Give the relation between standard free energy change and equilibrium constant.
Answers
Because ΔH° and ΔS° determine the magnitude of ΔG° and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. "Free Energy", ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. The general relationship can be shown as follow (derivation not shown):
ΔG=VΔP−SΔT(19.7.1)
If a reaction is carried out at constant temperature (ΔT = 0), then Equation 19.7.1 simplifies to
ΔG=VΔP(19.7.2)
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the V in Equation 19.7.2 by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express ΔG in terms of the initial and final pressures ( Pi and Pf , respectively):
ΔG=(nRTP)ΔP=nRTΔPP=nRTln(PfPi)(19.7.3)
If the initial state is the standard state with Pi = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows:
G−G∘=nRTlnP(19.7.4)
This can be rearranged as follows:
G=G∘+nRTlnP(19.7.5)
As you will soon discover, Equation 19.7.5 allows us to relate ΔG° and Kp. Any relationship that is true for Kp must also be true for K because Kp and K are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
aA+bB⇌cC+dD(19.7.6)
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG:
ΔG=∑mGproducts−∑nGreactants=(cGC+dGD)−(aGA+bGB)(19.7.7)
Substituting Equation 19.7.5 for each term into Equation 19.7.7 ,
ΔG=[(cGoC+cRTlnPC)+(dGoD+dRTlnPD)]−[(aGoA+aRTlnPA)+(bGoB+bRTlnPB)](19.7.8)
Combining terms gives the following relationship between ΔG and the reaction quotient Q:
ΔG=ΔG∘+RTln(PcCPdDPaAPbB)=ΔG∘+RTlnQ(19.7.9)
where ΔG° indicates that all reactants and products are in their standard states. For gases at equilibrium ( Q=Kp ,), and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and Kp for gases as follows:
0=ΔG°+RTlnKp(19.7.10)
ΔG°=−RTlnKp(19.7.11)
If the products and reactants are in their standard states and ΔG° < 0, then Kp > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then Kp < 1, and reactants are favored over products. If ΔG° = 0, then Kp=1 , and neither reactants nor products are favored: the system is at equilibrium.
For a spontaneous process under standard conditions, Keq and Kp are greater than 1.
This relationship allows us to directly relate the standard free energy change to the equilibrium constant. ... If delta G standard is between -20 kJ and +20 kJ, then there is an equilibrium, a mixture of both reactants and products. And when delta G = 0, K = 1 and there are equal amounts of both.
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