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:

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

  • 1
    $\begingroup$ Welcome to Chemistry.SE! Did you take the stoiochiometry for $\ce{C + 2F2 -> CF4}$ into account? $\endgroup$ Mar 16, 2015 at 8:19
  • $\begingroup$ @KlausWarzecha Yes, but I still couldn't get an answer. Am I taking the right approach by using Hess's Law? $\endgroup$
    – deusy
    Mar 16, 2015 at 8:48
  • $\begingroup$ Using Hess's Law is fine! Did you consider that you have 4 $\ce{C-F}$ bonds? $\endgroup$ Mar 16, 2015 at 11:39

1 Answer 1


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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.