Enthalpy of Formation vs. Bond Enthalpy
1. The problem statement, all variables and given/known data
Find the average bond enthalpy, εNF, 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) εFF = 155 kj/mol 1/2N2(g) → N(g) εFF = 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 εNF = 3336.0 kj εNF = 278.0 kj/mol Using bond enthalpies 4NF3(g)→ 2N2(g) + 6F2(g) 4(∆ƒHºm = 124.3 kj/mol) 6(F2(g) → F(g) εFF = 155 kj/mol) 2(N2(g) → N(g) εFF = 163 kj/mol) 4NF3(g)→ 4N(g) + 12F(g) ∆Hm = 1753.2 kj 12 εNF = 1753.2 kj εNF = 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 εNF vary that much? 
Quote:

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. 270289 & 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

Quote:
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? 
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 3εNF = 834 εNF = 278 kj/mol F2(g) → 2F(g) εFF = 155 kj/mol N2(g) → 2N(g) εNN = 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) εFF = 155 kj/mol) 1/2(N2(g) → 2N(g) εNN = 945 kj/mol) NF3(g)→ N(g) + 3F(g) ∆Hm = 829.3 kj 3εNF = 829.3 kj εNF = 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.

All times are GMT 5. The time now is 10:44 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums