# Enthalpy of Formation vs. Bond Enthalpy

1. Jul 30, 2007

### jbowers9

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. Jul 30, 2007

### Gokul43201

Staff Emeritus
What kind of bond exists between the N-atoms in N2: single, double, triple? Which one did you use above?

3. Jul 31, 2007

### jbowers9

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. Jul 31, 2007

### jbowers9

I "assumed" wrongly however. Dopey mistake, but the difference between 193 & 278 is still kind of large.

5. Jul 31, 2007

### chemisttree

never mind

Last edited: Jul 31, 2007
6. Jul 31, 2007

### Gokul43201

Staff Emeritus
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
7. Aug 1, 2007

### jbowers9

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. Aug 1, 2007

### Gokul43201

Staff Emeritus
Yip. Looks alright now.