- #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?