ELECTROCHEMISTRY Electrical and Chemical Energy Interconversion.
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THERMODYNAMICS- an area of physical chemistry • the interconversion of different forms of energy• the determination of the direction of spontaneous processes• consists of 3 fundamental laws using mathematical functions to describe the systems
bioenergetics- subsection within thermodynamics; providing a quantitative analysis of how organisms process energy
Review of basic concepts:system- part of universe we are interested in, has defined boundaries; e.g. cell, petri dish, earthsurroundings- all else outside the system
Types of systems:1) isolated system: unable to exchange mass or energy with surroundings2) closed system: energy flow, no mass flow with surroundings3) open system: can exchange both mass and energy
Internal Energy - represented by E - any system contains a certain amount of energy- kinetic and potential energy (all non-nuclear forms of E)- at atomic and/or molecular level:
- translational energy = mv2/2 = (3/2)RT- rotational - quantized E states > translational E- vibrational - quantized E states > rotational E- electronic - largest spacings between quantized
states (E in chemical bonds)- energy in noncovalent interactions; inter & intra
molecular as discussed in Ch 1
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Internal Energy - is a function of the state of a system
- the value of a “State Function” does not depend on the path of a process or on how a system arrived at that state, but depends only on the state of the system
e.g. Efinal – Einitial = E is independent of path
- the thermodynamic state of a system can be defined by the amount of all substances in the system and 2 out of 3 of the following variables: T, P, V (these are also state variables)
There are two main ways a system can exchange energy with its surroundings:• transfer of heat (q):
• (+) heat absorbed by the system; increase in E• (–) heat flows from the system; decrease in E
• work (w) done by or on the system• (+) work done by system on surroundings; decrease in E• (–) surroundings do work on the system; increase in E
• work can take many forms- • PV work- e.g. expansion of lungs against external P
work = PV •electrical work - pumping ions across a membrane
q and w are not state variables- can be thought of as energy in transit- we speak of the energy of a system, but not of the heat or work of a system
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1st Law of Thermodynamics - describes the “bookkeeping” involved when a system exchanges energy
E = q – w - conservation of energy- E can be gained and spent by a system but not destroyed
- E depends only on the initial and final states of a system- in contrast, the amount of q and w exchanged depends on the path taken to get from initial to final state, depends on conditions
e.g. of E in presence and absence of catalyst
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see e.g. from book p.59 - palmitic acid oxidation
under 2 different sets of conditions:1) reax carried out in bomb calorimeter immersed in water bath at constant volume - no PV work done -ignite mixture
-measure heat transferred from system to water bath as increase in T- we know mass of water and heat capacity of water
E = qv = –9941.4 kJ/mol (unit:Joules;1 cal = 4.184 J)
- we are measuring energy released (–) that was stored in the chemical bonds of the system
2) oxidation of palmitic acid under conditions of constant P - run at P = 1 atm- from balanced equation can see decrease in number of
moles of gas: ( n = 16final – 23initial)- measure decrease in V proportional to a decrease in the number of moles of gas present- decrease in V = work done on the system (-)
w = P V- assume initial and final temperatures are 25°C (298K)and the gases are ideal so we can use PV = nRT
V = n (RT/P) R = 8.314 J/K•moland
w = – 17.3 kJ/K•mol
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now consider the q evolved for constant P conditions:
-q is not a state function and depends on conditions
- from constant V conditions: E = –9941.4 kJ/mol- E is a state function; the sane for both sets of conditions - use 1st Law;
E = q – w = –9941.4
q = E + w = –9941.4 + (–17.3) = –9958.7 kJ/mol
- more heat is evolved under 2nd set of conditions (constant P);(in the 2nd case, the PV work done on the ends up as extra heat released by the system)
Essentially all biochemical processes take place under constant P conditions, (P = 1 atm )
Thus, another state function has been defined to describe the heat evolved under constant P conditions:
Enthalpy - represented by H; function of state H = E + PV = qp
H = E + PV
- the value of H is the same as the heat of a reaction measured at constant pressure
- the differences between H & E in biological systems are very small (because most reax are carried out in solution and amount
of PV work done is small relative to H & E)
as a result, H is often referred to as the energy change
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2nd Law of Thermodynamics -
- In addition to E and H, there are other factors which appear to be fundamental to biological processes
- consider the criterion for a favorable process. It was thought that
reactions with H < 0 used to all be spontaneous - there are many counterexamples.
- many processes without energy changes always go in a specific direction
e.g. 2 beakers with water, one at 75°, one at 25°, when brought into thermal contact, both end up at 50°. The reverse never happens.
-- Nothing in the 1st Law prevents the reverse from happening.
Two tendencies are at work in the universe:1) tendency towards lowest energy state2) tendency towards randomness or disorder
2nd tendency is described by the state function Entropy (S)
- we also must consider whether or not a process is reversible- the reversibility of a process depends on the given conditions
- reversible processes occur when conditions are close to the equilibrium state for the system- irreversible processes occur when conditions are far from equilibrium, then the process drives the system towards equilibrium
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Different terms are used- spontaneous vs favored direction of a reaction
- spontaneous implies “rapid” rate, but thermodynamics does not tell us anything about the rate of a reaction, only which direction is favored
Entropy (S) - is a measure of the degree of randomness of a system
- consider a system in a given thermodynamic state- this state may have many substates of equal energy
e.g. molecules making up a system with a particular amount of energy can be arranged in different ways (substates)
S = k ln W where k = Boltzmann’s constantk = R/ Avogadro’s numberW = number of substates
so, the molar entropy for a system is:S (molar) = R ln W
- as number of substates (W) increases, so does S
in addition:S ordered state < S disordered state
low S high S ice at 0°C water at 0°C protein in native, folded state protein unfolded(due to the number of substates that have equal energy)
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There are several ways to state the 2nd Law:1) Isolated systems tend to a state of maximum entropy2) The efficiency of a real process (fraction of heat that can be converted to work) < 1. There are always frictional losses.3) For an isolated system, an increase in entropy is the driving force towards equilibrium.
3rd Law of Thermodynamics creates a scale for entropies.Whereas the scale for energy is arbitrary, the scale for entropy is not.
S = 0 only for a perfect crystal at 0 K (-273°C)
Biochemists never deal with isolated systems. In open or closed systems, there are 2 opposing tendencies. Systems tend to - low energy and
- high entropy, and the physical state that dominates depends on T.
To encompass both factors, Gibb’s free energy (G) is defined
G = H – T S G is a state functionT is in K
At constant T, G = H – T S
Now we have a criterion for favorable (spontaneous) processes that encompasses both governing factors:
(G) T,P < 0 exergonic
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Overall, consider:
favorable process; (G) T,P < 0; exergonic
at equilibrium; G = 0; reversible process
reverse of reax is favorable; G T,P > 0; endergonic
There is a balance between S and H. Favorable processes typically occur when there is a decrease in energy (–H) and /or an increase in entropy (+ S).
At low T:TS term is small; solid is the stable state.
At high T:TS dominates, the gaseous state is stable.
(your book has some good examples of the interplay of H and S)
2 points that can be confusing:
1) whether or not a process is favorable has nothing to do with the rate; a process may have a large - G and be very slow
2) S of an open system can decreasethis often happens is biological systems (food -- becomes complex, ordered proteinswhich is accompanied by a large decrease in S)
however, energy is expended in the process to pay for this increase in organization; living organisms spend energy to overcome the decrease in entropy
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G called “free” energy because it represents the portion of H that is available to do useful work (beyond PV work)- the amount of E in TS is not available
in other words, we measure the efficiency of a process as:
work accomplished / maximum work that can be done by G
-the sign of G tells us whether a process or its reverse is favorable, and - the magnitude tells us how far a process is from equilibrium
-we want to express these ideas quantitatively for solution thermodynamics; to do so, we need to define a relationship between concentration and G:
Chemical Potential: partial molar free energy-represents the contribution per mole of a particular component to the total free energy of a system
-so for a mixture containing a moles of component A and b moles of component B, etc.
G = a GA + b GB + c GC
where GA or a represents the chemical potential per mole of A
-for dilute solution, chemical potentials depend only on the concentration of the substance but at higher concentrations we need to consider the activity of the substance
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activity- represented by a; a dimensionless quantity that measures the effective concentration of a substance
component A’s contribution to the free E of the system can be described by:
GA = G°A + RT ln aA T (K)R = 8.314 J / K mol
for dilute solutions: aA ~ [A]
[ ] represents molar concentrationmoles / liter
and:GA = G°A + RT ln [A]
when the concentration of substance A is 1 molar then:
GA = G°A
so you can see that:G°A represents a reference or standard state defined as:
- 1M concentration for each solute in solution and as - pure solvent for the solvent
- chemical potential is always expressed with respect to a standard state
-chemical potential is the driving force to equilibrium
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- chemical reactions have certain enthalpy and free energy changes associated with them- in combining chemical reactions and their enthalpies or free energies, we treat the products and reactants as algebraic quantities, for example
aA + bB cC + dD
H = Hproducts) - H (reactants)
H = cHc + dHd – aHa – bHb
H = enthalpy / mole
- we are concerned with H differences, the scale is arbitrary and we need to define one point- by convention a value has been assigned to enthalpies for elements in their most stable forms at 1 atm pressure- this condition is referred to as the standard state and
H° f = 0 for elements in the standard state(also referred to as heat of formation)
- then the enthalpies of compounds are related to this arbitrarily set zero value- the standard enthalpy / mole of a compound equals the enthalpy of formation of 1 mole of the compound at 1 atm from its elements in their standard states- many standard enthalpies / mole of a compound
H° (compound) have been tabulated at 25 C
- the same convention exists for standard free energies of elements and compounds in their standard states
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- now apply this algebraic treatment to chemical potentials to quantitatively describe the free energy changes accompanying chemical reactions- this enables us to predict the favored direction of a reaction under certain conditions
aA + bB cC + dD
G = Gproducts) - G (reactants)
write the free energies in terms of chemical potentials:
G = cGc + dGd – aGa – bGb
the driving force for the reaction is the total free energy of the products minus that of the reactants
- in order to calculate G, we substitute using:
GA = G° A + RT ln[A](and similarly for each component)
- we group terms and rearrange to yield:
G = G° + RT (ln[C]c + ln [D]d - ln [A]a - ln [B]b)
G° represents the standard state free energy change in the reaction which would be observed if a moles of A and b moles of B, at 1 molar concentration, formed c moles of C and d moles of D, each at 1 M.
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We can combine these terms one step further:
G = G° + RT ln ( [C]c [D]d / [A]a [B]b )
in general, this can be written as:
G = G° + RT ln ( [products] / [reactants] )
remember, this is with each concentration raised to the power equal to its stoichiometry in the reaction
- so when all concentrations are equal to 1 M:
G = G°
- suppose the reaction has come to equilibrium, then 2 things must be true:
1) the concentration terms must be at equilibrium concentrations, where the equilibrium constant (K),
K = ([products] / [reactants])(also raised to stoichiometric powers)
2) at equilibrium, G = 0
so, nowG = 0 = G° + RT ln K
or: - G° = RT ln K
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At equilibrium:– G° = RT ln K
can also be stated as:
K = e – G° / RT
-this last equation allows us to use tabulated standard state free energy changes to predict equilibrium constants
G° can be used as a reference value to compare intrinsic G° values for different reactions under equivalent conditions
BUT- it is G (in vivo conditions) that tells us if a reaction goes forward not G° (under standard state conditions)
Consider the following important biochemical reaction:
the isomerization of glucose-6-phosphate into fructose-6-phosphate (second step in glycolytic pathway)
G6P F6P G° = + 1.7 kJ/mol
- endergonic under standard condition- reverse reaction is favored
at equilibrium: [G6P] > [F6P]
- let’s express this quantitatively
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G6P F6P G° = + 1.7 kJ/mol
K = e –G° / RT at 25°C
K = e – (1700 J/mol) / (8.314 J/K•mol)(298K)
K = 0.504 = ([F6P] / [G6P])equilibrium conc
you can see K < 1, equilibrium lies to the left
let’s represent concentrations as fractions of total material
K =[F6P] / total concentration (f F6P) eq ---------------------------------- = --------------[G6P] / total concentration (f G6P) eq
Now, replace (f G6P) = 1– (f F6P) and express in terms of one component:
K = (f F6P) / 1 – (f F6P) = 0.504
solve for (f F6P): (f F6P) eq = 0.335
at equilibrium: 33.5% sugar exists as F6P66.5% as G6P
we are now able to calculate G at any concentration using:
G = G° + RT ln {(f F6P) / 1 – (f F6P)}
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Standard States: in biochemistry the standard state usually includes one additional condition: pH = 7; [H+] = 10-7 (which is much less than [H+] = 1M)
this is denoted by G°’ as opposed to just G° chemists’ standard state = G°biochemists standard state = G°’(the additional condition alters the value of K )
what is important is G which is independent of the standard state, this is what tells us if a reaction is favored
the difference in standard state becomes important in reactions where [H+] is produced or consumed by the reaction
consider the following where [H+] is produced (this is the case for hydrolysis reactions which are common in biochemistry):
A + B C + xH+
chemists’ standard state: K = [C] [H+] x / [A] [B]
G° for all [ ] = 1M
biochemist has G° ‘ for [A], [B], [C] = 1 M and [H+] = 10 -7 M
to compensate for the difference in [H+] , K’ (biochemist) > K (chemist)
can show that K / K’ = 10 -7x; G° – G°’ = 39.9 kJ/mole of H+
----low [H+] (pH~7) contributes enormously to the favorability of a reaction when a proton is being produced
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Coupling of reactions and spontaneity:-when looked at alone, many biochemical reactions do not have favorableG’ values (transport against a concentration gradient)
- these intrinsically unfavorable reactions are made thermodynamically favorable by coupling them to strongly favorable reactions
-as an example, one step on the glycolytic pathway;the conversion of glucose to glucose-6-phosphate (G6P) is unfavorable:
glucose G6P G°’ = + 3.2 kcal/mole
- this reaction won’t go under standard conditions- the reaction is driven by coupling it with ATP hydrolysis
ATP + H2O ADP + H+ + Pi G°’= –7.3
Overall,glucose + ATP + H2O G6P + ADP + H+ + Pi
G°’ = + 3.2 – 7.3 = – 4.1 kcal
(This reaction without the enzyme, hexokinase, is slow, in spite of –G°’. This enzyme aids in providing the means for the coupling to take place.)
Coupling is very common in biochemical processes. Cells contain a number of compounds that undergo reactions with large –G’s and often enzymes facilitate the coupling of these reactions with unfavorable reactions. Enzymes also accelerate reactions.
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Factors that contribute to large free energy changes for high energy phosphate compounds (see fig 3.7 in text):
1) resonance stabilization or tautomerization of product molecules:eg. phosphate products
Pi HPO4 2–
orthophosphate, inorganic phosphate
- resonance forms of equal energy (substates) contribute to the high entropy of the product, Pi
- so release of Pi results in an increase in entropy
2) additional hydration of charged hydrolysis products -recall hydration of ions in aqueous solution is favored
3) electrostatic repulsion between charged products - reaction is favored when repulsion between 2 anionic products takes place
4) release of a proton in buffered solution-we just saw that when [H+] is produced under conditions where [H+] is kept low (pH= 7);G°’ < G° by 39.9 kJ/mole- your text shows you how to arrive at this number from
G°’ = G° + RT ln (10 -7)