Enthelpy of Formation vs. Bond Enthelpy

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Discussion Overview

The discussion revolves around the calculation of the average bond enthalpy, εN-F, for the decomposition of NF3 into nitrogen and fluorine atoms. Participants explore the differences in calculated values using heats of formation and bond enthalpies, raising questions about the assumptions and methodologies used in their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants present calculations for εN-F using both heats of formation and bond enthalpies, leading to different results: 278.0 kJ/mol and 146.2 kJ/mol, respectively.
  • Some participants question the validity of their arithmetic and assumptions regarding the use of enthalpies of formation.
  • There is a discussion about the nature of the bond between nitrogen atoms in N2, with one participant initially assuming it to be a single bond, later correcting themselves.
  • One participant mentions a significant discrepancy in their calculations, noting a difference between 193 kJ and 278 kJ, and seeks clarification on the bond energy for the N~N triple bond.
  • Another participant reflects on confusion regarding the heats of formation for diatomic gases versus bond enthalpies, indicating a misunderstanding in their approach to rewriting the equations.
  • Participants share their recalculations and adjustments to their methodology, including addressing bookkeeping errors and clarifying the use of factors in their equations.
  • One participant confirms that their recalculated values align with another's, indicating progress in resolving discrepancies.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct value of εN-F, as multiple competing calculations and assumptions are presented. Disagreements exist regarding the nature of the N2 bond and the methodologies used in calculations.

Contextual Notes

Some participants express uncertainty about the appropriateness of their assumptions and the implications of using different bond enthalpy values. There are mentions of potential calculation errors and confusion regarding the treatment of diatomic gases in the context of heats of formation versus bond enthalpies.

Who May Find This Useful

This discussion may be useful for students and practitioners in chemistry who are exploring the concepts of bond enthalpy and heats of formation, particularly in the context of molecular decomposition reactions.

jbowers9
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Homework Statement



Find the average bond enthalpy, εN-F, for
NF3(g)→ N(g) + 3F(g)
Heats of formation
NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
1/2 F2(g) → F(g) ∆ƒHºm = 79 kj/mol
1/2N2(g) → N(g) ∆ƒHºm = 472.7 kj/mol

Bond Enthalpies
NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
1/2F2(g) → F(g) εF-F = 155 kj/mol
1/2N2(g) → N(g) εF-F = 163 kj/mol

Homework Equations



Using heats of formation
4NF3(g)→ 2N2(g) + 6F2(g) -4(∆ƒHºm = 124.3 kj/mol)
6F2(g) → 12F(g) 6*2*(∆ƒHºm = 79 kj/mol)
2N2(g) → 4N(g) 2*2(∆ƒHºm = 472.7 kj/mol)

4NF3(g)→ 4N(g) + 12F(g) ∆Hm = 3336.0 kj
12 εN-F = 3336.0 kj
εN-F = 278.0 kj/mol

Using bond enthalpies
4NF3(g)→ 2N2(g) + 6F2(g) -4(∆ƒHºm = 124.3 kj/mol)
6(F2(g) → F(g) εF-F = 155 kj/mol)
2(N2(g) → N(g) εF-F = 163 kj/mol)
4NF3(g)→ 4N(g) + 12F(g) ∆Hm = 1753.2 kj
12 εN-F = 1753.2 kj
εN-F = 146.2 kj/mol

The Attempt at a Solution



The text is the 6th Ed., Chem. Thermo. Basic Theory & Methods, Irving Klotz pg. 72 # 5. The data is from tables in the chapter that precede the problem set. Why, assuming that the arithmetic and my assumptions about being able to use the enthalpies of formation are correct, do the calculated values for εN-F vary that much?
 
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jbowers9 said:

Homework Statement



Find the average bond enthalpy, εN-F, for
NF3(g)→ N(g) + 3F(g)
Heats of formation
NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
1/2 F2(g) → F(g) ∆ƒHºm = 79 kj/mol
1/2N2(g) → N(g) ∆ƒHºm = 472.7 kj/mol

Bond Enthalpies
NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
1/2F2(g) → F(g) εF-F = 155 kj/mol
1/2N2(g) → N(g) εF-F = 163 kj/mol[/color]
...
Why, assuming that the arithmetic and my assumptions about being able to use the enthalpies of formation are correct, do the calculated values for εN-F vary that much?
What kind of bond exists between the N-atoms in N2: single, double, triple? Which one did you use above?
 
I've always "assumed" that the N2 bond was single sp3.
The table of bond enthalpies is from TL Cottrell, The Strengths of Chemical Bonds, 2nd ed. '58 pp. 270-289 & AG Gaydon, Dissociation Energies, 3rd ed. '68
 
I "assumed" wrongly however. Dopey mistake, but the difference between 193 & 278 is still kind of large.
 
never mind
 
Last edited:
jbowers9 said:
I "assumed" wrongly however. Dopey mistake, but the difference between 193 & 278 is still kind of large.
How did you get 193kJ? When I looked up the bond energy for the N~N triple bond, I get the correct answer.

Notice that the bond energy equation is nothing but twice the corresponding formation equation. So, since 155kJ is roughly twice of 79kJ, that part checks out okay. But clearly, 163kJ is nothing like twice of 472kJ. When you find the correct bond energy, make sure it's close to 944kJ.

Also, you've unnecessarily multiplied all equatins by an extra factor of 4 in the beginning, and then divided by 4 in the end - more room to make a calculation error. Why did you need to do that?
 
Last edited:
I recrunched the numbers. Is this what you got? I apologize for making stupid bookkeeping errors. I've been out of the loop for awhile, this is review for me. I graduated in '85. The concept that the heats of formation for the diatomic gases are for the unimolar product species and that for the bond enthalpies it's the bimolar species confused me too after rewriting the equations. I used the factor of four to try and remove the fractions, since I wrote the equations out as below originally, but used the wrong bond enthalpy for N2.

NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
1/2F2(g) → F(g) ∆ƒHºm = 79 kj/mol
1/2N2(g) → N(g) ∆ƒHºm = 472.7 kj/mol

NF3(g)→ N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
3/2F2(g) → 3F(g) 3(∆ƒHºm = 79 kj/mol)
1/2N2(g) → N(g) ∆ƒHºm = 472.7 kj/mol

NF3(g)→ N(g) + 3F(g) ∆Hm = 834 kj
3εN-F = 834
εN-F = 278 kj/mol

F2(g) → 2F(g) εF-F = 155 kj/mol
N2(g) → 2N(g) εN-N = 945 kj/mol
NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol

NF3(g)→ 1/2N2(g) + 3/2F2(g) -∆ƒHºm = 124.3 kj/mol
3/2(F2(g) → 2F(g) εF-F = 155 kj/mol)
1/2(N2(g) → 2N(g) εN-N = 945 kj/mol)
NF3(g)→ N(g) + 3F(g) ∆Hm = 829.3 kj
3εN-F = 829.3 kj
εN-F = 276.4 kj/mol

Just give me a heads up if your numbers agree. Thanks. I'll be making other posts.
 
Yip. Looks alright now.
 

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