Ideal Gas Law and Partial Pressures

In summary: No, 90% yield is the percentage of the theoretical amount of a substance that is actually produced in a chemical reaction.
  • #1
jackthehat
41
5
Homework Statement
Below is the reaction for the industrial synthesis of acrylonitrile (C3H3N) from propene (C3H6):

A 150.0 liter steel reactor at 25 ºC is filled with the following partial pressures of reactants: 0.500 MPa C3H6, 0.800 MPa ammonia, and 1.500 MPa oxygen gas.


The reaction equation that describes the process is as follows

2 C3H6 (g) + 2 NH3 (g) + 3 O2 ---------> 2 C3H3N (g) + 6 H2O (g)


If the reaction described in question part a proceeds to 90% yield, and no side reactions take place, what is the final total pressure in the reaction vessel at 400 ºC?
Relevant Equations
PV = nRT ; Total Pressure = P1 + P2 + ... P(n)
I had already found the Mass of the product (C3H3N) produced by this reaction (theoretical mass at 100% yield) in a previous problem. I did this by finding the Limiting Reagent (C3H6) in the reaction , calculating the number of moles of C3H6 and using the Molar Ratios in the balanced reaction equation to find the number of moles of product produced (C3H3N) and so from this .. calculating the product Mass, and that Mass was found to be 1603.8g (1.604Kg).
Now to solve this problem as described in Homework Statement section, do I now calculate the mass of the other product (H2O) and then get the number of moles of both these products and of the non-limiting reactants (NH3 and O2) then from here apply the Ideal Gas Law to find the Partial Pressures of each of the non-limiting reactants and products (NH3; O2; C3H3N and H2O) sum all of these Partial Pressures to get a Total final Pressure value then finally multiply the final sum of these pressures by 0.9 (ie. 90%) to find the correct total pressure ?
Is this correct or am I on the wrong track ? If so is there some alternative way of tackling this problem ?
 
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  • #2
Not sure why you use mass for anything, just follow the moles.

If C3H6 is the limiting reagent, how much of the initial amount was converted into the product? How much was left? How much ammonia and oxygen were consumed in the reaction? Basically: how many moles of each of these gases will be present in the final mixture?

Then, yes, ideal gas equation FTW ;)
 
  • #3
Hi Borek,
Thank you once again for taking the time to help me.
A couple of things .. Sorry should have explained more clearly why I calculated mass of product earlier. There were two parts to this question .. first part specifically asked for the mass of product synthesised by this reaction for 100% yield.
The problem above was the second part (part B). Two particular things confused me about this .. one was I was marked wrong for my answer but was never given the reason why my answer was wrong .. so I wasn't sure if it was my reasoning and method that was wrong or if it was just a careless mistake with my arithmetic (the caculations). So I wasn't sure whether I was on the right track or not with the method I used. Secondly I spoke to someone about this and he said the problem had to be solve using molar concentrations of the gases and that totally confused me as I have never come across that in any other problem of this type.

Just one other small query on this one ... since the limiting reagent (C3H6) I assume is consumed entirely in the reaction then and I don't include it in the calculations. Instead I include number of moles of the products (C3H3N and H2O) produced, together with the excess moles of the NH3 and the Oxygen left over at reaction's end .. calculate the partial pressures of each (using PV=nRT) .. and then sum these partial pressures to find a final total pressure value. If so do I multiply the mole amounts for each substance above by 0.9 (to satisy 90% yield condition) at the start of the calculations or wait until I have got the final total pressure and multiply that by 0.9 at the end ?

Regards,
Jack
 
  • #4
jackthehat said:
Just one other small query on this one ... since the limiting reagent (C3H6) I assume is consumed entirely in the reaction then and I don't include it in the calculations. Instead I include number of moles of the products (C3H3N and H2O) produced, together with the excess moles of the NH3 and the Oxygen left over at reaction's end .. calculate the partial pressures of each (using PV=nRT) .. and then sum these partial pressures to find a final total pressure value. If so do I multiply the mole amounts for each substance above by 0.9 (to satisy 90% yield condition) at the start of the calculations or wait until I have got the final total pressure and multiply that by 0.9 at the end ?
Sorry to say, but this is completely wrong.

What does 90% yield mean?
 
  • #5
Hi there,
Thank you for replying to my query ..
Now you say that my above method of solving the problem is completely wrong .. and you seem to infer that my understanding of 'yield' in a chemical reaction is not correct .. do you mean that 90% yield is not a reduced amount of product after a reaction has completed ?
I thought it to mean that the actual amount of the product produced was 90% (or 0.9) of the theoretical amount (of moles) of a substance expected to be produced in a 100% complete reaction of the same type.
Is this wrong ? If so I am, completely confused now ..
I took it to be that the amount of product expected if we go by the reaction equation is the theoretical (100%) yield .. and a 90% yield is therefore 9/10 of this theoretical yield (in number of moles of product) .. no ?
Could you please clear this up for me .. if you don't, mind .. because I not sure what to do now to start to solve this problem. Sorry but I am confused as hell now.
regards,
Jackthehat
 
  • #6
jackthehat said:
I thought it to mean that the actual amount of the product produced was 90% (or 0.9) of the theoretical amount (of moles) of a substance expected to be produced in a 100% complete reaction of the same type.

That's correct, but this:

jackthehat said:
since the limiting reagent (C3H6) I assume is consumed entirely in the reaction

doesn't follow. Mass is conserved, it can't just disappear.
 
  • #7
In a limiting reagent problem like this, it is probably easiest to start with that reagent. As long as there are no side reactions (which you are told is the case), 90% yield means 90% of the limiting reagent is consumed. So
how many moles of propene disappear, and how many moles are left?
how many moles of ammonia disappear, and how many moles are left?
how many moles of oxygen disappear, and how many moles are left?
how many moles of acrylonitrile are formed?
how many moles of water are formed?
 
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  • #8
Hi mjc123.
Thank you for that explanation .. I see where I am going wrong I'm think now. As I said earlier although I had been marked wrong for my solution I was given no feedback as to why and I wanted to understand the problem and solution more clearly.
Just one little thing I would like to get clear if you don't mind. as I said in an earlier post regarding this problem .. I spoke to a colleague about it and he said that because the temperature in the reaction chamber had changed in this scenario .. I had to solve this problem using the concentrations of the gases .. is this incorrect ?
Regards,
Jackthehat
 
  • #9
Hi again everyone,
I went over my problem again taking into account all of what I hope I have learned in this forum from all of you who have helped and tackled the problem again from the begining.
I calculated the number of mole of C3H6 (Limiting Reagent) that theoretically would react completely in the reaction using Ideal gas Law from the Partial Pressure, Volume and Temperature given in the problem description.
Then from the balanced equation's Molar Ratios I then calculated the number of Moles of the other reactants that would react with the Limiting Reagent to complete the reaction and the number of moles of product that would be produced.
Since we are told that the reaction has a 90% yield on completion then as explained by yourselves 90% of the C3H6 is used up so 10% remains. As regards the other reactants I subtracted their theoretical starting amounts (in moles) from the amount of Moles of C3H6 used (it's 90% amount) to obtain how much of these are left over at the end. Then coming to the products (C3H3N and H2O) I estimated the 90% amount for each of their theoretical values (in moles).
The 90% left over amounts for NH3 and O2 I got as 16.335 moles and 40.842 moles respectively.
The 10% amount left of C3H6 (Limiting Reagent) I got as 3.026 moles (10% of 30.26 moles theoretical).
The 90% values for the products (C3H3N and H2O) I got as 27.234 moles and 163.394 moles respectively.
I then used each of these Mole values separately in the equation P = nRT/V to find their Partial Pressures. Finally I summed all of these Partial Pressures to get the Final Total Pressure in the Reaction Chamber an I came up with the final value Total Pressure = 8.98 Mpa.
Does this look okay to you. Is it the correct way to go about solving this problem.
I think what I am asking is have I understood the problem and the advice you have given me .. or have I still not fully understood it yet ?
Regards,
Jackthehat
 
  • #10
jackthehat said:
The 90% left over amounts for NH3 and O2 I got as 16.335 moles and 40.842 moles respectively.
The 10% amount left of C3H6 (Limiting Reagent) I got as 3.026 moles (10% of 30.26 moles theoretical).
The 90% values for the products (C3H3N and H2O) I got as 27.234 moles and 163.394 moles respectively.
I then used each of these Mole values separately in the equation P = nRT/V to find their Partial Pressures. Finally I summed all of these Partial Pressures to get the Final Total Pressure in the Reaction Chamber an I came up with the final value Total Pressure = 8.98 Mpa.
Check the number in red. How did you get the values for NH3 and O2?
 
  • #11
Hi DrClaude,
Thanks for getting back to me .. I will list all of the amounts for each of the reactants and products below ..
The mole amounts I got (for a 90% yield reaction) are as follows ...
C3H6 = 16.335 moles
O2 = 40.082 moles
C3H6 = 3.026 moles
C3H3N = 27.234 moles
H2O = 163.394 moles

I will look at my calculation for the moles of H2O , the reason I got so big a value I think is because in the reaction equation we have 2 moles of C3H6 producing 6 moles of H2O so it is 3 times the molar ratio of the Limiting Reagent ... Ah I have just noticed my mistake , why I have so large a value .. I have mistakenly taken the molar ratio as 1:6 instead of 2:6 so it should actually be half of what I have written down above .. it should be ..
H2O = 81.697 moles.
Thank you for bringing this careless mistake to my attention.
This mistake apart is the method I have used above a valid one .. am I doing it the right way ?
Regards,
Jackthehat
 
  • #12
I'm sorry, but there are still some numbers I don't get, like
jackthehat said:
C3H6 = 16.335 moles
Can you please state the amount of each substance before the reaction, and the amount after the 90%-yield reaction?
 
  • #13
Hi again DrClaude,
I am sorry for the confusion I have caused on this .. the amounts before the reaction and the amounts after the 90% yield reaction that I estimated are as follows ...

Substance Before reaction After 90% yieldd
C3H6 30.255 moles 3.0255 moles (10% of limiting reagent left)
NH3 43.569 moles 16.309 moles
O2 81.693 moles 54.433 moles

Products
C3H3N 27.260 moles
H2O 81.780 moles

These are the amounts (in moles) I calculated after I went back to check things since your last but one reply.
I hope that this helps you understand what I was doing better. Once you have a look at the numbers can you let me know if you think I am on the right track to solvingthis problem correctly. Thanks.
Regards,
Jackthehat
 
  • #14
jackthehat said:
Substance Before reaction After 90% yieldd
C3H6 30.255 moles 3.0255 moles (10% of limiting reagent left)
NH3 43.569 moles 16.309 moles
O2 81.693 moles 54.433 moles
The numbers for NH3 and O2 are incorrect. I think you are applying the 90% too early.
 
  • #15
Hi again,
I applied the 90% immediately after I calculated the number of moles for each of the reactants .. the actual figures for each substance before the reaction .. the figures for NH3 and O2 before I applied the 90% were ..
NH3 = 48.42 moles
O2 = 90.79 moles
When do I apply the 90% then ?
Are the mole figures I gave for after the reaction correct .. that is
NH3 = 16.309 moles
O2 = 54.433 moles
Regards,
Jackthehat
 
  • #16
jackthehat said:
When do I apply the 90% then ?
That part you did correctly now. You had the initial amounts, you figured out that C3H6 was the limiting reagent, so calculated that you have 27.260 moles of C3H6 that reacted, or 90%. The following is also correct:
jackthehat said:
Products
C3H3N 27.260 moles
H2O 81.780 moles

The problem is with NH3 and O2. You wrote before that you had initially
jackthehat said:
NH3 43.569 moles
O2 81.693 moles
but you now have the correct numbers, which are
jackthehat said:
NH3 = 48.42 moles
O2 = 90.79 moles
Use these numbers to figure out how much NH3 and O2 are left after the reaction.
 
  • #17
Hi DrClaude,
Thank you for all the help you gave me .. they say that you only really learn by your mistakes and I think that I now have a better understanding of things. Better late than never I suppose. Thanks agin for all your help.
Regards,
Jackthehat
 
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FAQ: Ideal Gas Law and Partial Pressures

What is the ideal gas law?

The ideal gas law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of a gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

What is the significance of the ideal gas law?

The ideal gas law is significant because it helps us understand the behavior of gases under different conditions. It allows us to predict the changes in pressure, volume, temperature, and number of moles of a gas when one or more of these variables are altered.

How does the ideal gas law relate to partial pressures?

The ideal gas law can be used to calculate the partial pressure of a gas in a mixture of gases. The partial pressure of a gas is the pressure it would exert if it occupied the same volume on its own at the same temperature. The total pressure of a gas mixture is equal to the sum of the partial pressures of each gas in the mixture.

What factors affect the partial pressure of a gas in a mixture?

The partial pressure of a gas in a mixture is affected by the number of moles of the gas, the total pressure of the mixture, and the temperature. As the number of moles of a gas increases, its partial pressure also increases. Similarly, as the total pressure of the mixture increases, the partial pressure of each gas in the mixture also increases. Additionally, an increase in temperature will lead to an increase in the partial pressure of a gas in a mixture.

How is the ideal gas law used in real-world applications?

The ideal gas law is used in many real-world applications, such as in the design of gas storage tanks, the production of industrial gases, and the study of atmospheric gases. It is also used in the development of weather forecasting models and in the design of gas pipelines. Additionally, the ideal gas law is used in the analysis of gas mixtures in medical and scientific research.

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