If the inventory of the reactor of about 20,000kg

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Discussion Overview

The discussion revolves around a homework problem involving the oxidation of zirconium (Zr) by steam, focusing on the calculations of heat generated, hydrogen produced, energy released from combustion, and pressure changes in a reactor. Participants explore stoichiometric relationships, ideal gas behavior, and the implications of their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the heat generated from Zr oxidation as 13.5x10^3 MJ based on the given reaction and parameters.
  • Another participant suggests that the volume fraction of hydrogen should consider both remaining water and produced hydrogen, proposing a temperature of 100°C for calculations.
  • Discussions on the ideal gas law are raised, with participants confirming the total volume of the reactor as 90,000 m³.
  • Several participants debate the correct stoichiometric relationships, with one asserting that the mass of hydrogen produced should be calculated using stoichiometry rather than enthalpy change.
  • Participants express confusion over the correct number of moles of hydrogen produced, with varying interpretations of stoichiometric coefficients leading to different conclusions.
  • One participant correctly identifies the number of moles of Zr in 20,000 kg and calculates the corresponding moles of hydrogen produced, but others challenge the accuracy of their calculations and units used.
  • There is a recurring emphasis on the importance of using consistent units and correctly interpreting stoichiometric coefficients in the context of the reaction.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations, with multiple competing views on the correct approach to stoichiometry and the resulting quantities of hydrogen and energy. Disagreements persist regarding the interpretation of the reaction and the application of the ideal gas law.

Contextual Notes

Participants express uncertainty about the correct application of stoichiometric principles and the implications of their calculations, indicating a need for clarity on the relationships between mass, moles, and gas behavior in the context of the reaction.

Who May Find This Useful

This discussion may be useful for students studying chemical engineering, chemistry, or related fields, particularly those interested in stoichiometry and gas laws in reaction contexts.

matt222
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Homework Statement


If the inventory of the reactor of about 20,000kg of Zr given the reaction:
Zr+2H2O---> ZrO2+2H2, Q=616MJ/Kg-mole of Zr

1- find the 10% of the heat generated from Zr oxidation by steam in MJ
2- Find the 10% hydrogen generated from Zr oxidation by steam in kg and the volume fraction (given initial pressure 1 bar and volume 90,000m^3, treat hydrogen as an ideal gas.
3- what is the energy released from the combustion of H generated and what will be the power if it is released within 10s, assumption is the amount of Oxygen is enough to burn the total H generated, heat of combustion is 240MJ/Kg-mol
4- what will be the pressure increased by the H generation from 10% oxidation given initial pressure 1 bar and volume 90,000m^3, treat hydrogen as an ideal gas, ignorw the T rise.


Homework Equations





The Attempt at a Solution


1- it is simple :
molar mass for Zr is 91.22 kg/kmole

10%x 20000kgx 616/91.22 =13.5x10^3 MJ

2- it is by: 2kg/kmol of 2H, so 2x13.5x10^3/(616)=43.8kg
I didnt understand the second part of volume fraction?

3- energy release from the combustion of H:

H2+1/2O2---->H2O+ 240MJ/Kgmol

240x43.8=10.51x10^3MJ

the power is going to be:

10.51x10^3MJ/10=10.51x10^2MW

4-I have no idea but I need a hint

is my work true
 
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I guess they mean volume fraction in the mixture after the reaction - it still contains remaining water AND produced hydrogen. However, it is not clear to me what is the temperature. Perhaps just 100°C, as that's enough for water to be a steam at 1 bar.

Convert between mass/moles and volume using ideal gas equation, nothing more fancy.
 


so PV=nRT, so in this case 90,000m^3 going to be the total volume right?

n=2kg/kmol for hydrogen once we find the volume of hydrogen just divide it by the total volume of 90,000m^3 is it right?
 


the T=100c you are right
 


matt222 said:
n=2kg/kmol for hydrogen once we find the volume of hydrogen just divide it by the total volume of 90,000m^3 is it right?

Yes.

Note that this case is specific - total number of moles of gas doesn't change (one mole of H2O is replaced with one mole of H2 - this is in the stoichiometric coefficients). It doesn't have to be true, which may add an additional step to calculations.
 


so the is going to be V of hydrogen=8.32x10^3x373x2/1x10^5=31m^3

which is very small!

volume fraction=31/90,000=3.45x10^-4!

What do you think
 


There is in general something wrong with your numbers.

matt222 said:
2- it is by: 2kg/kmol of 2H, so 2x13.5x10^3/(616)=43.8kg

Why do you mix 616 (which is enthalpy change) with masses, when calculating stoichiometry?
 


oops yes you are right its mistake it should be 13.5x10^3/616=21.9kg>>what about now?
 


No, it shouldn't be. You can't calculate mass produced using enthalpy change (which would be obvious if you were paying attention to the units). You start with masses, you end with masses. This is a simple stoichiometry.
 
  • #10


I confused really because this is not my background field!
can you refer me to a wiki link for understanding more because I really lost
 
  • #12


Zr+2H2O---> ZrO2+2H2
91.1kg/kmol for Zr...2kg/kmol of 2H
10%x20,000kg of Zr...Y

in this case Y is the required mass for this problem! which is equal to 43.91kg! almost the same as what Igot before!
 
  • #13


No, it is still wrong. 2 kg is a correct mass of a kilomole of hydrogen, but you are not producing 1 kilomole of hydrogen (H2) per kilomole of Zr.
 
  • #14


matt222 said:
Zr+2H2O---> ZrO2+2H2
91.1kg/kmol for Zr...2kg/kmol of 2H
10%x20,000kg of Zr...Y

in this case Y is the required mass for this problem! which is equal to 43.91kg! almost the same as what Igot before!

91.1kg/kmol for Zr...4kg/kmol of 2H
10%x20,000kg of Zr...Y

87.82kg what about now? this is what i understand from the link? really almost will be crazy from chemistry lol
 
  • #15


can you give me example to understand this problem please
 
  • #16


matt222 said:
4kg/kmol of 2H

Not 4kg/kmol of 2H but 2 times 2kg/kmol of H2. Think what each 2 in 2H2 means.

87.82kg what about now?

That's the correct mass of hydrogen which is a first part of the answer. Not you have to convert it to the volume.
 
  • #17


so the is going to be V of hydrogen=8.32x10^3x373x4/1x10^5=124m^3

volume fraction=124/90,000=1.38x10^-3! what about this one?
 
  • #18


Units please - I am not going to guess what you are doing.

I think you are wrong, but without knowing what each number represents I can't be sure.
 
  • #19


so the is going to be V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K) x4 (Kmol)/1x10^5(pa)=124m^3

volume fraction=124/90,000=1.38x10^-3! what about this one?

this is after showing the units!
 
  • #20


Why 4 kmols? Is that amount of hydrogen produced?
 
  • #21


sorry i mixed up it should be 2kmol which is the number of hydrogen mol

so the is going to be V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K) x2 (Kmol)/1x10^5(pa)=61m^3

volume fraction=62/90,000=6.89x10^-4!

I think you will be happy right
 
  • #22


No, you are not producing 2 kmols of hydrogen. You took both numbers - 4 kmols and 2 kmols - from the stoichiometric equation, ignoring mass of Zr present. Following this line of thinking amount of hydrogen produced will not depend on the amount of Zr and it will be always 2 (or 4) kmols (even if there is no zirconium present). That's wrong - that's not what stoichiometric coefficients say.

Note that you have already earlier calculated (correctly!) amount of hydrogen produced using stoichiometry - you just calculated its mass, not number of moles. Now either convert this mass to moles or - better - use similar approach to calculate number of moles of hydrogen produced. THEN put this number of moles of hydrogen into ideal gas equation.
 
  • #23


Zr+2H2O---> ZrO2+2H2
1 : : 2

0.5 : : x

x=1


so the is going to be V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K) x1 (Kmol)/1x10^5(pa)=31m^3

volume fraction=31/90,000=3.44x10^-4

what about now
 
  • #24


You don't have 1 Zr (in whatever units), you have 20,000 kg of Zr.

Please reread the section on "how to read balanced reaction equation" from the lectures. You are right about the ratio, but you are all the time doing the same mistake - you are treating stoichiometric coefficients as if they were amounts of substance. THEY ARE NOT.
 
  • #25


Zr+2H2O---> ZrO2+2H2
1 mole of Zr: : 2 mole of 2H2

1 mole : : x

x=2

so the is going to be V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K) x2 (Kmol)/1x10^5(pa)=62m^3


sorry but i am not professional on these kind of material this is my first time in my life to do this!
 
  • #26


Sorry, all I can do is to repeat - you don't have 1 moles of Zr, you have 20,000 kg of Zr.

How many moles (or kmoles) is 20,000 kg of Zr? That will be the number that will be the basis of your calculations:

Zr+2H2O---> ZrO2+2H2

1 mole of Zr : 2 mole of 2H2 - this is ratio given by the stoichiometric coefficients

number of moles of Zr in 20,000 kg : x moles of H2 produced - substances in your reaction must be in the same ratio
 
  • #27


1 mole of Zr=91.2kg/mol, so total for one 1 of Zr we have= 10%x20,000/91.2=21.93mol total moles

Zr+2H2O---> ZrO2+2H2

1 mole of Zr : 2 mole of 2H2

21.93mol : x moles of H2 produced

x=43.86moles total


V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K)x 43.86/1x10^5(pa)=1359.5m^3
 
  • #28


Omitting units and making two errors you managed to get the correct answer, as your errors canceled:

matt222 said:
1 mole of Zr=91.2kg/mol, so total for one 1 of Zr we have= 10%x20,000/91.2=21.93mol total moles

Should be 21.93 kmol, as molar mass of Zr is 92.1g (not kg) per mole.

x=43.86moles total

You started with kmoles, so you should get 43,86 kmol

V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K)x 43.86/1x10^5(pa)=1359.5m^3

You didn't write 43.86 of what - you wrote moles above, would you put moles here, you would be off by the factor of 1000. Accidentally equation expects you to put kmoles, as these are units of the R (ideal gas constant) in the version you used. That was the second error - and it put you back on track.
 
  • #29


1 mole of Zr=91.2g/mol, so total for one 1 of Zr we have= 10%x20,000/91.2=21.93kmol total moles

Zr+2H2O---> ZrO2+2H2

1 mole of Zr : 2 mole of 2H2

21.93kmol : x moles of H2 produced

x=43.86 kmoles total


V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K)x 43.86 kmoles/1x10^5(pa)=1359.5m^3

what about the final new look
 
  • #30


Much better, although you are still doing tricks (perhaps you are not even aware you are doing them):

matt222 said:
1 mole of Zr=91.2g/mol

OK

so total for one 1 of Zr we have

No idea what it means "one 1 of Zr".

= 10%x20,000/91.2=21.93kmol total moles

You are again mixing units - getting correct result. I am not convinced you know what you are doing, I have a feeling result is correct accidentally.

20,000 of what? 91.2 of what?
 

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