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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 ??

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  • $\begingroup$ 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
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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).

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