{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Note 5 - Note 5 Equilibrium I Free Energy 5.1 Helmholtz...

This preview shows pages 1–3. Sign up to view the full content.

1 Note 5. Equilibrium I: Free Energy 5.1 Helmholtz free energy 5.1.1 Derivation In the previous section, I alluded to a new state function which will help us sort out the total entropy change. This new state function is called the free energy . When we created the enthalpy state function, we said that we wanted something to reflect the change in heat at constant pressure. Here, we have a very different goal. In the previous section, we saw that using just entropy alone required us to consider the entropy changes of both the system and its surroundings. This is often difficult to do and best to avoid if possible. But how can we avoid this? One way is to build a state function which reflects the nature of equilibrium. To find such a state function, we return to the idea of maximum work. In Note 2, we learned that a system can do maximum work in a reversible case (i.e. equilibrium along the path). We will now introduce thermodynamic functions which can indicate whether one can do the maximum amount of work and thus tell us whether we are at equilibrium. In this section, we only consider systems at constant temperature. For a reversible reaction, we have rev rev dq dw dU + = Since we can relate the change in reversible heat to entropy by T dq dS rev = We can write TdS dw dU rev + = rearranging, we get const T at TS U d TdS dU dw rev = = = ) ( Thus, it is natural to define a new state function, called the Helmholtz free energy A TS U A = since (at constant temperature) rev dw TS U d dA = = ) (

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 One way to think of the free energy is that it is a measure of the maximum work the system can do on its surroundings . Just as mechanical systems work to decrease U , molecular systems (at constant T and V ) try to reduce A to reach equilibrium. What does this mean? In mechanical systems, we expect spontaneous processes to occur only if dU < 0 . Apples spontaneously fall from trees, but they never spontaneously leap from the ground into the sky. Similarly, spontaneous processes in molecular systems occur only if dA < 0 . If there is a state with lower free energy, then the system will eventually get there. What happens at equilibrium? At equilibrium, we expect that the system will do no work and thus dA = 0. This naturally parallels dU = 0 in mechanical systems. 5.2 Another way to look at free energy and equilibria In the Note 3, we showed that the 2 nd law is T dq dS and said that dS > dq/T for spontaneous processes and dS = dq/T for reversible processes. Now, we’ll see what the interpretations of this are for other state functions. Let’s take dU for example. From the first law, we have dU = dq + dw and thus dq=dU - dw . For a gas, we have dw = -PdV and thus dq = dU + PdV . Putting this into the inequality above, we get TdS dU + PdV . Rearranging terms, we get PdV TdS dU Thus, we see that at constant S and V ( i.e. dS = 0 and dV = 0) 0 dU Thus, under constant S and V , spontaneous processes must have dU < 0 . Note that this is what we expect for the mechanical examples – you only have to monitor U to determine the equilibrium.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 24

Note 5 - Note 5 Equilibrium I Free Energy 5.1 Helmholtz...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online