# Determining the Average Bond Enthalpy for the C-F bond

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 8:19 • @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 8:48 • 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 ## 1 Answer 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:

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