# Regarding enthalpy of a reaction

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 ??

• 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. – 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).