Thermodynamics

Thermodynamics is the study of energy and its transformations. Before launching into it, it's worth briefly considering what energy is, as it's not as obvious as you might think. As Richard Feynmann put it, you can think of energy as something akin to children's building blocks. Say you have ten bricks. You let the kid play with them, and then notice that there are only eight left. You're slightly worried until you notice that there is a suspicious lump under the sheets of his bed, which looks just like two blocks. Ah ha, you think, I still have ten blocks. Whilst doing this, three more blocks are secreted by the kid. You hunt about, then notice that the level of water in the bath is slightly higher, just sufficient to account for the volume of the three blocks. Then you notice a smell of burning and another missing block, but by checking the carbon dioxide concentration in the air, you deduce that the little brat has burnt one of the blocks. And so on. Energy is rather like the building blocks, in that you know you have a certain amount of it that never changes, BUT, in reality there are NO literal, concrete building blocks in nature: energy never exists in some 'pure' unadultered form: it always exists as clever mathematical trickery of the volume displacement, air composition and suspicious lumpiness varieties.

Energy therefore can take many forms, none of them more fundamental than the others. It can usually be expressed as the multiplication of a potential factor by a capacity factor. For example, the energy E, required to compress a gas by a certain volume (capacity factor) ∆V (where ∆ means 'a change in' ) is simply E = p∆V where p is the pressure of the gas you are trying to compress (potential factor). Similarly, the energy you can release from moving a certain amount of charge ∆Q towards an opposite charge producing an electric field of strength E, is E∆Q. The osmotic equivalent is to move a certain mass (M) of molecules with an osmotic potential (C) is E = C∆M; the heat equivalent is to change the entropy (S) of a gas at a certain temperature (T) is E = T∆S; and so on. These are the 'block counting' mechanisms we have. Ultimately, as Einstein showed, energy and mass are same thing: the energy extractable from a system is equal to its mass multiplied by the speed of light squared, E = m c². However, even when considering quite large energy changes, the change in mass is so small as to be negligible, which is why the conservation of energy (the fact that the energy extractable from a system equals the energy originally put into it), and the conservation of mass (mass in = mass out) have traditionally been considered separate laws. In fact they are both approximations to the conservation of mass-energy.

Energy can be loosely defined as the ability of a system to do work on its surroundings. Work means moving things, deforming things, breaking things, etc.

An important distinction in thermodynamics is between the different types of system. This is where the idiots who think evolution defies the second law of thermodynamics slip up in a frighteningly stupid way. There are three sorts of system:

An isolated system cannot exchange energy or matter with the universe. They are dull.
A closed system cannot exchange matter with the universe, but can exchange energy. Most biochemical systems can be approximated as closed systems, as the maths is easier, but in reality biology deals with:
An open system, which can exchange both matter and energy with the universe.

Thermodynamics involves transferring matter or energy (we'll only really consider the latter) from the surroundings to the system, which is a little bit of universe we are interested in. There is a convention in thermodynamics that we must remember, so as not to get into mathematical tangles later.

If energy enters the system from the universe, its sign is positive. If work (W) is done on the system, its sign is positive.
If energy leaves the system, its sign is negative. If work is done by the system, its sign is negative.

These are the conventions generally used by chemists. However, physicists are excited by steam engines, where heat is added to a cylinder of steam to make it expand and do work, so they often use the opposite convention, where W is the work done by the system. The moral of the story is to make sure you know what convention is being used for the sign of W.

Thermodynamics has four important laws, which we will take on at a time. The first is actually called the zeroeth, as it got tacked on as an after-thought:

Zeroeth law

If a system A is in thermodynamic equilibrium with system B, and system B is in thermodynamic equilibrium with system C, then A is also in equilibrium with C. Put more plainly, if A = B and B = C then A = C too.

Well, durrr.

First law

Energy can be neither created nor destroyed: it can only change form. You cannot win by making energy from nothing: you can only break even.

The are several ways to express the first law, and physicists and chemists often use different conventions. The prevalent mathematical way, and the one most important to thermodynamics, which deals primarily with work and heat exchanges, is:

U = Q + W

The internal energy (U) of a system is the heat in the system (Q) plus the work (W) previously done on the system. Using 'delta' notation for changes in energy, where

∆X = X_after − X_before

changes in internal energy can be expressed as:

∆U = ∆Q + ∆W

The change in internal energy is the heat supplied to the system, plus the work done on the system (note the positive signs). This is more convenient, since the absolute value of the internal energy is extremely difficult to measure with any accuracy (ultimately, it is mc², but you'd need to know the mass in kilograms to an accuracy of about fifteen decimal places to get the answer in Joules to any reasonable degree of accuracy).

Physicists, with their "W is work done by the system" convention may write the first law as:

∆U = ∆Q − ∆W

The internal energy is termed a state function: this means that its value is independent of the history of the system. That is to say, it doesn't matter how the energy was supplied (heat or work), a system that has had 10 J (Joules, the SI energy unit) of heat energy supplied to it will experience the same change in internal energy as a system that has had 5 J of heat supplied and 5 J of work done on it. Other thermodynamic functions of state include entropy, enthalpy and free energy, which we will cover shortly.

A concept closely related to internal energy is that of enthalpy, H. When a reaction is done on Earth in an open vessel (like a test tube, conical flask, etc.), the reactants may well have a different volume to the products. If the products are more voluminous than the reactants, a certain amount of energy will be 'wasted' pushing the atmosphere away; and if the volume decreases, a certain amount of work will be done on the system as the atmosphere squashes it. The change in enthalpy of the system is the change in internal energy plus the work done on the system by the atmosphere (if it is squashed), or less the work done on the system (if it expands). That is to say, enthalpy change is the energy exchanged as heat (∆Q) between system and surroundings once you've accounted for any energy lost by shifting the atmosphere out of the way. The mechanical work done on the atmosphere is −p∆V (using our convention that work done by the system is negative), i.e. atmospheric pressure (p) times the increase in volume of the system (∆V). If we define ∆H as ∆Q under constant pressure, then:

∆H = ∆U + p∆V

A reaction that leads to a decrease in enthalpy (∆H < 0) will release heat to the surroundings, and is termed exothermic. A process that increases the enthalpy of the system (∆H > 0) will be positive in sign, and hence will take up energy from the surroundings, and is termed endothermic. Exothermic reactions tend to be more favoured than endothermic reactions because low energy states are more stable. Note that for a reaction to be exothermic, the system must lose internal energy (∆U is negative) and/or contract in volume (∆V is negative). No heat will be evolved during a chemical reaction if the liberated internal energy is entirely used up in pushing the atmosphere away.

Second law

Heat is a specially degraded form of energy, and is difficult to convert to useful, workable energy. The efficiency of a process measures how well it converts forms of energy amongst each other. In particular, the efficiency of converting heat to work is related to the temperature of the heat sink to which the heat gets dumped. You can only break even in heat/work conversion if your heat sink is at absolute zero. Another way to express the second law is that the entropy (i.e. the amount of disorder) of the universe is increased by all energy-conversion processes: they all produce some amount of heat, which cannot be converted back to useful energy without an absolute zero (0 Kelvin) heat sink.

The efficiency, η, of converting heat to work is related to the temperatures of the heat source (e.g. a plane engine) and heat sink (e.g. the atmosphere). This is maximally:

η = ( T_source − T_sink ) ⁄ T_source

Where T is the absolute temperature, in Kelvin. Note that if the sink is at absolute zero, (0 K, or −273.15°C), the efficiency will be 1, i.e. 100%. This is a convenient way to understand why the second law says you can only break even (i.e. convert heat to work with 100% efficiency) with a sink at absolute zero. The commoner way of expressing the second law is in terms of entropy. Entropy, S, is the amount of disorder in a system, technically:

S = k ln W

Where W is the number of ways of arranging the heat energy in the system, and k is Boltzmann's constant (1.380650 × 10⁻²³ J K⁻¹). The amount of disorder in the universe is always increased by physical processes:

∆S_universe = ∆S_system + ∆S_surroundings > 0

This is intuitively obvious: if we separate chlorine and oxygen gases into two halves of a fish-tank with a glass partition, then remove the partition, it is overwhelmingly likely that within a short period of time, the two gases will be intimately mixed, and the amount of order in the system hugely decreased. Although every individual arrangement of the molecules has a similar probability, those in which there is no apparent order in the system for the vast majority of them.

When deciding whether a reaction is favoured, i.e. spontaneous, we need to consider both the energy (enthalpy) change, and the entropy change. Reactions which release energy, and increase disorder will be spontaneous, feasible, and will occur (even if this is extremely slowly). Those which both increase order and take up energy will never occur spontaneously: they must be coupled to some other spontaneous process in order to be made to occur. The other two combinations require a more mathematical treatment involving free energy to determine whether they are spontaneous. The most well known of the free energy equations is that of Gibbs:

∆G = ∆H − T∆S

∆G is the (change in) Gibbs free energy of the system, which can be used to describe changes in free energy in constant pressure systems (such as biological systems), and is the most important term for biologists, as it determines whether a reaction is feasible. Helmholtz free energy (A for Arbeit) is the constant volume version (∆A=∆U −T∆S). A reaction with negative ∆G is termed exergonic, and will be spontaneous, although it may not occur at any appreciable rate. One with positive ∆G is termed endergonic, will not be spontaneous, and will not go ahead, unless coupled to another reaction with negative ∆G (biology does this all the time, by coupling the hydrolysis of ATP (very exergonic) to thermodynamically unfavourable reactions).

Note that you could write out Gibbs energy in full as:

∆G = ∆U + p∆V − T∆S

∆G is the maximum quantity of useful work that a chemical reaction can perform on its surroundings (∆G = −W_max). ∆U can be thought of as the total energy liberated by a chemical reaction. Remembering that ∆U = U_after − U_before, if internal energy is liberated, it has a negative sign, so let's arbitrarily say this is −200 J. At constant (atmospheric) pressure, p∆V of this energy is involved in expansion/contraction of the system. If there is an increase in volume, p∆V could have a value of +50 J, hence some of the liberated internal energy is 'wasted' in expansion against the atmosphere. Likewise, some of the energy may have to be 'wasted' in increasing the order of the system: if there is a decrease in entropy, then −T∆S might also take a (positive) value of +50 J. After accounting for these two factors, we are left with −100 J of energy that is 'free' to be used to do more exciting things.

Another way to think about this equation is that G is the total energy required to create a system from nothing:

G = U + pV − TS

If we were to create a chemical system from nothing (an infinitesimal speck at 0 K), we would need to add energy, the value of which is U. Let's say this is 250 J. However, the creation of systems does not usually occur 'in vacuum' (literally or metaphorically). If we were creating a system on Earth, it would be created in an atmosphere at a pressure of about 0.1 MPa and temperature of about 298 K. Because the system is being created from with a certain non-zero volume, we will need some extra energy (pV) to push the atmosphere away and make space for it, let's say another 50 J, for a total cost of 300 J. However, because the surroundings have a temperature of 298 K, we can expect some spontaneous energy flow (TS) into our newly created system from the surroundings. If our new system is at 0 K, and the atmosphere is at 298 K, entropic considerations would suggest heat would flow from the hot body to the cold one until thermal equilibrium is achieved. This heat flow into the new system would help offset the energy requirement. If say, 100 J were to flow from the atmosphere into our system, then the total 'free' energy we would need to supply would only be 200 J.

Back to the point. The actual value of ∆G is dependent on the concentrations of products and reactants in a reaction. This should be clear if you consider that if the reaction of lots of A to form B is spontaneous, then the reaction of lots of B to form A will not be: the relative concentrations of A and B determine the spontaneity of the reaction or its reverse.

The standard free energy change, ∆G⁰ is defined as the free energy change associated with e.g. the reaction:

A + B ⇌ C + D

when the temperature is 298 K (25°C), the pressure 1 atmosphere (101 325 Pa), and A, B, C and D are all at 1 M concentrations. This is actually a bad idea for biology, since water (55.5 M) and proton concentration (10⁻⁷ M at pH 7) are almost never at this standard value. Hence, ∆G⁰′ has been defined to account for this: it assumes water as a 55.5 M solvent, at pH 7.

If we set up a reaction under ∆G⁰ (or ∆G⁰′) conditions, and we find it has a positive sign, we know the forward reaction will not be energetically favoured with these concentrations of products and reactants. However, the reverse reaction has the same value of ∆G⁰ but the opposite sign, hence the reverse reaction will be favoured.

So, if ∆G⁰ has negative sign, we know that (under standard conditions) the forward reaction is feasible, and A and B will convert (at some rate) to C and D. This means the product and reactant concentrations will change over time. To work out the ∆G under these new concentration conditions, we need to use the formula:

∆G = ∆G⁰ + R T ln Q

where R is the universal gas constant (8.314472 J K⁻¹ mol⁻¹), T the absolute temperature, and Q is the mass action ratio. This is calculated by multiplying together the molar concentrations of the products and dividing them by the molar concentrations of the reactants multiplied together (i.e. product of the products divided by the product of the reactants):

A + B ⇌ C + D

Q = [C] [D] ⁄ [A] [B]

The [square brackets] mean 'molar concentration of'. For more complex reactions, we get different values of Q:

A + A ⇌ C + D + D

Q = [C] [D] [D] ⁄ [A] [A] = [C] [D]² ⁄ [A]²

Generally, for the reaction:

ρ₁R₁ + ρ₂R₂+ ρ₃R₃ + … ⇌ π₁P₁ + π₂P₂+ π₃P₃ + …

the mass action ratio is:

Q = ∏ [P]_i ^π_i ⁄ ∏ [R]_i ^ρ_i

If you understand big-pi notation ☺

Let's start with the ∆G⁰ reactant and product concentrations, all 1 M, and let the reaction run. As the reactant and product concentrations change during the course of the reaction (starting from the ∆G⁰ conditions), ∆G will change as Q changes, and will head towards zero. If ∆G⁰ is negative, the forward reaction is favoured, and Q will increase (products produced at expense of reactants), so making ∆G less negative. If however ∆G⁰ is positive, the backwards reaction is favoured, and Q will decrease, so making ∆G less positive. At some combination(s) of product and reactant concentrations, a dynamic equilibrium will be achieved, where neither net forward nor net backwards reaction is favoured, and ∆G = 0. Under these conditions:

∆G = ∆G⁰ + R T ln Q = 0

∆G⁰ = − R T ln K_eq

where the value of Q is a special one called the equilibrium constant, K_eq:

K_eq = Q_eq = [C]_eq [D]_eq ⁄ [A]_eq [B]_eq

At this equilibrium point, no net reaction will occur. A very high K_eq indicates that the reaction proceeds to completion, i.e. the concentration of products at equilibrium vastly exceeds the residual reactant concentrations.

Note that we can combine the two equations above:

∆G = ∆G⁰ + R T ln Q

∆G⁰ = − R T ln K_eq

∆G = R T ln Q − R T ln K_eq = − R T ln ( K_eq ⁄ Q )

This shows clearly that the free energy change is negative (spontaneous reaction) only if Q is less than K_eq, i.e. when there is an excess of product over reactant (w.r.t. the equilibrium values). We can also write this as:

∆G = R T ln ( Q ⁄ K_eq )

Third law

The entropy of a perfect crystal at absolute zero is 0. There is no physical process whereby absolute zero can be achieved, hence you cannot have a heat sink at 0 K. Therefore you cannot break even: you can only lose.

This one doesn't have too much relevance to biology, beyond nailing down the fact that the second law hinted strongly at: the first law says you can't win (i.e. make energy from nothing), the second law says you can't break even except with a heat sink at 0 K, and the third law says that you can't even do that, since 0 K is unattainable. Rather a depressing triplet.

The application of thermodynamics to biology has important implications in determining which reactions are feasible (i.e. spontaneous) and which can be made spontaneous by coupling them to another process (e.g. the exergonic hydrolysis of ATP).

Electrochemistry

An important application of thermodynamics is that of measuring redox potentials. Redox reactions are those reactions where electrons are transferred from one chemical species to another. For example, Cu²⁺ + 2e⁻ → Cu is a reduction, as the oxidation number of copper is reduced from +2 to 0. Remember OILRIG (oxidation is loss of electrons, reduction is gain). Zn → Zn⁺ + 2e⁻ is an oxidation therefore.

These two redox half reactions can be carried out in half cells: a piece of copper metal suspended in a solution of copper (II) sulfate is a Cu/Cu²⁺ half cell. We can connect an electrode to the copper, and dangle another one in the solution (usually made of platinum), but no electrons will flow out of (or into) this cell unless we connect it to another half cell of a different kind. The redox potentials of these two cell determines in which direction electrons will flow. For example, if we connect a copper and a zinc half cell:

Cu²⁺ + 2e⁻ → Cu

Zn → Zn²⁺ + 2e⁻

Electrochemical cells are formed by linking two half cells with one another, usually by platinum electrodes and/or a salt bridge.

We will be able to couple the redox reactions, and observe electron flow between the two cells. The salt bridge is a saturated solution of potassium chloride, which is just a smart-arse sort of electrode connection that prevents the copper and zinc half cells from mixing. The coupled reactions will be:

Cu²⁺ + Zn → Cu + Zn²⁺

Cu(+2) + Zn(0) = Cu(0) + Zn(+2)

The copper ions are reduced by the reaction, and hence are termed oxidising agents, because they oxidise the zinc metal. The zinc metal is oxidised by the reaction, and is itself therefore a reducing agent because it reduces the copper ions. Note that if we just mixed the solutions together we'd get the same reaction, but by separating them, we can force the electrons to take a long route via something useful (like a bulb), and put their energy to a more useful use than making the reaction vessel hot.

In this case we have 'assumed' that the electrons will flow from the copper half cell to the zinc half cell. Where did we get this information from? We got it by doing a calculation using a table of standard redox potentials. We measure the standard redox potential, E⁰, of a half-cell against a standard hydrogen half cell.

Electrochemical cells are formed by linking two half cells with one another, usually by platinum electrodes and/or a salt bridge.

Both the hydrogen half cell and the other half cell are standard half cells, which it will not surprise you to know are those where all solutions are at 1 M, the temperature is 298 K, pressures 1 atm, etc. For the copper half cell, this means a lump of copper in a 1 M solution of copper (II) sulfate. For the hydrogen half cell, this means hydrogen gas at 1 atm in contact with a 1 M solution of hydrochloric acid, etc. For an NADH half-cell, you would need an NADH/NAD⁺ solution, both at 1 M.

The potential of two coupled cells such as:

Zn (s) | Zn²⁺ (aq) ¦ Cu²⁺ (aq) | Cu (s)

is given by:

E⁰ = E⁰_{reduction
process} − E⁰_{oxidation process}

In cell diagrams such as the one above, the spontaneous reaction runs from left to right (zinc metal becomes ions; copper ions become copper), and the oxidation reaction is conventionally written on the left hand side.

The potential of the hydrogen half-cell (E⁰_H) is defined as zero, hence for:

Zn (s) | Zn²⁺ (aq) ¦ H⁺ (aq) | H₂ (g)

the redox potential is given by:

E⁰_cell = E⁰_H − E⁰_Zn = − E⁰_Zn

The E⁰ of half-cells coupled to the hydrogen half-cell can be arranged in an electrochemical series, i.e. a list of cells from those generating a very negative E⁰ such as K/K⁺ through intermediate ones like Zn/Zn²⁺, to the hydrogen half cell (H₂/H⁺), with E⁰ of 0 V, then to those with very positive potentials like Cl₂/Cl⁻. Note that the E⁰ is the opposite of what you might expect: although chlorine is very electronegative, this means it is desperate to gain electrons, hence its electrode potential is very positive, and highly attractive to electrons. Likewise, iron slowly rusting into a solution of Fe³⁺ ions is a good source of electrons, so has a negative E⁰. Electrons will flow from the more negative half-cell to the more positive half-cell. To actually calculate the voltage generated by coupling a copper and a zinc half cell to one another, we need to get the standard potentials from a table:

Redox couple	E⁰ (V)
Zn²⁺ + 2e⁻ → Zn	−0.76
Cu²⁺ + 2e⁻ → Cu	+0.34

The reaction clearly favours the reduction of copper ions by the oxidation of zinc metal. Hence the voltage generated by the two standard half-cells above would be:

E⁰_cell = E⁰_Cu − E⁰_Zn = 0.34 − (−0.76) = +1.10 V

Note that the values in the table conventionally show the reduction reaction: this is the reason for the minus sign in the equation to calculate the overall cell potential: the reaction occuring in the zinc couple is the reverse of that in the table.

Electrochemical cells are formed by linking two half cells with one another, usually by platinum electrodes and/or a salt bridge.
In an fully aqueous half cell, a platinum electrode is used, dangling into a solution of e.g. 1 M NADH and 1 M NAD. Comparison to a hydrogen half-cell would measure the E₀ of the NADH/NAD redox couple.

The standard redox potential of two coupled half-cells generating a voltage E⁰ is related to the free energy change of the reaction, by the following equation:

∆G⁰ = − z F E⁰

Where F is Faraday's constant and z is the number of electrons transferred. Note that any couple with a positive value will therefore be exergonic (spontaneous), as it will have a negative ∆G⁰. We saw this above: the reduction of copper ions to copper metal has a E⁰ of +0.34 V, and is spontaneous.

These series have relevance to electron transport chains: the NAD/NADH couple is more negative (−0.32 V) than that of the cyt-c_ox/cyt-c_red couple (+0.22 V), hence electrons will flow spontaneously from the former to the latter in the respiratory chain of mitochondria. Note that just like conventional reactions, we rarely have a cell that contains 1 M NADH or 1 M cytochrome-c. Hence, we use a very similar equation to that used to calculate ∆G from ∆G⁰, to calculate the voltage, E generated by a non-standard half-cell:

E = E⁰ − ( R T ⁄ z F ) ln ( [reduced form] ⁄ [oxidised form] )

The equation above is one form of the famous Nernst equation. Note the negative sign: if the reduced (electron rich) form is at a high concentration, it will be difficult to reduce any more of the oxidised form, hence the voltage generated by the half-cell will be lower. Here are some examples:

E = E⁰ − ( R T ⁄ z F ) ln ( [NADH] ⁄ [NAD] )

E = E⁰ − ( R T ⁄ z F ) ln ( [cyt-c_red] ⁄ [cyt-c_ox] )

If we couple these two as non-standard half-cells:

NADH (aq) | NAD⁺ (aq) ¦ cyt-c_ox (aq) | cyt-c_red (aq)

we use the equation from above:

E⁰ = E⁰_{reduction
process} − E⁰_{oxidation process}

And substitute in the two definitions given above to yield:

E = (E⁰_cyt-c − RT ⁄ zF ln [cyt-c_red] ⁄ [cyt-c_ox]) − (E⁰_NADH − RT ⁄ zF ln [NADH] ⁄ [NAD] )

E = (E⁰_cyt-c − E⁰_NADH) − RT ⁄ zF ( ln [cyt-c_red] ⁄ [cyt-c_ox] − ln [NADH] ⁄ [NAD] )

E = (E⁰_cyt-c − E⁰_NADH) − RT ⁄ zF ( ln [cyt-c_red] [NAD] ⁄ [cyt-c_ox] [NADH] )

E = ∆E⁰ − RT ⁄ zF ln Q

Note the similarity between this equation and that for calculating the the free energy change under non-standard conditions.

Another application of the Nernst equation is in determining the electrical potential of cell membranes. If we consider a strong solution of potassium chloride, we know that the solution will contain potassium ions and chloride ions. If we separate this solution from one of weak potassium chloride with a membrane, the second law of thermodynamics tells us that the ions will try to flow across the membrane into the more dilute solution.

However, an interesting thing happens if we make the membrane freely permeable only to the potassium ions but not to the chloride ions. In this case, the potassium ions will attempt to flow across the membrane, but as soon as any of them have done, they will be attracted to the negatively charged chloride ions left behind. This means that the flow will stop after only a tiny number of ions have crossed the membrane. This tiny separation of charge will generate an appreciable voltage across the membrane, which is unsurprisingly called the membrane potential. Note that the potential is caused only by those ions associated very closely with the membrane: the potassium ions that have crossed the membrane are so strongly attracted to the chloride ions left behind, that the separation of charge only occurs over just a few tens of nanometres.

If a membrane is only permeable to one ion (here the K⁺ cations), then the membrane potential can be calculated using another form of the Nernst equation, which we will derive now. For any ion crossing a membrane, across which there is an electrical gradient, from the 'outside' to the 'inside', there are two forces in action: the concentration gradient, and the electrical potential.

Firstly, the concentration (or 'chemical') effect: the potassium ions will generally distribute themselves so that their concentrations on either side of the membrane are the same. From the definition of chemical potential:

G = G⁰ + R T ln [X]

we can see that the free chemical energy of a species is the sum of the free energy of the species under standard conditions, plus a factor that varies with the concentration of the species. For potassium ions, we can write:

G_in = G⁰ + R T ln [K⁺]_in

G_out = G⁰ + R T ln [K⁺]_out

The free energy associated with the potassium ions moving from the outside ('before') to the inside ('after') is the difference between these (after minus before, i.e. inside minus outside):

∆G = G⁰ + R T ln [K⁺]_in − ( G⁰ + R T ln [K⁺]_out )

Which simplifies to:

∆G = R T ln [K⁺]_in ⁄ [K⁺]_out

Note how similar this is to the equation governing the free energy change of a chemical reaction:

∆G = R T ln ( Q ⁄ K_eq )

which we derived above. As far as our potassium ions are concerned, K_eq is 1 (concentration of 'products', i.e. ions on the inside, will be equal to the concentration of 'reactants', i.e. ions on the outside, at equilibrium). If there are more ions outside than inside, then [K⁺]_in ⁄ [K⁺]_out will be less than one, and the logarithm will be negative, indicating that the movement will be exergonic (i.e. will occur spontaneously by diffusion). If there are more ions inside, the movement will require energy (active transport).

However, there is a second factor we need to consider. This is the electrical effect, i.e. the charge on the potassium ions. Positively charged ions will generally migrate to the negatively charged side of the membrane, and vice versa. This may significantly affect our naïve assumption that ions flow down their concentration gradients until they are equally distributed across the membrane. The component of ∆G associated with this can be derived by considering the electrical potential of the ions with respect to the ground.

G_in = z F E_in

G_out = z F E_out

In each case, the electrical potential, E, is measured with respect to the ground. F is Faraday's constant (the charge on a mole of protons). F = N_Ae = 96485.34 C mol⁻¹ (these units can alse be written equivalently as J V⁻¹ mol⁻¹). N_A is Avogadro's constant (6.02214199 × 10²³ mol⁻¹), and e is the charge on a proton (1.602176 × 10⁻¹⁹ C). It is often more convenient to use the value F = 96.48534 J mV⁻¹ mol⁻¹, as we usually work in millivolts rather than volts. z is the charge on the ion (for K⁺, this is +1).

The ∆G associated with these charged ions moving from the outside to the inside is the difference between these two values:

∆G = G_in − G_out

∆G = − z F ( E_in − E_out)

∆G = z F ∆E

∆E is defined as the difference between the electrical potentials of the two compartments. In biological systems, these compartments will be separated by a membrane, so ∆E is termed the membrane potential, and is often written as ∆E_m or ∆Ψ. The inside of the cell is generally negative compared to the outside. This means that ∆E is also generally negative (if the cell is −50 mV w.r.t. ground inside, and +100 w.r.t. ground on the outside, then the membrane potential is −50 − (+100) = −150 mV).

Note that if the ion has a positive charge like K⁺, and there is a negative membrane potential (one that is more negative on the inside), then ∆G for the movement of ions to the inside of the cell will be negative, which is exactly what we expect, since these cations will be attracted to the negative charge inside.

The combination of these (possibly opposing) 'forces' is:

∆G = R T ln [K⁺]_in ⁄ [K⁺]_out + z F ∆E

If we leave the potassium ions to reach equilibrium, then ∆G will equal 0, and therefore, by rearrangement, we get:

∆E = − ( R T ⁄ z F ) ln [K⁺]_in ⁄ [K⁺]_out

or equivalently:

∆E = ( R T ⁄ z F ) ln [K⁺]_out ⁄ [K⁺]_in

This describes the membrane potential we would expect due to a K⁺ at dynamic equilibrium (and obviously, if we replace K⁺ with any other ion, 'X', the membrane potential we would expect from X alone). If we actually measure the values of ∆E and K⁺ concentration and find they do not match these values, we should expect that there is some other process (generally membrane impermeability, etc.) at work too. The Nernst equation is often useful in biology, even though it is only really applicable in the case where a membrane is permeable to a single ion, and not to any other species. This is because the permeability of a biological membrane is often high to just a single ion (often just potassium or sodium), and consequently the membrane potential is dominated by the effect of just this one ion. In cases where this does not hold, the Goldman equation can be used instead. The 'P' values refer to the permeability of the membrane to the ion in question. Note that the Goldman equation collapses to the Nernst if the permeability to one ion species is 1, and to all other ions, 0:

∆E = ( R T ⁄ F ) ln ( P_K[K⁺]_out + P_Na[Na⁺]_out + P_Cl[Cl⁻]_in)/ ( P_K[K⁺]_in + P_Na[Na⁺]_in + P_Cl[Cl⁻]_out)

The simpler Nernst equation may be looked at in two ways. Firstly, it tells you that if you measure the concentrations of a single ion across a membrane, and the potential across that membrane, you know that the ion is being actively excluded or pumped by the membrane if the concentrations differ significantly from those predicted by the equation. Secondly, it tells you that if the membrane potential is very similar to that predicated by the equation for a particular ion, then the membrane is probably freely permeable to that ion, and to no others. This is seen in animal nerve cells, where at rest the cell is essentially freely permeable to potassium (and no other ions), and during the peak of an action potential, freely permeable to sodium (and to no other ion).

One final application of this is to calculate the 'proton motive force' (PMF), generated by organelles such as chloroplasts and mitochondria, which pump protons out of themselves using the energy of light or NADH. The proton motive force is merely the ∆G for protons, expressed in millivolts.

∆G = R T ln [H⁺]_in ⁄ [H⁺]_out + z F ∆E

We can simplify this is a number of ways. Firstly, z equals 1 (protons have a charge of +1), and we can divide through by F to convert the free energy value (in J mol⁻¹ to a voltage (usually written ∆p for the PMF) in J C⁻¹:

∆p = ( R T ⁄ F ) ln [H⁺]_in ⁄ [H⁺]_out + ∆E

Note that the PMF is simply the free energy expressed in volts (joules of energy per coloumb of charge) instead of the more usual joules of energy per mole of ions. All that is needed to convert between the two is Faraday's constant. This is very similar to the definition of water potential, which is the free energy expressed in units of pressure (joules per cubic metre of water), with the molar volume of water as the conversion factor.

From the definition of pH:

pH = − log₁₀ [H⁺]

and knowing that we can convert base-e logs to base-10 logs using the factor 2.3 (ln 10), we can simplify this to:

∆p = ∆E + ( 2.3 R T ⁄ F ) ( log₁₀ [H⁺]_in − log₁₀ [H⁺]_out)

∆p = ∆E + ( 2.3 R T ⁄ F ) ( − pH_in + pH_out)

∆p = ∆E − ( 2.3 R T ⁄ F ) ( pH_in − pH_out)

If we define ∆pH as pH_in − pH_out, we get:

∆p = ∆E − ( 2.3 R T ⁄ F ) ∆pH

This is often written as:

∆p = ∆E − 59 ∆pH

The factor 59 is simply ( 2.3 R T ⁄ F ), set up so that the value for the PMF will be in millivolts. Typical PMF values are about −200 mV.

You should also note that membrane potentials are usually negative (on the inside), and that ∆pH will be positive if protons are expelled to the outside (∆pH = pH_in − pH_out. Hence according to the calculation above, ∆p will be negative. However, you will often (usually) see ∆p quoted as a positive value (the negative of the value derived above). This is because the PMF is the free energy that would be released as protons diffuse down their electrochemical gradients. This has a negative ∆G, hence is a negative voltage. We can then use this energy to power endergonic processes with positive ∆G, hence positive voltage.

The two components of the PMF are expended by different processes. Positively charged ions can diffuse into a mitochondrion down their electrochemical gradients through ion channels. This diffusion relies only on the ∆E component. Negatively charged ions such as phosphate and pyruvate can be taken into the mitochondrion by symport with protons (in fact, they're generally taken up by antiport with hydroxide, but this is essentially equivalent). Electroneutral exchanges such as these can use both the ΔpH and ΔE components. Both components can also be used to generate ATP with the F_OF₁-ATPase.

A brief calculation. Mitochondria tend to expel protons. Typically, mitochondria have a membrane potential of −150 mV, and maintain a ∆pH of 0.5 (approximately 7.8 on the inside, and 7.3 on the outside), yielding a ∆pH component of −29.5 mV, for a total PMF of −179.5 mV.

If the PMF were used to symport pyruvate into the mitochondrion from a surrounding solution at 10 mM, what would the pyruvate concentration in the mitochondrion be? Using the equation:

∆G = F ∆p = − R T ln [Pyr]_in ⁄ [Pyr]_out

We fill in the numbers, and rearrange:

F = 96.48534 J mV⁻¹ (note that PMF is in mV, not V, so we need to be careful of the units of F).
R = 8.314472 J K⁻¹ mol⁻¹
∆p = −179.5 mV (note the minus sign in the equation: we are using the free energy given up by the pH gradient to power a process with a positive ∆G, so we need to reverse the sign of ∆p).
[Pyr]_out = 10 mM.

This calculations will show that [Pyr]_in will equilibrate to about 10.8 M. Note that this result is insanely high, and other processes would act to down-regulate such an enormous uptake.

Test yourself

"You can't win, you can't even break even, you can only lose". How is this a useful summary of the laws of thermodynamics?
Why are living things described as open systems, but usually modelled as closed systems?
The reaction between sodium bicarbonate (baking soda) and hydrochloric acid, to make sodium chloride, carbon dioxide and water, is spontaneous under standard conditions. However, the enthalpy change is +28.5 kJ mol⁻¹, indicating that the reaction needs to take up heat energy from the surroundings. How can this be the case?
What is ∆G⁰ at 298K for the Haber process reaction, given that ∆H⁰ = −91.8 kJ mol⁻¹ and ∆S⁰ = −199 J K⁻¹ mol⁻¹?
N₂ + 3H₂ ⇌ 2NH₃
∆G for the reaction diamond → graphite is about -3 kJ mol⁻¹, which indicates it should be spontaneous. Why do we not see diamonds turning into pencil leads then?
In animal nerve cells, a potassium/sodium pump maintains a gradient of sodium and potassium ions across the membrane. Typically, the concentration of sodium ions is 10 times higher on the outside of the cell than on the inside, and vice versa for the potassium ions. Other pumps equalise the long-range on the two sides of the membrane by pumping other ions, so that the cell membrane essentially delimits two compartments, the cytosolic one containing 150 mM KCl / 15 mM NaCl; the extracellular one containing 15 mM KCl / 150 mM NaCl. The membrane potential of a typical resting nerve cell is −70 mV (negative on the inside). Is this more consistent with the membrane also possessing ion channels that are free permeable to potassium or sodium ions?

Why do electrons flow from NADH to oxygen in the respiratory chain? Use the table to help you answer.

Reaction	E⁰′ (V)
NAD⁺ + 2e⁻ + H⁺ → NADH	-0.32
UQ + 2e⁻ + 2H⁺ → UQH₂	0.10
cyt_c³⁺ + e⁻ → cyt_c²⁺	0.22
Fe³⁺ + e⁻ → Fe²⁺	0.77
½O₂ + 2e⁻ + 2H⁺ → H₂O	0.82

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