# How can change in Enthalpy for a reaction be written like this?

If a gas expands in a container with constant external pressure then:

$$W=-P_{ext}\Delta V$$

Instead of $$P_{ext}$$ if we use $$P_{int}$$ then we will not get the same value because $$P_{int}$$ is constantly decreasing from a certain value until equilibrium is attained, while the change in volume remains the same.

This is the reason why we use $$P_{ext}$$ here:

$$\Delta H= \Delta U +P_{ext} \Delta V$$

But then, $$\Delta H$$ for a reaction is:

$$\Delta H= \Delta U +\Delta n_{g}RT$$ ($$n_{g}$$ is for change in number of gaseous moles as the solids and liquids are ignored)

where $$PV=nRT$$ is used. However, the $$P$$ in the ideal gas equation is $$P_{int}$$.

So my question is how can $$P_{int}$$ be substituted in the place of $$P_{ext}$$ if the values of work done change when we do so?

• Who says the change in enthalpy is $\Delta U+P_{ext}\Delta V$? Please provide a reference. This is not correct. Mar 20 at 12:30
• I've seen $\Delta H=\Delta U +P \Delta V$ in text books and when showing that $\Delta H=q$ under constant pressure the work from $\Delta U = q+w$ gets canceled out with $P \Delta V$ from $\Delta H$ so I thought that for it to cancel out with work the $P$ had to be $P_{ext}$.Hence, the $\Delta H=\Delta U +P_{ext} \Delta V$ Mar 20 at 12:58
• How would that work for an adiabatic irreversible expansion at constant external pressure (for which q = 0)? Mar 20 at 13:17
• @ChetMiller I don't see why it woudn't work. Wouldn't $\Delta H=0$ for adiabatic irreversible expansion. Mar 20 at 14:13
• Does the temperature change? Is the enthalpy of an ideal gas a function of temperature? Mar 20 at 14:26

Your claim that $$\Delta H = \Delta U + P_{ext} \Delta V$$ is incorrect. It is only correct when the system is always in mechanical equilibrium with the external pressure, or in other words irreversible expansion against constant pressure. The more general statement is $$\Delta H = \Delta U + \Delta (PV)$$. Here as you can see even when you compare it with $$\Delta H = \Delta U + \Delta n_gRT$$ there is no problem as we don't have to replace $$P_{in}$$ with $$P_{ext}$$.