1. Feature Article: SDN Announces Medical School Admissions Workshop
SDN Mobile: Available for both iPad/iPhone iOS7 and Android.
SDN Shop: The SDN Shop is now open! All items sold at cost. Student groups, contact us for additional discounts.

# Internal Energy/Heat/Work

Discussion in 'MCAT Study Question Q&A' started by cgbb, Jun 10, 2012.

1. This forum made possible through the generous support of SDN members, donors and sponsors. Thank you.
1. ### cgbbNew Member

Joined:
Jun 10, 2012
Messages:
18
Status:
Pre-Medical

When an exothermic reaction is carried out in a closed system with constant pressure, is the heat released due to a decrease in the internal energy of the molecules?

I know that this is a basic question, but EK Chemistry has led me to second guess myself.
2. ### jcf44New Member

Joined:
May 4, 2012
Messages:
4
Status:
Pre-Medical
I'll take a shot at this, someone correct me if I'm wrong. The internal energy of a system can change through two ways: work or heat, where work is generally defined by pressure* change in volume. Assuming your system does not change in volume(which I can't tell from your question), then no work is done and the heat released must be due to a decrease in internal energy of the molecules. Hopefully that helps.
3. ### cgbbNew Member

Joined:
Jun 10, 2012
Messages:
18
Status:
Pre-Medical
Are you sure that internal energy is = W +q ?

I think that H= PV + U (U=change in energy) is similar to E= W + q. In this case the change in the internal energy would be analogous to the transfer of heat. I'm just not certain that this is a fair assumption.
4. ### cgbbNew Member

Joined:
Jun 10, 2012
Messages:
18
Status:
Pre-Medical
I forgot to say thanks for the feedback!
5. ### jcf44New Member

Joined:
May 4, 2012
Messages:
4
Status:
Pre-Medical
So, I think different books define it differently. Depending on how you define work in the equation affects what sign you put on it. One thing that is universal is that internal energy increases if heat is gained by the system or if work is done ON the system and that internal energy decreases if heat is lost or if work is done BY the system.

For this reason, I think it's better to define changes in internal energy like this:
Change in internal energy = (heat gained- heat lost) + (Work done on system- work done by system)

Or, you could also define it in its most basic terms like change in internal energy = energy gained - energy lost. In this case change in internal energy = (heat gained + work done on system) - (heat lost + work done by system)

Also, I think you're right when you say H = PV + U is analogous to E = q + w because the heat content doesn't really change, only gets converted between work and internal energy, just like energy is conserved and just converted between heat and work in the other equation.

Pretty cool is if you consider that U = q + PV, then if no work is done U = q or internal energy change = heat change. Then substitute it into enthalpy equation(H = PV + U) you get H = PV + (q + PV), where if no work is done then both the PV's drop and enthalpy change = heat change. Similarly, in the other equation E = q + w, if you assume that no work is done than change in Energy = heat change also.

Conceptually, I think it also makes sense that an exothermic reaction would lead to a decrease in internal energy of the molecules, because exothermic reactions are more likely to be spontaneous(depending on temperature and entropy change) and in physics and chemistry molecules or what not always want to exist at the lowest possible internal or potential energy.

Hopefully, that helps and I didn't mess anything up. This stuff was a headache to learn, and I think I only got it about a week ago.