Enthelpy of Formation vs. Bond Enthelpy

In summary, the average bond enthalpy, εN-F, for NF3(g)→ N(g) + 3F(g) is 276.4 kj/mol. The differences in calculated values for εN-F can be attributed to incorrect assumptions and errors in calculations.
  • #1
jbowers9
89
1

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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  • #2
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
...
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?
 
  • #3
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
 
  • #4
I "assumed" wrongly however. Dopey mistake, but the difference between 193 & 278 is still kind of large.
 
  • #5
never mind
 
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  • #6
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?
 
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  • #7
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.
 
  • #8
Yip. Looks alright now.
 

What is the difference between Enthalpy of Formation and Bond Enthalpy?

The enthalpy of formation is the energy change that occurs when one mole of a compound is formed from its constituent elements in their standard states. Bond enthalpy, on the other hand, is the energy required to break one mole of a specific type of bond in a molecule in the gas phase.

Which one is a measure of the stability of a compound?

The enthalpy of formation is a measure of the stability of a compound. A lower enthalpy of formation indicates a more stable compound.

How do these two concepts relate to each other?

Bond enthalpy is a component of the enthalpy of formation. The enthalpy of formation takes into account all of the bonds in a molecule, while bond enthalpy only considers a specific type of bond.

Can bond enthalpy be measured experimentally?

Yes, bond enthalpy can be measured experimentally through techniques such as calorimetry or spectroscopy.

Which one is more useful in predicting the energy changes in a chemical reaction?

The enthalpy of formation is more useful in predicting the energy changes in a chemical reaction. It takes into account all of the bonds broken and formed in a reaction, while bond enthalpy only considers a specific type of bond.

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