Janiceleong26 said:
Half of the total KE, but why? I thought they are asking for the sample? Why do we need to consider KE of just one of them?

i think the average velocity of a particle in an enclosure at Temp. T may be taken as v(rms) and
1/2 . m(i). v(i)^2 = 3/2 k T ; now one has to check how the total KE will be shared.
 
drvrm said:
now one has to check how the total KE will be shared.
Not sure what you mean by that. Shared between what?
 
haruspex said:
Shared between what?

shared between the two kinds of particles namely deuteron and protons.
 
drvrm said:
shared between the two kinds of particles namely deuteron and protons.
Doesn't the equation in your post answer that?
 
  • #10
haruspex said:
Doesn't the equation in your post answer that?
well it does -but the question is being asked as to why half of the total KE of the two (given in the question) is being used for estimating T.
 
  • #11
haruspex said:
They are asking for the temperature at which you would expect the reaction to occur. It doesn't have to occur for all particle pairs immediately. Leaving aside that some will have greater than the average KE, they will be moving in various directions. From the temperature you can find the average KE of any given particle, but the reaction involves two particles. Their relative directions of travel matter. Which case makes them most likely to reach the required KE level?
When they collide head-on?
Does it mean that we have to supply 1.2x10^-14 J of energy to both deuterium and the proton for the reaction to start?
 
  • #12
Janiceleong26 said:
When they collide head-on?
Does it mean that we have to supply 1.2x10^-14 J of energy to both deuterium and the proton for the reaction to start?
It means that given those average KEs each, you can expect the reaction to start.
 
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  • #13
haruspex said:
It means that given those average KEs each, you can expect the reaction to start.
Ok, got it, thanks very much!
 

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