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$$\ce{N2 + 3 F2 -> 2 NF3} \quad ΔH = \pu{-263 kJ mol-1}$$

The bond dissociation enthalpies of the nitrogen triple bond and nitrogen–flourine bond are $\pu{946 kJ mol-1}$ and $\pu{272 kJ mol-1}.$ Find the bond dissociation enthalpy of the $\ce{F-F}$ bond.

A. $\pu{-423 kJ mol-1}$
B. $\pu{-393 kJ mol-1}$
C. $\pu{-141 kJ mol-1}$
D. $\pu{+141 kJ mol-1}$
E. $\pu{+423 kJ mol-1}$

I know that since bond dissociation enthalpy is endothermic it has to be one of the last two.

I tried substracting bond dissociation enthalpies of products from reactants and equating that to the enthalpy change but my answer is nowhere near the given ones. Is my method wrong?

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If you can't wrap your head around getting the answer directly, just calculate the enthalpy of reaction for the two bond dissociation energies you are considering.

Assume bond dissociation energy is 141 kJ/mol

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{141 kJ/mol} - 6 \cdot \pu{272 kJ/mol} = \text{negative value}$$

Assume bond dissociation energy is 423 kJ/mol

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot \pu{423 kJ/mol} - 6 \cdot \pu{272 kJ/mol} = \text{positive value}$$

So it can't be 423 kJ/mol because the reaction is exothermic. If you complete the math for a bond energy of 141 kJ/mol, the numerical answer is -263 kJ/mol for the reaction enthalpy.

What if you actually wanted to estimate a bond energy?

In this case, you could call the bond energy $x$, and solve for it:

$$\Delta H_r = \pu{946 kJ/mol} + 3 \cdot x - 6 \cdot \pu{272 kJ/mol} = \pu{-263 kJ/mol}$$

Solving for $x$ would give you the bond energy.

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