Question

Hello I was thinking about two thermodynamics problems and I wanted to get some insights into how to solve them.

The first problem was: Suppose we have one mole of ideal gas under constant external pressure (1 atm) conditions, and let it conduct a reaction (for example a photoinduced isomerization) where there is no net change in number of molecules. This reaction for example has an enthalpy change which is not known. We can measure the total heat Q evolved due to the reaction (like in a constant pressure calorimeter).

A\to B,\Delta H=?

So because there is no change of molecules this can be treated as a formal isobaric change (can it?). As enthalpy change is equal to heat Q in isobaric processes, we can say that the measured heat when reaction goes to completion is equal to the reaction enthalpy change, and therefore we can calculate the temperature change.

\Delta T=\frac{Q}{C_{p}}

My doubt is: Is the work the system performs on its surroundings included in ΔH?

How can we determine the temperature change then if we have an reaction of the type:

2A\to B

Also at constant external pressure? Here, the particle number changes.

Answer 1

The heat Q measured in an isobaric process is equal to the enthalpy change ΔH, which includes the work done by the system. For reactions with changing number of molecules.

Step-by-step explanation:

When we say that there is no change in temperature despite energy going into a system, it's often because the energy is being used to change the state of the substance, such as ice melting into water, which is a change in entropy without a temperature change.

However, when dealing with reactions that occur under constant pressure such as enthalpy changes in a calorimeter, we can measure the heat released or absorbed.

The enthalpy change (ΔH) includes the work performed by the system on its surroundings due to expansion or contraction, which means for an isobaric, or constant-pressure, condition, the measured heat change is equal to the enthalpy change of the reaction

In the given reaction A → B, assuming it's an isobaric process, the enthalpy change is indeed equal to the heat Q absorbed or released by the system.

If the reaction were 2A → B, taking place at constant external pressure, you still treat the heat absorbed or released as ΔH, but due to the change in the number of gas molecules.

You would need to account for the work done by or on the system, which is part of the enthalpy change calculation.