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?