Biological Physics/Free Energy

Motivation for Gibb's Free Energy

We have now encountered both enthalpy and entropy separate from each other. Now, we want to be able to combine them together to give us a full description of energy for a given system. At this time, we introduce a new concept called Gibb's free energy $\Delta G$. In order to understand the elusive and mysterious free energy term $\Delta G$, let's first start out with a simple and concrete example. The table below shows the enthalpy and entropy for water freezing.

$H_{2}O(l){\overset {\Delta }{\rightarrow }}H_{2}O(s)$

Temperature (ºC)	∆H (J/mol)	∆U (J/mol)	∆S (J/mol*K)	T∆S (J/mol)	∆G (J/mol)
-10	-5619	-5619	-20.555	-5406	-213
0	-6008	-6008	-22.007	-6008	0
10	-6397	-6397	-23.406	-6624	227

Recall from earlier that ${\frac {1}{T}}={\frac {dS}{dU}}$ or $dU=TdS$. Also, remember that $\Delta U=\Delta H+P\Delta V$. Since $\rho _{liquid~water}=1000{\frac {kg}{m^{3}}}$ and $\rho _{ice}=920{\frac {kg}{m^{3}}}$ then the volume for one mol of water is $V_{liquid~water}=({\frac {1000~kg}{m^{3}}}*{\frac {1~mol}{18.02~g}}*{\frac {1000~g}{1~kg}})^{-1}=1.8\times 10^{-5}{\frac {m^{3}}{mol}}$and for ice $V_{ice}=({\frac {920~kg}{m^{3}}}*{\frac {1~mol}{18.02~g}}*{\frac {1000~g}{1~kg}})^{-1}=1.956\times 10^{-5}{\frac {m^{3}}{mol}}$.

Assuming a constant atmospheric pressure of $P=1.013\times 10^{5}{\frac {N}{m^{2}}}$, then $P\Delta V=(1.013\times 10^{5}{\frac {N}{m^{2}}})(1.956\times 10^{-5}-1.8\times 10^{-5}{\frac {m^{3}}{mol}})$ giving $P\Delta V=0.158{\frac {J}{mol}}$.

Since $Work=-P\Delta V$, then $W=-0.158{\frac {J}{mol}}$. This value is very small compared to the enthalpy; therefore, the change in internal energy ∆U is $\Delta U\approx \Delta H$.

Latent Heat is defined as the heat that is released or absorbed during a constant temperature process. Since the process is moving from a liquid to a solid, then the latent heat is the amount of energy that is leaving. This means that at 0 ºC, $\Delta H=H_{ice}-H_{water}<0$. What is really happening is that there is a difference in heat capacities between liquid water and ice at 0 ºC. This This energy change is about 38.9 J/mol*K.

The rest of the Delta H portion needs to be written

Now let's look at the ∆S term. Recall that $S={\frac {Q}{T}}$ $\Delta S={\frac {\Delta H}{T}}$ for a phase transition at constant temperature.

Electrolysis

Now let's look at Electrolysis: the splitting of a water molecule:

$H_{2}O\rightarrow H_{2}+{\frac {1}{2}}O_{2}$

This reaction is an important reaction both in biology and fuel cell applications. Below is a table that that gives us pertinent information about the molecules. The data is given for $T=298K$ and $P=1.013\times 10^{5}$.

Molecule	∆H (kJ/mol)	S (kJ/mol*K)	TS (J/mol)	G (J/mol)
$H_{2}O(l)$	-285.83	0.06991	20.833	264.9982
$H_{2}O(g)$	0	0.13068	38.943	38.94264
$O_{2}(g)$	0	0.20514	61.132	61.13172

Since $\Delta G=\Delta H-T\Delta S$, then by substitution from the above data yields $\Delta G=285.83-(61.132+38.943-20.833)$ so $\Delta G=237{\frac {kJ}{mol}}$. This is the amount of energy that we would need to be put into the system to drive electrolysis forward. Also, the amount of energy ∆ST would be put into the system as heat input (the system is going to a more disorderly state). Some of this energy will be used for expanding the gas into a larger volume. We know that for an ideal gas $V={\frac {nRT}{P}}$. Since there is 1 mol of hydrogen and 1/2 mol of oxygen, $V_{gas}={\frac {(1.5)(8.31)(298)}{1.013\times 10^{5}}}=0.03675~m^{3}$. Liquid water has a density $\rho =1{\frac {g}{cm^{3}}}$. Then $V_{liquid}={\frac {1~cm^{3}}{g}}*{\frac {18~g}{mol}}*{\frac {1~m^{3}}{(100~cm^{3})^{3}}}=1.8\times 10^{-5}~{\frac {m^{3}}{mol~H_{2}O}}$. Then $\Delta V=0.03675-0.000018=0.036732~m^{3}$. Therefore, the work it takes to expand the gas is $W=-P\Delta V=-(1.013\times 10^{5})(0.036732)=-3.721{\frac {J}{mol}}$. This is the amount of energy that is lost in the system. Thus, the total internal energy would be $\Delta U=\Delta G+T\Delta S-P\Delta V=282.109~{\frac {kJ}{mol}}$.

Fig. 1 displays water splitting electrolysis and Fig.2 displays fuel cell water production

Typically, electrolysis is driven in fuel cells by using electrical energy. The below setup is used to produce the necessary electrical energy required to push the reaction forward. Two electrodes are stuck in a beaker of water (typically a salt water), one connected to the positive side of a battery and the other connected to the negative side. Electrons leave the negative electrode (cathode) and react with the free positively charged hydrogen atoms to give hydrogen gas $2H^{+}+2e^{-}\rightarrow H_{2}(g)$. Water is split and the hydroxide is electrically attracted to the anode. At the anode, the hydroxide undergoes an oxidation $2H_{2}O~(l)\rightarrow O_{2}~(g)+4H^{+}+4e^{-}$. This produces oxygen gas, hydrogen ions that are then attracted to the cathode, and free electrons that are absorbed by the anode and complete the electrical circuit. Thus, oxygen gas is produced at the anode and hydrogen gas is produced at the cathode. For redox half reactions to balance, twice as many hydrogen than oxygen gas will be produced.

Caption text

To electrically push the reaction forward, a certain voltage needs to be produced. If 237 kJ/mol are needed, then ${\frac {237~kJ}{mol}}*{\frac {1000~kJ}{J}}*{\frac {1~mol}{6.02\times 10^{23}~rxn}}*{\frac {1~rxn}{2e^{-}}}*{\frac {1~e^{-}}{1.602\times 10^{-19}~C}}=1.23{\frac {J}{C}}=1.23~V$. So if 1.23 V are applied to the above system, then electrolysis occurs.

Chemical Potential

Up to this point, we have looked at what happens when energy packets can move around in a system, but we have ignored the notion that particles can be exchanged in a system. This new notion of particles moving around in a system is called Chemical Potential Energy. Now, we have to modify our definition of calculating free energy

$\Delta G=\Delta H-T\Delta S+\mu \Delta N$ for constant temperature and pressure

N stands for the number of particles and µ is known as the chemical potential. To understand what chemical potential is, let's start off with an isolated system consisting of 2 subsystems: A and B (Figure 3)

Exchange of particles between two systems

A and B have certain attributes: the internal energy U, the volume V, the number of particles N, and the entropy S. There is some sort of divider between A and B that only allows one of these characteristics of A and B to change. For instance, the divider may be some sort of slider if the volume between A and B was allowed to change. The divider may be a permeable membrane that only allows the number of particles to move between A and B. Our goal is to describe all of these characteristics in terms of entropy and ultimately maximize entropy for the whole system.

We have already encountered systems that only exchange energy between A and B (look back at the entropy section). We showed that at thermal equilibrium, ${\frac {dS_{total}}{dU}}=0$ or ${\frac {dS_{A}}{dU}}=-{\frac {dS_{B}}{dU}}$. At thermal equilibrium (max total entropy), the temperatures of A and B are also equal.

Now, there's also a mechanical equilibrium of the pressures as a change in volume that's being applied to both sides. So what exactly is ${\frac {dS}{dV}}=?$. Well, as the volume increases in a system, there are more ways to rearrange the particles. Doing a little dimensional analysis will help bridge the gap between S and V. $S={\frac {J}{K}}$ and $V=m^{3}$. So, ${\frac {S}{V}}={\frac {J}{K~m^{3}}}={\frac {N~m}{K~m^{3}}}={\frac {N}{K~m^{2}}}={\frac {P}{K}}={\frac {Pressure}{Temperature}}$. So then it makes sense that ${\frac {dS}{dV}}={\frac {P}{T}}\rightarrow TdS=PdV$.

Finally, let's allow the two systems to echange particles and understand how entropy and the number of particles are related. ${\frac {dS}{dN}}={\frac {T/K}{\#}}={\frac {J}{K}}$ is known as Diffusive Equilibrium and, if temperature is multiplied to this value, an energy value is obtained $-T{\frac {dS}{dN}}=\mu$. This is known as the Chemical Potential. But why is there a negative sign? A larger, stable system has more free energy.

To sum up, the following expression exemplifies how entropy is related to internal energy, volume, and the number of particles by using a mathematical concept of partial derivatives. A partial derivative of a function with multiple variables is when the derivative of the function is taken with respect to only one of its variables while the others are held constant. It shows how the function changes when only one variable is allowed to change. The symbol for partial derivative is the lowercase delta $\delta$. The subscripts indicate what is being held constant.

$({\frac {\delta S}{\delta U}})_{V,N}={\frac {1}{T}}$ and $({\frac {\delta S}{\delta V}})_{U,N}={\frac {P}{T}}$ and $({\frac {\delta S}{\delta N}})_{U,V}={\frac {-\mu }{T}}$

To account for all the ways that entropy can change, all of the partials are added together to give a total differential of entropy. This idea of adding the partial derivatives together is a fundamental theorem of multivariable calculus. Essentially, the $\delta$ on bottom and the $d$ on top "cancel each other out, leaving just the $dS$ on top.

$dS=({\frac {\delta S}{\delta U}})_{V,N}dU+({\frac {\delta S}{\delta V}})_{U,N}dV+({\frac {\delta S}{\delta N}})_{U,V}dN$.

So by some substitutions,

$dS={\frac {dU}{T}}+{\frac {PdV}{T}}+{\frac {-\mu dN}{T}}$ and some rearranging $TdS=dU+PdV-\mu dN$ which implies

$dU=TdS-PdV+\mu dN$

Internal Energy = Heat + Work + Diffusion

Now, starting with $G=H-TS\rightarrow G=U+PV-TS$ then by the product rule, $dG=dU+PdV+VdP-TdS-SdT$ . With a substitution of what was solved for above, $dG=TdS-PdV+\mu dN+PdV+VdP-TdS-SdT$ or with some cancellations $dG=VdP-SdT+\mu dN$. This equation illustrates how equilibrium would change if the other variables change accordingly. This is different from $\Delta G=\Delta H-T\Delta S$; this equation is a measure of energy difference from a non equilibrium and equilibrium state at a constant T,P,N. In fact, dG is a Thermodynamic Identity

Example 1

Graphite has a density of $2.2{\frac {g}{cm^{3}}}$ and 1 mol of graphite (carbon) is 12.01 g. So the volume from 1 mol of graphite is $V={\frac {1~cm^{3}}{2.2~g}}\times {\frac {12.01~g}{1~mol}}\times {\frac {1~m^{3}}{100^{3}~cm^{3}}}=5.4\times 10^{-6}m^{3}$. Diamond has a density of $3.5{\frac {g}{cm^{3}}}$ and an identical calculation yields the volume of 1 mol of diamond to be $V=3.4\times 10^{-6}m^{3}$. At $298~K$ and $1.013\times 10^{5}~Pa$, $G_{graphite}=0{\frac {kJ}{mol}}$ and $G_{diamond}=2900{\frac {kJ}{mol}}$ (the energy required to form diamond at STP). So $\Delta G=2900{\frac {kJ}{mol}}$. Now, let's only allow pressure to change. Then $dG=VdP-SdT+\mu dN$ reduces to just $dG=VdP$. By plotting G vs P, $G_{0}$ is the free energy at STP and the slope of the line is ${\frac {dG}{dP}}$ or since $dG=VdP/rightarrow{\frac {dG}{dP}}=V$. So by plugging in the volumes as the slope of the lines for free energies of graphite and diamond $G=VP+G_{O}$ yields

$G_{graphite}=(5.4\times 10^{-6})(P)+0$

$G_{diamond}=(3.4\times 10^{-6})(P)+2900$.

Insert Graph of the G curves

$G_{eq}$ is the value of free energy where diamond and graphite are equally abundant. This is the intersection of $G_{graphite}$ and $G_{diamond}$. By solving the above system of equations, it can be shown that $P=1.45\times 10^{9}Pa\times {\frac {1~atm}{1.013\times 10^{5}~Pa}}=14000~atm$!

Example 2

Before the specifics of the example are given, let's look at dG when temperature and pressure are held constant. By the thermodynamic identity $dG=-SdT+VdP+\mu dN$, and since temperature and pressure are held constant, $dG=\mu dN$. By integrating both sides, $\int dG=\int \mu dN\rightarrow G=\mu N\rightarrow \mu ={\frac {G}{N}}$ and now allow the pressure to change ${\frac {d\mu N}{dP}}={\frac {d}{dP}}G\rightarrow N{\frac {d\mu }{dP}}={\frac {dG}{dP}}$ and since it has been shown that $V={\frac {dG}{dP}}$. so $N{\frac {d\mu }{dP}}=V\rightarrow {\frac {d\mu }{dP}}={\frac {V}{N}}$. From the Ideal Gas Law, $PV=Nk_{B}T\rightarrow {\frac {V}{N}}={\frac {k_{B}T}{P}}$ so by substitution ${\frac {d\mu }{dP}}={\frac {k_{B}T}{P}}$. Now, the equation is all in terms of pressure. By multiplying both sides by $d\mu$ and then taking the antiderivative, $\int d\mu =\int {\frac {k_{B}T}{P}}dP\rightarrow \Delta \mu =k_{B}Tln(\Delta P)\rightarrow \Delta \mu =k_{B}Tln({\frac {P}{P_{0}}})$. Also, it was previously shown that $d\mu ={\frac {dG}{dN}}\rightarrow \Delta \mu ={\frac {\Delta G}{\Delta N}}$ so by substitution ${\frac {\Delta G}{\Delta N}}=k_{B}Tln({\frac {P}{P_{0}}})$.

In the lungs, there exists a partial pressure of oxygen of $P_{O}=0.2*1.013\times 10^{5}Pa=2.026\times 10^{4}Pa$. In the venous blood that gets pumped to the lungs, $P=0.25(lungs)=0.25*2.026\times 10^{4}Pa=5.06\times 10^{3}Pa$, so the ratio of partial pressure of oxygen in the air compared to the blood is just 0.25. So, the change in chemical potential can be calculated as $\Delta U=(1.381\times 10^{-23})(310)ln(0.25)=-5.93\times 10^{-21}{\frac {J}{molecule~O_{2}}}=-3.57{\frac {kJ}{mol}}$. This is the reason oxygen moves from the air in the lungs to the deoxygenated blood and why we keel over.

Caption text

A picture illustrating what's going on

Example 3

Once again, going back to the Ideal Gas Law$P=({\frac {n}{V}})RT$. So by substitution into the above equation for the change in chemical potential, $\Delta \mu =ln({\frac {{\frac {n}{V}}RT}{{\frac {n_{0}}{V_{0}}}RT}})\rightarrow \Delta \mu =k_{B}Tln({\frac {\frac {n}{V}}{\frac {n_{0}}{V_{0}}}})$. Since ${\frac {n}{V}}=concentration~[A]$, then by substitution $\Delta \mu =k_{B}Tln({\frac {[A]}{[A_{0}]}})$.

The typical glucose level of our body when we first wake up is $[A]=3.5~{mM}$; this level is usually called the fasting level because it has been several hours since our last meal. However, a typical glucose level after eating a meal, say breakfast, is $[A_{0}]=7~mMol$. From the above equation, the change in chemical potential can be calculated as $\Delta \mu =(8.315)(310)ln({\frac {3.5}{7}})=-1787~{\frac {J}{mol}}$.

Example 4

Imagine a charged membrane, like the membrane of a cell, with a certain charged particle that needs to be pumped across other ions of like charge.

Caption text

Insert Picture of the charged membrane

The charged particle must work against both the electric field and concentration gradient to successfully move through the membrane. Recall from introductory physics that the work a charged particle exerts in an electric field is $W=qEd$ or in terms of the potential $W=qV$. Tacking this energy as a portion of the total chemical potential energy, $\Delta \mu =qV+k_{B}Tln({\frac {[A]}{[A_{0}]}})$. To rewrite this equation in terms of mols, let's define a variable F where $F=q*mols$ so then the new equation would be $\Delta \mu =nFV+RTln({\frac {[A]}{[A_{0}]}})$.

Let's look at a sodium ion trying to be pumped across a membrane and outside the cell where there exists a lot more sodium ions. At 100 mV, what concentration gradient for the sodium ion exists? Well, when $\Delta \mu =0$, $qV=-k_{B}Tln({\frac {[A]}{[A_{0}]}})\rightarrow {\frac {[A]}{[A_{0}]}}=e^{\frac {-qV}{k_{B}T}}\rightarrow {\frac {[A]}{[A_{0}]}}=e^{\frac {-1.602\times 10^{-19}(0.1)}{1.381\times 10^{-31}(310)}}=0.0237$. So then ${\frac {[A_{0}]}{[A]}}=42.2$. Therefore, there's 42.2 times more sodium ions outside than inside the cell.