Chemical Equilibrium - Part 2: Free Energy - Biology

Chemical Equilibrium - Part 2: Free Energy - Biology

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In a previous section we began a description of chemical equilibrium in the context of forward and reverse rates. Three key ideas were presented:

1. At equilibrium the concentrations of reactants and products in a reversible reaction are not changing in time.

2. A reversible reaction at equilibrium is not static - reactants and products continue to interconvert at equilibrium, but the rates of the forward and reverse reactions are the same.

3. We were NOT going to fall into a common student trap of assuming that chemical equilibrium means that the concentrations of reactants and products are equal at equilibrium.

Here we extend our discussion and put the concept of equilibrium into the context of free energy, also reinforcing the Energy Story exercise of considering "Before/Start" and "After/End" states of a reaction (including the inherent passage of time).

Reaction coordinate diagram for a generic exergonic reversible reaction. Equations relating free energy and the equilibrium constant. R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1; T is temperature in Kelvin.

Attribution: Marc T. Facciotti

The figure above shows a commonly cited relationship between ∆G° and Keq: ∆G° = -RTlnKeq. Here, G° indicates the free energy under standard conditions (e.g 1 atmosphere of pressure, 298K). In the context of an Energy Story, this equation describes the change in free energy of a reaction whose starting condition is out of equilibrium, specifically all matter at the "start" is in the form of reactants and the "end" of the reaction is the equilibrium state. Implicit is the idea that the reaction can theoretically proceed to infinite time so that no matter the shape of its energy surface it can reach equilibrium. One can also consider a reaction where the "starting" state is somewhere between the starting state above and equilibrium and perhaps where the reaction is not at equilibrium. In this case one can examine the ∆G (not standard conditions) between the "intermediate" starting state and equilibrium by considering the equation ∆G = ∆G° + RTlnQ, where Q is called the reaction quotient. From the standpoint of Bis2a we will use a simple (a bit incomplete but functional) definition for Q = [Products]st/[Reactants]st at a defined non-equilibrium "starting" condition st. The equation ∆G = ∆G° + RTlnQ can therefore be read as the as the free energy of the transformation being equal to the free energy associated with the free energy difference for ideal standard condition plus the contribution of free energy that represents the deviation away from the "ideal" starting state represented by the actual starting state and conditions. In both cases, the "final" condition is still equilibrium we are just changing starting points. One can extend this idea and calculate the free energy difference between two non-equilibrium states profided they are properly defined, but that's for your Chemistry instructor to bother you with. The key point here is that there is a way to conceive of and compute free energy changes between specifically defined states and not just the standard initial state and equilibrium as the end state.

This takes us to the core summary point. In many Biology books, the discussion of equilibrium includes both the discussion of forward and reverse reaction rates but also a statement that ∆G = 0 at equilibrium. This often confuses some students because they are also taught that a non-zero ∆G can be associated with a reaction going to equilibrium. We do this each time we report the ∆G of a reaction or examine a reaction coordinate diagram. So, students tend to memorize the "∆G=0 at equilibrium" statement without appreciating where it comes from. The key to closing the apparent disconnect for many is to appreciate that the interpretation of the sometimes seemingly contradictory statements depend a lot on the definition of the starting and ending states used to calculate ∆G. In the case of reporting ∆G for a reaction the starting state was described in the paragraphs above (in one of two ways - either standard conditions or non-standard out-of-equilbrium state) and the ending state is some time later once the reaction has reached equilibrium. Since the starting and ending states are different ∆G can be non-zero positive or negative. By contrast, the statement that concluds "∆G=0 at equilibrium" is considering a different starting state. In this case the starting state is the system already at equilibrium. The ending state is considered to be sometime later, but still at equilibrium. Since the starting and ending states are ostensibly the same, ∆G = 0.

AP Chemistry Spontaneity, Entropy, & Free Energy, Part II

In order to determine whether a process will be spontaneous, both the entropy of the system and of the surroundings must be considered. The entropy of a system is usually a fixed amount and can be looked up in thermodynamic tables, so the ΔS across a reaction can be found by subtracting the entropy of the reactants from the entropy of the products. Endothermic processes have a negative entropy because heat is leaving the surroundings to enter the system. Exothermic processes have a positive entropy because they increase the entropy of the surroundings. The magnitude of ΔS depends on temperature (-ΔH/absolute temperature in Kelvin). This lecture explains the derivation of that equation as well as introducing the Gibbs Free Energy (ΔG=ΔH-TΔS).

AP Chemistry PowerPoint: Thermochemistry, Thermodynamics, and Gibbs Free Energy

This 76 slide PowerPoint presentation covers, in detail, all of the following AP chemistry concepts in Thermodynamics:

- Definition of thermochemistry

- Exothermic versus endothermic reactions

- Stoichiometry with enthalpy including problem calculations

- Calculating delta H using calorimetry including problem calculation

- Using heats of formation to determine delta H

- Definition of entropy and conceptual examples thereof

- Calculating delta G given delta H and delta S

- Predicting the spontaneity of a reaction

- Calculating delta G given Gibbs free energies of formation

- Predicting temperatures when a reaction shifts from spontaneous to non-spontaneous

- Relating Gibbs free energy to the equilibrium constant

- Calculating delta G given the equilibrium constant

- Calculating new values for delta G given non-equilibrium conditions or Q

Includes sample AP problems within the PowerPoint presentation worked out in detail!

Post Assessment Chemical Equilibrium Part 2 #29

Post by Kelly Kiremidjian 1C » Fri Nov 24, 2017 10:00 pm

Can anyone explain #29 from the Chemical Equilibrium Part 2 Post Assessment? Here it is

29. A researcher fills a 1.00 L reaction vessel with 1.84 x 10-4 mol of BrCl gas and heats it to 500 K. At equilibrium, only 18.3 % of the BrCl gas remains. Calculate the equilibrium constant, assuming the following reaction is taking place.
2BrCl(g) ⇌ Br2 (g) + Cl2(g)

Re: Post Assessment Chemical Equilibrium Part 2 #29

Post by Ammar Amjad 1L » Sat Nov 25, 2017 10:51 pm

So in this problem you're given the initial concentration of BrCl, 1.84x10-4 mol/L. The question also tells you that at equilibrium the molar concentration is 18.3% of the initial concentration, which would be 1.84x10^-4 mol/L times .183 which gives you 3.36x10^-5. Now that you have the initial and equilibrium concentrations of BrCl, you would set up the ICE table and solve for X to find the equilibrium constant.

2BrCl(g) ⇌ Br2 (g) + Cl2(g)
1.84x10^-4 0 0
-2x +x +x
3.36x10^-5 x x

Then you can find the value of x to find the equilibrium constant. X = 7.52x10^-5.

From there you can plug the values in, and you end up with the final answer as C.

Gibbs Free Energy And Keq

Remember from our look at the gibbs free energy that if the formation of products is favored then the dg is negative. Calculating k when you know the standard free energy of reaction.

Calculations Of Free Energy And Keq Ck 12 Foundation

I understand the reasoning behind deltah 0 deltas 0 and deltag0 but i dont understand how you can determine that keq is 1.

Gibbs free energy and keq. An interaction between two subunits of a protein was determed to have a delta g of 5705 kjmol. 071 free energy and the equilibrium constant in this video paul andersen explains how thermodynamic and equilibrium reasoning can be related through change. This means the equilibrium constant for transesterification is approximately 1.

Gibbs free energy calculating keq. If you look up or calculate the value of the standard free energy of a reaction you will end up with units of kj mol 1 but if you look at the units on the right hand side of the equation they include j not kj. Gibbs free energy and chemical equilibrium r how to predict chemical reactionso without doing experiments ocn 623 chemical oceanography reading.

The relationship between standard gibbs free energy change and the equilibrium constant k. Calculating k when you know the standard free energy of reaction. What is the keq for the reaction at 25 degrees celsius.

This implies that the change in free energy for transesterification is around 0. Calculating an equilibrium constant from the free energy change. On the other hand if the reaction is at equilibrium when there are still more reactants left over than products lies to the left then the concentrations of a and b will always be larger than the concentrations of c and d.

First half of chapter 3 snoeyink and jenkins 1980. Rearrangement gives in this equation. Ad and ae are the energy and entropy of the body in its initial state ab and ac its available energy gibbs free energy and its capacity for entropy the amount by which the entropy of the body can be increased without changing the energy of the body or increasing its volume respectively.

The relationship between standard gibbs free energy change and the equilibrium constant k. You must convert your standard free energy value into joules by multiplying the kj value by 1000. If we know the standard state free energy change g o for a chemical process at some temperature t we can calculate the equilibrium constant for the process at that temperature using the relationship between g o and k.

Gibbs free energy g is defined as. Gibbs free energy is a measure of the potential for reversible or maximum work that may be done by a system at constant temperature and pressure. It is a thermodynamic property that was defined in 1876 by josiah willard gibbs to predict whether a process will occur spontaneously at constant temperature and pressure.

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Step 1: Combine and balance the two half-reactions.

The oxidation half-reaction produces 2 electrons and the reduction half-reaction needs 6 electrons. To balance the charge, the oxidation reaction must be multiplied by a factor of 3.
3 SO2(g) + 6 H20(ℓ) → 3 SO4 - (aq) + 12 H + (aq) + 6 e -
+ Cr2O7 2- (aq) + 14 H + (aq) + 6 e - → 2 Cr 3+ (aq) + 7 H2O(ℓ)
3 SO2(g) + Cr2O7 2- (aq) + 2 H + (aq) → 3 SO4 - (aq) + 2 Cr 3+ (aq) + H2O(ℓ)
By balancing the equation, we now know the total number of electrons exchanged in the reaction. This reaction exchanged six electrons.

Step 2: Calculate the cell potential.
This electrochemical cell EMF example problem shows how to calculate cell potential of a cell from standard reduction potentials.**
cell = E°ox + E°red
cell = -0.20 V + 1.33 V
cell = +1.13 V

Step 3: Find the equilibrium constant, K.
When a reaction is at equilibrium, the change in free energy is equal to zero.

The change in free energy of an electrochemical cell is related to the cell potential of the equation:
ΔG = -nFEcell
ΔG is the free energy of the reaction
n is the number of moles of electrons exchanged in the reaction
F is Faraday's constant (96484.56 C/mol)
E is the cell potential.

The cell potential and free energy example shows how to calculate free energy of a redox reaction.
If ΔG = 0:, solve for Ecell
0 = -nFEcell
Ecell = 0 V
This means, at equilibrium, the potential of the cell is zero. The reaction progresses forward and backward at the same rate, meaning there is no net electron flow. With no electron flow, there is no current and the potential is equal to zero.
Now there is enough information known to use the Nernst equation to find the equilibrium constant.

The Nernst equation is:
Ecell = E°cell - (RT/nF) x log10Q
Ecell is the cell potential
cell refers to standard cell potential
R is the gas constant (8.3145 J/mol·K)
T is the absolute temperature
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant (96484.56 C/mol)
Q is the reaction quotient

**The Nernst equation example problem shows how to use the Nernst equation to calculate cell potential of a non-standard cell.**

At equilibrium, the reaction quotient Q is the equilibrium constant, K. This makes the equation:
Ecell = E°cell - (RT/nF) x log10K
From above, we know the following:
Ecell = 0 V
cell = +1.13 V
R = 8.3145 J/mol·K
T = 25 &degC = 298.15 K
F = 96484.56 C/mol
n = 6 (six electrons are transferred in the reaction)

Solve for K:
0 = 1.13 V - [(8.3145 J/mol·K x 298.15 K)/(6 x 96484.56 C/mol)]log10K
-1.13 V = - (0.004 V)log10K
log10K = 282.5
K = 10 282.5
K = 10 282.5 = 10 0.5 x 10 282
K = 3.16 x 10 282
The equilibrium constant of the cell's redox reaction is 3.16 x 10 282 .


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Watch the video: Equilibrium HL 2 - Position of Equilibrium and Gibbs Free Energy (July 2022).


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