Enthalpy of Formation vs. Bond Enthalpy

  1. 1. The problem statement, all variables and given/known data

    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

    2. Relevant 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

    3. 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?
     
  2. jcsd
  3. Gokul43201

    Gokul43201 11,141
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    What kind of bond exists between the N-atoms in N2: single, double, triple? Which one did you use above?
     
  4. 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
     
  5. I "assumed" wrongly however. Dopey mistake, but the difference between 193 & 278 is still kind of large.
     
  6. chemisttree

    chemisttree 3,721
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    never mind
     
    Last edited: Jul 31, 2007
  7. Gokul43201

    Gokul43201 11,141
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    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: Jul 31, 2007
  8. 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.
     
  9. Gokul43201

    Gokul43201 11,141
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    Yip. Looks alright now.
     
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