WebLecture

Energy changes and entropy changes are *state* changes. We are concerned with the conditions of the initial and final states (which we define); we are not concerned with *how* the state change occurs.

When we consider energy changes in reactions, we take into account enthalpy and entropy changes separately. We can combine these two types of change to produce a third energy amount, Gibbs free energy, which we define as the change in enthalpy energy, minus that change in energy due to entropy changes: $$\Delta G\text{}=\text{}\Delta H\text{}-\text{}T\Delta S$$

If the reaction occurs at standard conditions, we use H° and S°.

As with most state functions, we look at the situations where

- the change is positive (greater than zero), that is, the end state has more than the initial state
- the change is zero and there is no difference between the end state and initial state
- the change is negative, and the end state amount is less than the initial state

Determining how Gibbs free energy reaches these states requires us to pay attention to the negative, zero, or positive states of ΔH and ΔS (T is always positive).

Consider a reaction that is taking in heat from its environment: an endothermic reaction, with ΔH > 0. In most cases, not all of the heat will, however, go to further the reaction itself. Some of the heat will be lost to random motion of molecules in the system: to increase entropy, so ΔS will also be postive. If the energy absorbed by the system from its environment is greater than the amount lost to entropy, the remaining amount is "free" to do the work of the system, and to run the reaction. ΔG will have a postive amount, measuring the energy actually available from the environment for the reaction. But the reaction will be endothermic, and not spontaneous, because it requires constant energy input to continue.

If the reaction is exothermic, ΔH is negative. It is also likely, again, that the reaction will have an increasing entropy and ΔS will be positive. Since both terms ΔH and TΔS are negative, ΔG will be negative as well. This reaction *produces* free energy which can run the reaction or heat up the environment. Such reactions where ΔG is negative are spontaneous.

As with enthalpy and entropy, we can determine ΔG° values at standard temperatures, but it is useful to consider ΔG at other temperatures. It turns out that ΔG is a function not only of its base value at STP, but also of the reaction quotient and temperature: $${\Delta}_{r}G\text{}={\Delta}_{r}G\text{\xb0}+\text{}\mathrm{RT}\text{}\mathrm{lnQ}\text{}$$

This relationship allows us to determine the Gibbs free energy available for a reaction with any concentration of products and reactants at any temperature. At equilibrium, we have a special condition: *there is no change to Gibbs free energy*.
$${\Delta}_{r}G\text{}=\text{}0\text{}={\Delta}_{r}G\text{\xb0}+\text{}\mathrm{RT}\text{}\mathrm{ln}\text{}K\text{}$$

We can rewrite the equation: $${\Delta}_{r}G\text{\xb0}=\text{}-\mathrm{RT}\text{}\mathrm{ln}\text{}K\text{}$$

if we substitute this expression back into our equation for Q, we have $$\begin{array}{c}{\Delta}_{r}G\text{}=-\mathrm{RT}\text{}\mathrm{ln}\text{}K\text{}+\text{}\mathrm{RT}\text{}\mathrm{lnQ}\text{}\\ {\Delta}_{r}G=\text{}\mathrm{RT}(\mathrm{ln}\text{}Q\text{}-\text{}\mathrm{ln}\text{}K)\text{}\\ {\Delta}_{r}G=\text{}\mathrm{RT}(\mathrm{ln}\text{}\frac{Q}{K})\end{array}$$

If Q < K, ln Q/K will be negative, Δ_{r}G will be negative, and the reaction will be spontaneous in the forward direction.

If Q = K, ln Q/K = ln 1 = 0. Δ_{r}G will be zero, and the reaction is at equilibrium, because the energy state is not changing.

If Q > K, ln Q/K will be positive, Δ_{r}G will be positive, and the reaction will not be spontaneous in the forward direction, but it will be spontaneous in the reverse direction.

Consider again our basic definition of Gibbs free energy: ΔG = ΔH - TΔS. If a reaction is endothermic (ΔH is positive) and T is low and ΔS, though positive, is small, ΔG will be positive and the reaction will not be spontaneous. However, if we increase T, the negative TΔS factor will eventually be large enough to overcome the positive ΔH value and make ΔG negative. The reaction will then be spontaneous, because the energy change due to increasing entropy (say, in the production of a gas at high temperature) becomes the dominant factor in determining whether or not the reaction is product-favored.

- What are common "entropy-favored" situations (i.e., where entropy is increasing)?
- Can entropy increase in an endothermic reaction?

- Take a look at Gibbs Free Energy concepts at the Chemistry Department of Purdue University. This site includes a number of practice problems.

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