(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Enthalpy of Formation vs. Bond Enthalpy

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