Enthalpy of a reaction $$\ce{2H2 (g) + O2 (g)-> 2H2O(g)} \Delta H_1$$

$$\ce{2H2 (g) + O2 (g)-> 2H2O(l)} \Delta H_2$$

Now I was asked to compare the enthalpy.

If I subtract reaction 2 from 1, I get $$\ce{2H2O(g)-> 2H2O(l)} \Delta H_2 - \Delta H_1$$

Since this is exothermic enthalpy is negative and $$ \Delta H_2 - \Delta H_1 < 0 $$ or $$ \Delta H_2 < \Delta H_1 $$

But if I consider

Enthalpy of $\ce{2H2O(l)}$ is less than enthalpy of $\ce{2H2O(g)}$ therefore more energy will be released in the formation of $\ce{2H2O(l)}$ and thus $$ \Delta H_2 > \Delta H_1 $$

So which one is it, where am I going wrong ??

  • $\begingroup$ Negative enthalpy means energy released to surroundings, positive enthalpy means surroundings transfers heat to system. You might be confusing the sign and magnitude of the enthalpies. A smaller absolute (unsigned) number is not generally the same as a more negative number. $\endgroup$ – Buck Thorn Jun 20 at 9:55

You should be careful to distinguish between the magnitude of enthalpy (or some other change in energy) and the sign of the property:

  • the magnitude tells you how large the change was (yes, self-evidently)
  • the sign tells you in which direction the change occurred:

    a) $\text{system}\rightarrow \text{surroundings}$, $\Delta H<0$, or exothermic
    b) $\text{surroundings}\rightarrow \text{system}$, $\Delta H>0$, or endothermic

This is the typical convention used (not the only one).

In the case of the condensation of a gas, the magnitude can vary depending on the strength of intermolecular interactions in the liquid compared to the gas. The sign is negative because condensation at a constant temperature requires dissipation of heat to the surroundings.

So which one is it, where am I going wrong ??

That the enthalpy of $\ce{H2O(l)}$ is less than the enthalpy of $\ce{H2O(g)}$ means in the scenario you present that $$\Delta H_2 < \Delta H_1$$ not the other way around. You could also start from the fact that condensation is exothermic (or vaporization endothermic) so that $H_g-H_l=-\Delta_{cond}H =\Delta_{vap}H>0$$^\dagger$, from which $$\begin{align}H_g&>H_l\\ H_g-H_{reagents}&>H_l-H_{reagents}\\ \Delta H_1&>\Delta H_2\end{align}$$

$\dagger$ (Writing "$H_g$" is a bit "dangerous" because it suggests there is an absolute enthalpy scale (rather than relative) but you can assume that here you are using a common reference state for water).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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