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Hi I am obviously not an engineer but I need help with a grade 12 chemistry question. I am having extreme difficulty finding a few pieces of information regarding the Haber process for ammonia production. One is the Arrhenius factor for the reaction
N2(g) + 3H2(g) ---> 2NH3(g)
Perhaps it is inapplicable for this reaction? I have searched the net and looked in several libraries for hours and hours and e-mailed a number of companies with no success. I would hugely appreciate it if someone could tell me the Arrhenius factor or something that will allow me to graph the equilibrium constant against temperature.
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Haber Process
Started by Guest_Guest_*, Jul 30 2004 10:52 PM
11 replies to this topic
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#2
Art Montemayor
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Posted 30 July 2004 - 11:32 PM
Guest:
If you are looking for the values of the equilibrium constant for the synthesis of Ammonia (Haber Process) at various temperatures, then go back to the library and get a copy of the following book:
The Technology and Manufacture of Ammonia
Samuel Strelzoff
John Wiley & Sons
1981; 283 pages
Go to page 65 where you have the data graphed out with the ordinate being the log Kp. On the next page Strelzoff gives you a table of the Kp values for temperatures 298 K up to 1500 K. He also gives you his reference as being:
Comings, E.W.; "High Pressure Technology", McGraw-Hill; NY, 1956; pp 358-359.
I hope this gives you what you want.
Art Montemayor
Spring, TX
If you are looking for the values of the equilibrium constant for the synthesis of Ammonia (Haber Process) at various temperatures, then go back to the library and get a copy of the following book:
The Technology and Manufacture of Ammonia
Samuel Strelzoff
John Wiley & Sons
1981; 283 pages
Go to page 65 where you have the data graphed out with the ordinate being the log Kp. On the next page Strelzoff gives you a table of the Kp values for temperatures 298 K up to 1500 K. He also gives you his reference as being:
Comings, E.W.; "High Pressure Technology", McGraw-Hill; NY, 1956; pp 358-359.
I hope this gives you what you want.
Art Montemayor
Spring, TX
#3
Guest_Guest_*
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Posted 31 July 2004 - 01:37 AM
I am not sure if I'll find the book/s but thankyou very much for giving me some sort of direction.
#4
Art Montemayor
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Posted 31 July 2004 - 10:29 AM
Guest:
I presume you are a young student; and I can understand the frustration of trying to look for data in a location outside the USA - which is where I presume you are located. But there is a bright side to the position you've now arrived at.
You now have two concrete and specific references where the actual data is located - all thanks to your own initiative, ingenuity, and inquiry. You may be in a place that is devoid of technical libraries or sources of information, but you nevertheless have succeeded (through this forum) in obtaining the identity of the references that hold the answer. This is, in my opinion, 99% of the answer already. I recommend you approach your instructor or professor with this answer and kindly request his/her assistance in obtaining the referenced material. Engineering is not a clerical profession; physically obtaining the actual book or copies of it is a clerical job and shouldn't be part of your assignment. Engineers are paid to use their minds - not their feet.
Having emphasized the above, I will also add that if needed, yes we engineers can also do clerical work (& well, too); but it is analogous to using a race horse to plow a field. It doesn't make sense and is wasteful of time and effort. Engineers are not (& shouldn't be) above stooping to doing a lot of clerical, mundane work and effort to obtain their data and basic problem inputs. We will do it if there is no one else or if there is no time to lose. My point is that if your instructor(s) are worthy of their title and the respect you give them, then they should perfectly understand your polite request to assist in obtaining the references for you. They should recognize that you have, essentially, done 99% of the data retrival.
I don't know if the above will work in your case, but while I believe in being strict and hard with engineering students - I also maintain that if they respond with hard, dedicated mental work, they deserve the attention and assistance that they politely request. They earn it by proving their mental prowess and competitiveness.
Good Luck.
Art Montemayor
Spring, TX
I presume you are a young student; and I can understand the frustration of trying to look for data in a location outside the USA - which is where I presume you are located. But there is a bright side to the position you've now arrived at.
You now have two concrete and specific references where the actual data is located - all thanks to your own initiative, ingenuity, and inquiry. You may be in a place that is devoid of technical libraries or sources of information, but you nevertheless have succeeded (through this forum) in obtaining the identity of the references that hold the answer. This is, in my opinion, 99% of the answer already. I recommend you approach your instructor or professor with this answer and kindly request his/her assistance in obtaining the referenced material. Engineering is not a clerical profession; physically obtaining the actual book or copies of it is a clerical job and shouldn't be part of your assignment. Engineers are paid to use their minds - not their feet.
Having emphasized the above, I will also add that if needed, yes we engineers can also do clerical work (& well, too); but it is analogous to using a race horse to plow a field. It doesn't make sense and is wasteful of time and effort. Engineers are not (& shouldn't be) above stooping to doing a lot of clerical, mundane work and effort to obtain their data and basic problem inputs. We will do it if there is no one else or if there is no time to lose. My point is that if your instructor(s) are worthy of their title and the respect you give them, then they should perfectly understand your polite request to assist in obtaining the references for you. They should recognize that you have, essentially, done 99% of the data retrival.
I don't know if the above will work in your case, but while I believe in being strict and hard with engineering students - I also maintain that if they respond with hard, dedicated mental work, they deserve the attention and assistance that they politely request. They earn it by proving their mental prowess and competitiveness.
Good Luck.
Art Montemayor
Spring, TX
#5
Guest_Guest_*
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Posted 01 August 2004 - 02:38 AM
You're correct that I have very limited access to technical information. Thanks very much for your encouragement and advice and although I hope to study in engineering next year, I will repeat that I am only a year 12 (highschool) chemistry student whose job is to do clerical work.
#6
Guest_Guest_*
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Posted 01 August 2004 - 02:53 AM
Also I have read that the Arrhenius equation is:
k = Ae^(-E/(RT))
where:
k = the equilibrium constant of the reaction
e = Euler's number
R = the universal gas constant
T = the temperature of the reaction in degrees Kelvin
A = the Arrhenius factor = pZ
where:
p = the steric factor
Z = a constant related to the collision frequency
I have also read that:
f = e^(-E/(RT))
where f = the fraction of particles with sufficient kinetic energy to successfully react.
This would mean that as temperature decreases, the power of e decreases and so does this fraction. Therefore the rate of both the forward and reverse reactions would decrease. Apparently this was one of the problems faced by Fritz Haber who wanted a low temperature to 'increase' the equilibrium constant and the concentration of ammonia produced. If was the case why does the equilibrium constant also decrease as temperature decreases when considering the Arrhenius equation. That's why I asked is it applicable to this reaction?
k = Ae^(-E/(RT))
where:
k = the equilibrium constant of the reaction
e = Euler's number
R = the universal gas constant
T = the temperature of the reaction in degrees Kelvin
A = the Arrhenius factor = pZ
where:
p = the steric factor
Z = a constant related to the collision frequency
I have also read that:
f = e^(-E/(RT))
where f = the fraction of particles with sufficient kinetic energy to successfully react.
This would mean that as temperature decreases, the power of e decreases and so does this fraction. Therefore the rate of both the forward and reverse reactions would decrease. Apparently this was one of the problems faced by Fritz Haber who wanted a low temperature to 'increase' the equilibrium constant and the concentration of ammonia produced. If was the case why does the equilibrium constant also decrease as temperature decreases when considering the Arrhenius equation. That's why I asked is it applicable to this reaction?
#7
gvdlans
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Posted 02 August 2004 - 03:51 AM
I don't want to do your homework, but maybe following hints will help you to get on:
- Yes, the Arrhenius equation is applicable for both forward and reverse reactions
- You seem to assume that A and E (the activation energy) are same for both forward and reverse reactions. This is not true! Maybe you should write:
k1 = A1e^(-E1/(RT))
for forward reaction, and
k2 = A2e^(-E2/(RT))
for reverse reaction
and then express Keq as a function of k1 and k2...
By the way, the fact that activation energies for forward and reverse reactions are not equal becomes quite clear when you look at the energy profile for the reaction as shown on http://www.chemguide...troduction.html
- Yes, the Arrhenius equation is applicable for both forward and reverse reactions
- You seem to assume that A and E (the activation energy) are same for both forward and reverse reactions. This is not true! Maybe you should write:
k1 = A1e^(-E1/(RT))
for forward reaction, and
k2 = A2e^(-E2/(RT))
for reverse reaction
and then express Keq as a function of k1 and k2...
By the way, the fact that activation energies for forward and reverse reactions are not equal becomes quite clear when you look at the energy profile for the reaction as shown on http://www.chemguide...troduction.html
#8
Guest_Guest_*
Guest_Guest_*
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Posted 03 August 2004 - 07:23 AM
Do you mean that the activation energy of the reverse reaction is the negative of that of the first reaction?
I was recently given the equation:
lnk = -13.75 +6235/T
and I interpreted it as:
k = e^(-13.75 + 6235/T)
= e^-13.75 * e^6235/T
= Ae^(- -6235R/RT)
Is a negative activation energy for the forward reaction correct? If so what does this mean in terms of reaction mechanisms and why is a catalyst so important?
what is the function you referred to.
Thanks
I was recently given the equation:
lnk = -13.75 +6235/T
and I interpreted it as:
k = e^(-13.75 + 6235/T)
= e^-13.75 * e^6235/T
= Ae^(- -6235R/RT)
Is a negative activation energy for the forward reaction correct? If so what does this mean in terms of reaction mechanisms and why is a catalyst so important?
what is the function you referred to.
Thanks
#9
Guest_._*
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Posted 03 August 2004 - 07:24 AM
PS Sorry if I'm a bit hard to get through to. We don't do stuff in much depth at school.
#10
gvdlans
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Posted 03 August 2004 - 07:40 AM
Again, I am not going to do your homework!
The energy profile on the webpage I referred to in my previous post shows you the difference in activation energy for forward and reverse reactions if you realize that for the reverse reaction the products of the forward reaction become the reactants and vice versa. In other words, for the forward reaction you go "over the hill" from left to right, where for the reverse reaction you go from right to left.
You ask about the role of a catalyst. The website I referred to explains this in quite some depth, please look at that first.
The energy profile on the webpage I referred to in my previous post shows you the difference in activation energy for forward and reverse reactions if you realize that for the reverse reaction the products of the forward reaction become the reactants and vice versa. In other words, for the forward reaction you go "over the hill" from left to right, where for the reverse reaction you go from right to left.
You ask about the role of a catalyst. The website I referred to explains this in quite some depth, please look at that first.
#11
Guest_Guest_*
Guest_Guest_*
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Posted 07 August 2004 - 03:04 AM
I have found out that k should be K which is not the equilibrium constant but the rate constant!
#12
gvdlans
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Posted 08 August 2004 - 03:48 AM
Your last post is a bit confusing, but maybe you mean that you made a mistake in your post on Aug 1 2004, 08:53 AM. There you wrote:
"k = Ae^(-E/(RT))
where:
k = the equilibrium constant of the reaction"
This should have been:
k = the reaction rate constant
"k = Ae^(-E/(RT))
where:
k = the equilibrium constant of the reaction"
This should have been:
k = the reaction rate constant
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