Calculating Final Composition of Compounds in Haber's Process with 50% Yield

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Discussion Overview

The discussion revolves around calculating the final composition of compounds in the Haber process given a 50% yield of ammonia (NH3) from a starting mixture of nitrogen (N2) and hydrogen (H2). Participants explore the implications of yield on the reactants and products, including the concept of limiting reagents and conservation of volume in gas reactions.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents an initial calculation for the final composition of N2, H2, and NH3 based on a 50% yield, leading to a proposed final state of 20L N2, 0L H2, and 20L NH3.
  • Several participants question the assumption that the amounts of reactants and products can be directly equated, particularly in the context of yield definitions.
  • There is a discussion about the conservation of volume in gas reactions, with references to other chemical reactions to illustrate the concept.
  • Participants debate whether all of the limiting reagent (H2) is consumed when only 50% of NH3 is produced, leading to confusion about the final amounts of reactants left.
  • A later reply suggests a recalculation of the amounts based on the yield, proposing that if 50% of the theoretical yield is achieved, then 10L of NH3 is produced, with 15L of H2 and 25L of N2 remaining.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and assumptions regarding limiting reagents and yield. There is no consensus on the final amounts of reactants and products, as multiple interpretations of the yield and stoichiometry are presented.

Contextual Notes

Limitations in the discussion include unclear definitions of yield and limiting reagents, as well as unresolved calculations regarding the final volumes of reactants and products.

Who May Find This Useful

Students studying chemical reactions, particularly in the context of stoichiometry and gas laws, may find this discussion relevant.

Suyash Singh
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Homework Statement


30 litres of N2 and 30 litres of H2 are taken in Habers process with 50% yield.What is the final composition of the compunds in litres?

Homework Equations


N2+3H2->2NH3

The Attempt at a Solution


N2 + 3H2-> 2NH3
intial: 30 30 0
final: 30 - x/2 30-(3x)/2 (2x)/2

so,
30 - x/2 + 30-(3x)/2 = (2x)/2

this gives x=20
N2=20L
H2=0L
NH3=20L

Am i right?
 
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Suyash Singh said:
so,
30 - x/2 + 30-(3x)/2 = (2x)/2
Why do you expect these two to be equal?

How is the yield defined? Using all hydrogen with just 50% yield is odd.
 
mfb said:
Why do you expect these two to be equal?

Howis the yield defined? Using all hydrogen with just 50% yield is odd.
litres are conserved
50 percent yield means half of theoretical value of NH3
 
Suyash Singh said:
litres are conserved

2 L of hydrogen plus 1 L of oxygen produce 2 L of gaseous water, 2+1 yields 2, what is conserved here?

Suyash Singh said:
50 percent yield means half of theoretical value of NH3

If only 50% of ammonia was produced, was all the limiting reagent consumed?
 
Borek said:
2 L of hydrogen plus 1 L of oxygen produce 2 L of gaseous water, 2+1 yields 2, what is conserved here?
If only 50% of ammonia was produced, was all the limiting reagent consumed?
Ok i understand the limiting concept now.

NH3 produced=2/3 x 30 =20 L

H2 left= 0 L

N2 left= 30- (1/2 x 20) =20 L
 
So which one is the limiting reagent here? And how much was left if only 50% reacted?
 
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Borek said:
So which one is the limiting reagent here? And how much was left if only 50% reacted?
H2 the limiting reagent

all the elements react only 50 %?
 
Suyash Singh said:
H2 the limiting reagent

Right.

all the elements react only 50 %?

Stop guessing, use stoichiometry to calculate.
 
Suyash Singh said:
50 percent yield means half of theoretical value of NH3
But you got all of the maximum yield of NH3: You used up all hydrogen.
Based on that you can figure out how much NH3 is produced if you use only half of it.
 
  • #10
NH3 produced = 50/100 x (2/3 x 30) = 10 L
H2 left
= 30 - (3/2 x 10) = 15 L
N2 left
= 30 - (1/2 x 10)= 25 L
 
  • #11
That looks good.
 
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