The $\ce{H-H}$ bond enthalpy is $436 \mathrm{kJ/mol}$:

$$\ce{H2 (g) -> 2H (g)} \qquad \Delta H = +436 \mathrm{kJ/mol}$$

Does this only apply at the standard temperature of $298 \text{K}$, i.e. would the bond enthalpy vary if I decrease or increase the temperature?

And is it practically just so that, for a temperature $T$, as long as you pump the bond enthalpy’s worth of kilojoules into the substance, it will be atomized?

  • 3
    $\begingroup$ A bond dissociation enthalpy is the energy needed to break one generic bond isothermically. That's because when the reaction takes place in gaseous phase, the molecules will tend to react dependently on their average speed, and since $KE=f(T,...)$ in a directly proportional way, this parameter has to be fixed to eliminate its dependency on the energy needed for the process. In fact, one could derive empirically $\Delta\text{U}_{\text{bond}}$ if the volume is fixed. $\endgroup$
    – TheVal
    Commented Dec 20, 2013 at 20:55
  • $\begingroup$ As temperature increases, the bond enthalpy will usually decrease (at least in the way you're thinking about it). Remember that temperature is a measurement of energy. $\endgroup$
    – jeremy
    Commented Dec 21, 2013 at 3:08
  • $\begingroup$ This page describing Kirchoff's law might point you in the right direction. $\endgroup$
    – Tyberius
    Commented Apr 1, 2017 at 21:17

1 Answer 1


The bond dissociation enthalpy, like any enthalpy of reaction, will have some variation with temperature. This change in enthalpy can be determined using Kirchoff's Law: $$H_{T_f}=H_{T_i}+\int_{T_{f}}^{T_{i}}\Delta c_p(T')dT'$$ Where $$\Delta c_p(T')=\sum_i\nu_ic_{p,i}(T')$$ so that $c_{p,i}$ is the constant pressure heat capacity of the $\mathrm{i^{th}}$ compound in a reaction and $\nu_i$ is the the reaction coefficient of the $\mathrm{i^{th}}$ compound ($\nu_i$ is positive for products and negative for reactants).

If none of the $c_{p,i}$ have a temperature dependence, we can rewrite the $1^{st}$ equation above as $$H_{T_f}=H_{T_i}+\Delta c_p\cdot (T_f-T_i)$$ If they do have a temperature dependence, you will have to know the functional form of $c_{p,i}$ with respect to temperature to integrate them.

So, in regards to your specific question, the bond enthalpy for an $\ce{H-H}$ bond will be different at different temperatures. However, it is worth noting that for highly energetic reactions like this, the contribution due to $c_{p,i}$ is usually fairly small in comparison.


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.