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Jetzt kostenlos anmeldenBack in the Middle Ages, people believed in spontaneous generation. Frogs would erupt from wet fields and flies would be birthed from rotting meat. We know now, of course, that these species aren't actually being born from inanimate objects, but the concept of "spontaneity" in creation lives on in chemistry. The phenomenon of spontaneous reactions is explained by Gibb's Free Energy.
In this article, we will be diving into the relationship between Free Energy and equilibrium.
So, what is Gibb's free energy anyway?
Gibb's free energy, or just Free energy, is the amount of available energy in a system to do work. The formula for the change in free energy (ΔG) is:
$$\Delta G=\Delta H-T\Delta S$$
Where:
If Δ G is positive, that means that energy needed to be added to the system, so the reaction is non-spontaneous. This means that the reaction cannot happen unless external energy is added. If Δ G is negative, that means that energy was consumed by the system, so the reaction is spontaneous. This means that the system doesn't need more energy for the reaction to happen.
Free energy also relates to the concept of equilibrium
At equilibrium, the Concentration of reactants and products remains the same. The reaction hasn't "stopped", it just means that the rate of the forward reaction and the rate of the reverse reaction is the same. This is called Dynamic Equilibrium. The general form of an equilibrium equation is:
$$A + B \rightleftharpoons C + D$$
Where:
At equilibrium, either the reactants or products are "favored". This means that there will be a greater Concentration of one or the other. We measure this favoritism using the equilibrium constant.
The equilibrium constant (Keq) notes whether the reactants or products are favored. The formula for K for a general reaction is:
$$aA +bB \rightleftharpoons cC + dD$$
$$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$
Where the brackets mean "concentration of" and the lowercase letters (a, b, c d) are the stoichiometric coefficients for each species.
If Keq < 1, then the reactants are favored
If Keq > 1, then the products are favored
As a reaction proceeds, the Free energy (G) is going to decrease. This is because the "available energy" of the system is being used up. In experiments, we cannot measure the Free energy directly, but we can only measure the change in Free energy, ΔG. The change in Free energy will continue to decrease until it reaches a minimum, ΔG = 0. In order for the reaction to continue, ΔG would have to increase, which isn't possible unless energy is being added to the system.
Fig.1-As the reaction progresses, it will get to a point where the change in free energy is 0, which is the equilibrium point.
The blue portion of the graph is the forward reaction. The free energy will decrease as it progresses until it reaches equilibrium. If we were following the reverse reaction (purple side), the same thing happens. The reverse reaction will continue (and free energy will decrease) until equilibrium is reached and the change in free energy is zero, ΔG = 0.
Have you ever played with a music box? You turn the crank and the music will play until the gears stop. The music won't play again until you turn the crank again. This is what is happening with free energy. The system initially has some available energy and will continue until there is no more. There is also a relationship between the standard free energy change, ΔG°, and equilibrium.
ΔG° is the standard free energy change. It is the change in free energy that happens when a compound is formed from its elements at 1 atm and 298 K (25° C). While the non-standard free energy change, ΔG, changes over time, ΔG° is a constant.
The formula for the standard free energy change is given as:
$$\Delta G^\circ=\Delta G^\circ_{products}-\Delta G^\circ_{reactants}$$
We can use ΔG° to calculate Keq and vice-versa. Before we get started, we first need to understand the concept of the Reaction Quotient.
The Reaction Quotient (Q) is the ratio of products to reactants not at equilibrium. Here is the general formula for a gaseous reaction:
$$A + B \rightarrow C + D$$
$$Q=\frac{(P_D)(P_C)}{(P_A)(P_B)}$$
Where P is the Partial Pressure of the species. At equilibrium, Q=Keq.
To set up our equation, we have to assume that the reaction is happening in the gas state. The change in free energy of species, A, would be:
$$\Delta G_A=\Delta G^\circ_A + RT*ln(P_A)$$
Where R is the Gas Constant, T is the temperature and the symbol for the natural logarithm is, ln. Given this equation, we can solve for the change in the free energy of the overall process, ΔG:
1. First we note that the standard free energy change is given by:
$$\Delta G^\circ=(\Delta G^\circ_C+\Delta G^\circ_D)-(\Delta G^\circ_A +\Delta G^\circ_B)$$
Where C/D are products and A/B are reactants
2. Next, we calculate the change in free energy for the overall process:
$$\Delta G=[\Delta G_{products}]-[\Delta G_{reactants}]=[\Delta G_C + \Delta G_D]-[\Delta G_A + \Delta G_B]$$
$$\Delta G=[(\Delta G^\circ_C + RT*ln(P_C)+(\Delta G^\circ_ D+RT*ln(P_D))]-=[(\Delta G^\circ_A + RT*ln(P_A)+(\Delta G^\circ_ B+RT*ln(P_B))]$$
Or, by rearranging terms we get:
$$\Delta G=[(\Delta G^\circ_C+\Delta G^\circ_D)-(\Delta G^\circ_A+\Delta G^\circ_B)]+RT*[(lnP_C+lnP_D)-(lnP_A+lnP_B)]$$
3. Substituting in the for the standard free energy change, we get:
$$\Delta G=\Delta G^\circ + RT*[(lnP_C+lnP_D)-(lnP_A+lnP_B)]$$
and combining the natural logarithms, we get:
$$\Delta G=\Delta G^\circ + RT*ln(\frac{P_CP_D}{P_AP_B})$$
Where, we used the rule: \(ln(x) + ln(y) - ln(a) - ln(b)=ln(\frac{x+y}{a+b})\)
4. Then, we can substitute in the reaction quotient to get:
$$\Delta G=\Delta G^\circ + RTlnQ$$
Lastly, we know that at equilibrium, Q = Keq, and that, ΔG, the non-standard free energy difference is equal to zero, ΔG = 0, thus:
$$0=\Delta G^\circ + RTlnK_p$$
$$\Delta G^\circ=-RTlnK_p$$
The P is to indicate that Keq is being measured in terms of pressures. However, this equation is valid for other forms of K such as Kc (K in terms of concentration).
Now that we have our equation, what does it tell us? It shows us that the relationship between free energy and Keq is inversely proportional. This means that if one increases, the other decreases. We can rearrange our equation to solve for Keq:
$$\begin {align}\Delta G^\circ&=-RTln(K_{eq}) \\\frac{\Delta G^\circ}{-RT}&=ln(K_{eq}) \\e^{\frac{\Delta G^\circ}{-RT}}&=e^{ln(K_{eq})} \\e^{\frac{\Delta G^\circ}{-RT}}&=K_{eq} \\\end {align}$$
We can graph this equation to get a clearer sense of this relationship: (Note: in what follows we will write, K, when we really mean, Keq, so that you can get used to this practice as it is often used, remember K here is not to be confused with the symbol for Kelvins, K)
Fig.2-As K decreases, the change in free energy increases.
As we saw from our equation, K and ΔG° are inversely proportional. This makes sense when we consider what each value means. A, -ΔG°, means that the reaction is spontaneous in the forward direction while a, K > 1, means that products are favored (i.e. forward reaction is favored).
Now that we understand our equation, we can use it to calculate, K.
Calculate the equilibrium constant for the reaction using the thermodynamic data listed below: (Note: the uppercase, K, used as the dimension for temperature stands for Kelvins)
$$2SO_{2\,(g)} + O_{2\,(g)} \rightleftharpoons 2SO_{3\,(g)}$$
$$\Delta H_f^\circ (\frac{kJ}{mol}):\,2SO_2=-593.66\,\,O_2=0\,\,2SO_3=-791.44$$
$$\Delta S_f^\circ(\frac{J}{K*mol}):\,2SO_2=496.44\,\,O_2=205.138\,\,2SO_3=513.52$$
$$T:\,298K$$
We first need to calculate the standard enthalpy, ΔH°, and the standard entropy, ΔS°, of the reaction:$$\begin {align}\Delta H^\circ&=\Delta H^\circ_{products}-\Delta H^\circ_{reactants} \\\Delta H^\circ&=(-791.44\frac{kJ}{mol})-(-593.66\frac{kJ}{mol}+0\frac{kJ}{mol}) \\\Delta H^\circ&=-197.78\frac{kJ}{mol}\end {align} $$
$$\begin {align}\Delta S^\circ&=\Delta S^\circ_{products}-\Delta S^\circ_{reactants} \\\Delta S^\circ&=(513.52\frac{J}{K*mol})-(496.44\frac{J}{K*mol}+205.138\frac{J}{K*mol}) \\\Delta S^\circ&=-188.058\frac{J}{K*mol}=-0.188\frac{kJ}{K*mol} \\\end {align} $$
Now that we have these values, we can calculate ΔG°
$$\begin {align}\Delta G^\circ&=\Delta H^\circ-T\Delta S^\circ \\\Delta G^\circ&=-197.78\frac{kJ}{mol}-[(298\,K)(-0.188\frac{kJ}{K*mol})] \\\Delta G^\circ&=-141.756 \frac{kJ}{mol} \\\end {align} $$
The last step is to calculate K: (Recall, this "K" really stands for the equilibrium constant, Keq)$$ \begin {align}K&=e^{\frac{\Delta G^\circ}{-RT}} \\R&=8.314\frac{J}{mol*K}=8.314x10^{-3}\frac{kJ}{mol*K} \\T&=298\,K \\K&=e^{\frac{-(-141.756\frac{kJ}{mol})}{(8.314x10^{-3}\frac{kJ}{mol*K})(298\,K)}} \\K&=1.418x10^{25} \\\end {align} $$
Here we see that this reaction is spontaneous since we have a -ΔG° and that the reaction is heavily product favored since K is so large.
To add another variable into the mix, let's look at entropy.
Entropy (ΔS) is the measure of "disorder" in a system. It refers to the number of combinations or "microstates" that the atoms within a system can have.
If you are anything like me, you probably throw your clothes on the floor instead of putting them away. Since it doesn't take any energy, it is preferable from taking the time to properly sort them. As time passes, more and more clothes pile up. The "disorder" is only going to decrease when energy is added (i.e. actually putting your clothes away).
So back to equilibrium, If K > 1, then the products have higher entropy. However, if K < 1, then the reactants have higher entropy. The system is going to go towards whatever side that will increase entropy, but there will be a limit, just like with spontaneity. In our clothes example, you would eventually run out of clothes to throw on your floor unless you started borrowing someone else's. This would be our "equilibrium" point.
The equilibrium constant (K) and Free energy change (delta G) have an inverse relationship. The equation relating to two is: K=e(deltaG*/(RT)) Where R is the gas constant and T is temperature.
At equilibrium, there is no net change in concentration. Therefore, there is also no net change in free energy.
Yes, the change in free energy is 0 at equilibrium.
The change in Gibb's energy (delta G) is 0, however the value of the Gibb's energy (G) is dependent on the reaction.
The equilibrium constant (Kp) is based on the partial pressures of the reactants and products. The free energy change is dependent on the equilibrium constant, so it is therefore also dependent on pressure.
Flashcards in Free Energy and Equilibrium10
Start learningWhat is (Gibb's) Free energy?
Gibb's free energy, or just Free energy, is the amount of available energy in a system to do work.
What is equilibrium?
At equilibrium, the concentration of reactants and products remains the same. The reaction hasn't "stopped", it just means that the rate of the forward reaction and the rate of the reverse reaction is the same. This is called dynamic equilibrium.
Which of the following statements are correct?
If K>1, the products are favored
What is the standard free energy change?
ΔG° is the standard free energy change. It is the change in free energy that happens when a compound is formed from its elements at 1 atm and 298 K (25° C). While ΔG changes over time, ΔG° is a constant.
What is the reaction quotient?
The reaction quotient (Q) is the ratio of products to reactants not at equilibrium.
True or False: The relationship between the equilibrium constant and free energy is inversely proportional.
True
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