Determining the Average Bond Enthalpy for the C-F bond

Question

Given the following data, how would you work out the average bond enthalpy for $\ce{C-F}$ bond. I've tried setting up the chemical equations and applying Hess's Law, but that's not getting me anywhere.

$\Delta H_\mathrm f^\circ(\ce{CF4(g)})=-680~\mathrm{kJ~mol^{-1}}$

Bond enthalpy, $\ce{F2(g)}=+158~\mathrm{kJ~mol^{-1}}$

$$\ce{C(s) -> C(g)}\quad \Delta H=+715~\mathrm{kJ~mol^{-1}}$$

EDIT: These are the equations I used:

$$\begin{align} \ce{C(s) + 2F2(g) &-> CF4(g)}\\[6pt] \ce{F2(g) &-> 2F-(g)}\\[6pt] \ce{C(s) &-> C(g)} \end{align}$$

Welcome to Chemistry.SE! Did you take the stoiochiometry for $\ce{C + 2F2 -> CF4}$ into account? — Klaus-Dieter Warzecha, Mar 16 '15 at 08:19
@KlausWarzecha Yes, but I still couldn't get an answer. Am I taking the right approach by using Hess's Law? — deusy, Mar 16 '15 at 08:48
Using Hess's Law is fine! Did you consider that you have 4 $\ce{C-F}$ bonds? — Klaus-Dieter Warzecha, Mar 16 '15 at 11:39

score 3 · Accepted Answer · answered Mar 16 '15 at 11:37

Your approach to use Hess's Law is reasonable!

\[\Delta H_r = -680 - (715 + 2\cdot158) = -1711\ \mathrm{kJ\cdot mol^{-1}}\]

That's the enthalpy for $\ce{CF4}$ - a molecule with four $\ce{C-F}$ bonds.

The average $\ce{C-F}$ bond enthalpy is smaller:

\[\frac{1711}{4}\ \mathrm{kJ\cdot mol^{-1}} \approx 427\ \mathrm{kJ\cdot mol^{-1}}\]

Determining the Average Bond Enthalpy for the C-F bond

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