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Change in enthalpy of the system is equal to its temperature multiplied by its change in entropy (as change in enthalpy is equal to heat absorbed /given out at constant pressure and change in entropy is heat absorbed/given out divided by its temperature). So, through the Gibbs free energy equation, the change in Gibbs free energy should be zero?

This question might seem messy, as I don't know how to write equations on Stack Exchange.

Please help. The problem is in the elementary concepts, I think.

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    $\begingroup$ Not exactly sure what you have problems with but a common source of confusion is $\Delta G$ vs $\Delta G^\circ$. At equilbrium $\Delta G = 0$ and $\Delta G^\circ = -RT \ln(K)$ $\endgroup$ – Jan Jensen Jun 25 '15 at 8:54
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The change in Gibbs free energy for a closed system is given by:

$$\Delta G = \Delta H - T \Delta S$$

What you are describing is:

$$\Delta H = T \Delta S$$

which is only true for systems at equilibrium, or when $\Delta G = 0$

Why this is true is a little bit difficult to explain. It is essentially a form of the Clausius Inequality at constant temperature and pressure. The Clausius Inequality is really just a statement of the second law of thermodynamics, and it says that at best, for any possible process you can get a zero change in the entropy of the universe, and usually it will be a positive change. From the perspective of the system you can write that as:

$$\Delta S \ge \oint \frac{dQ}{T}$$

which is basically just saying the entropy change of the system is equal to or greater than the sum of the negative entropy changes of the rest of the universe, for each step in a closed cycle.

If you integrate that at constant temperature you get

$$\Delta S \ge \frac{\Delta Q}{T}$$

At constant pressure (and in the absence of non-expansion work) $\Delta Q = \Delta H$, so we can write:

$$\Delta S \ge \frac{\Delta H}{T}$$

...and a little rearranging gives:

$$\Delta H - T \Delta S \leq 0$$

So this says that any process which can occur on its own (without adding work or heat to it) will obey this inequality. That's a pretty useful thing to know, so we let

$$\Delta G = \Delta H - T \Delta S$$

to give us an easier way to talk about it. Processes will happen (on their own) if $\Delta G$ is negative, they won't happen if it's positive, and when they are at equilibrium, $\Delta G$ will be zero.

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if the second equation of thermodynamics holds good then dH-dST is greater than or equal to zero. A system cant have negative entropy because of the internal energy of a system, intermolecular forces of attraction aswell as the due to the fact that stable systems have minimum energy(fermi energy) and zero forces. futher explanation can be given when we analyse the atomic structure of atoms interms of quantum attraction forces and the forces of repulsion between atoms. so the energy released when a molecule moves from a high chemical potential region to a lower one is the same for the simultaneous phase transitions but the energy of the two systems is never reduced to zero.

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