The thermodynamics of a reaction is governed by both enthalpy and entropy. The Gibbs Free Energy (`G`) is a state function that takes both enthalpy and entropy into account in order to function as an indicator for the direction of a reaction. `G` is defined as such:
`G=H-TS`
`H` is enthalpy
`T` is temperature
`S` is entropy
Sometimes there are reactions that are favored by entropy (`DeltaS>0`) but not favored by enthalpy (`DeltaH>0`). The opposite also exists: some reactions are favored according to enthalpy (`DeltaH<0`) but not favored by entropy (`DeltaS<0`). The Gibbs Free Energy is used because it takes into account both enthalpy and entropy.
When we use the Gibbs energy to examine a system, we look at the change in Gibbs Free Energy (`DeltaG`). The expression for `DeltaG` is written as such:
`DeltaG=DeltaH-TDeltaS`
From there, we can assert that a negative `DeltaG` indicates an increase in the entropy of the universe. In other words, whenever `DeltaG` is negative for a process, the process is spontaneous. The correlary is true as well: whenever `DeltaG` is positive for a process, the process is not spontaneous. When `DeltaG=0` , the system is at equilibrium. Note that magnitude of `DeltaG` is not important. Only the sign of `DeltaG` is significant.
`DeltaG` |
Indicates |
`+` |
Process is not spontaneous |
`-` |
Process is spontaneous |
`0` |
Process is at equilibrium |
Notice that `DeltaG` has a temperature dependence. This means that `DeltaG` will be different at different temperatures. This explains why some reactions happen at certain temperatures but not others.
It turns out that the sign of `DeltaG` can explain, for all purposes, all chemical processes from melting to mixing to direction of reaction. Let's do a couple problems to illustrate this.
#1. Consider the following reaction:
` N_2O_(4(g))↔2NO_(2(g)) `
The reaction has `DeltaH=57 "kJ"` and `DeltaS=0.176 ("kJ")/"K"` .
a) Is the reaction favored according to enthalpy?
b) Is the reaction favored according to entropy?
c) Calculate `DeltaG` of the reaction at room temperature. Is this process spontaneous at room temperature?
d) Calculate the temperature at which the reaction is at equilibrium.
a) The reaction is not favored according to enthalpy. In general, exothermic reactions are favored because they involve releasing energy in the form of heat. In this case, the reaction is endothermic.
b) The reaction is favored by entropy. The change in entropy is positive, indicating that entropy is increasing.
c) `DeltaG=DeltaH-TDeltaS`
Plugging in the given values for `DeltaH` and `DeltaS` , with `T=298K` , we get:
`DeltaG=(57 "kJ")-(298K)(0.176 ("kJ")/"K")`
`DeltaG=4.55 "kJ"`
`DeltaG` is positive, therefore the reaction is not spontaneous at room temperature.
d) When the reaction is at equilibrium, `DeltaG=0` . Solve for `T` .
`0=DeltaH-TDeltaS`
`T=(DeltaH)/(DeltaS)`
`T=(57 "kJ")/(0.176 ("kJ")/"K")`
`T=324 "K"`
From analyzing the `DeltaG` of the reaction, we learned that the reaction is not spontaneous at room temperature, but is spontaneous at higher temperatures. This reaction occurs around us everyday, especially those of us in congested cities (LA, Beijing). `NO_2` is the brown smog that's commonly seen in these cities, and the reaction above is one that's responsible for its prevalence. Most of the time, the `NO_(2(g))` is in the form of `N_2O_(4(g))` , but when temperatures increase, the reaction becomes more and more favored to the right. This results in more `NO_(2(g))` being released, and subsequently more smog.
#2. Calculate the boiling point of `Br_(2(l))` given that `DeltaH=31 "kJ"` and `DeltaS=0.093 "kJ"/"K"`
The reaction being examined is the vaporization of liquid bromine:
`Br_(2(l)) ↔ Br_(2(g))`
Recall that the boiling point is the point at which both phases are in coexistence. Another way of saying this is that both phases are in equilibrium. This corresponds to `DeltaG=0` .
Setting `DeltaG=0` , we get the same expression for temperature as in #1: `T=(DeltaH)/(DeltaS)`
`T_"boiling"=(31 "kJ")/(0.093 "kJ"/"K")`
`T_"boiling"=333 "K"`
According to a quick Google search, the boiling point of liquid bromine is given as `331.8 "K"` , which is essentially the same taking into account rounding during calculation. The purpose of these two problems was to demonstrate how `DeltaG` governs chemical processes.
Summary
1. The Gibbs Free Energy is a thermodynamic function that takes into account enthalpy, entropy, and temperature.
2. When the change in Gibbs Free Energy (`DeltaG`) is negative, the process is spontaneous. When `DeltaG` is positive, the process is not spontaneous.
3. The magnitude of `DeltaG` is not important. Only the sign of `DeltaG` is examined at this point.
4. The Gibbs Free Energy can be used to describe essentially all chemical processes.