Chapter 18 Worksheet 3
G under non-standard conditions, relationship between Go and K
Consider the generalized reaction: aA + bB = cC + dD
Recall that if you mix A, B, C, and D together, you can decide which direction the reaction will proceed
by comparing the reaction quotient (Q) to the equilibrium constant (K) OR by determining the sign of
G.
Q
[C]c [D] d
[A] a [B] b
K
[C]ceq [D] deq
b
[A] aeq [B] eq
If Q < K or G < 0 then the forward reaction is spontaneous (forward rate > reverse rate).
If Q > K or G > 0 then the reverse reaction is spontaneous (forward rate < reverse rate).
If Q = K or G = 0 the reaction is at equilibrium (forward rate = reverse rate).
If the reaction is not at equilibrium, it continues until Q = K and G = 0.
Reactions proceed toward minimum free energy just as a ball rolling downhill proceeds toward
minimum potential energy.
Figure legend: The y-axis is G; the x-axis represents
the progress of the reaction.
The sign of G is given by the slope of the curve.
When the slope is negative, the forward reaction is
“downhill”. When the slope is positive, the reverse
reaction is “downhill”. At the minimum, G = 0 and
the reaction is at equilibrium.
This discussion suggests that there should be a mathematical relationship between
G and Q.
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The relationship between G and Q
G = Go + RTlnQ
(When using this equation, the unit on G must be J/mol or kJ/mol)
We will not derive this equation, but we can analyze it to see that it makes sense.
1. If a reaction is set up under standard conditions what is the value of Q and G (use the equation
above)?
Q = ____1 (by definition)___
G = ____Go (as expected!)___
2. As Q increases, what happens to the value of G? Is this as expected?
As Q increases, G becomes more positive as expected!
If Go < 0, then the reaction proceeds in the forward direction when Q = 1. However, if Q is
sufficiently large (Q > Keq), G becomes positive and the reaction proceeds in the reverse direction.
If Go > 0, then the reaction proceeds in the reverse direction when Q = 1. However, if Q is
sufficiently small (Q < Keq), G becomes negativee and the reaction proceeds in the forward direction.
(Remember that the log of a number less than one is negative!)
The relationship between Go and Keq and between G and Q
3. If a reaction is at equilibrium,
Q = ____Keq___
G = ___0____
4. Insert the values from question 3 into the equation for G above. Rearrange the equation to produce
an expression for Go. Rearrange this expression to solve for K.
Go = -RTlnKeq
Keq = e-Go/RT
5. Substitute your expression for Go into the original equation for G.
G = -RTlnKeq + RTlnQ = RTln(Q/Keq) (G tells you how far you are from equilibrium!)
If Q = K, what is the value of G? ____0 (as expected at equilibrium!)_____
If Q < K, what is the sign of G? ______negative (as expected!)___________
If Q > K, what is the sign of G? ______positive (as expected!)___________
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Experimental determination of Ho and So (expt. 12H)
Recall that:
G = H – TS
Therefore:
Go = Ho – TSo
From the previous page, there is a second way to calculate Go:
Go = -RTlnK
Equate these two expressions for Go and solve for lnK.
-RTlnK = Ho – TSo
 H o  1  S o
ln K 
 
R T 
R
This is the equation you used in experiment 12H!
A plot of lnK vs. 1/T produces a straight line with:
slope = –Ho/R
y-intercept = So/R
(This is called “C” in expt. 12H)
6. According to the equation above, how does K depend on temperature for an endothermic or
exothermic reaction?
For an endothermic reaction (Ho >0), K increases with temperature.
For an exothermic reaction (Ho <0), K decreases with temperature.
AS EXPECTED!
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Summary
G and Q provide the same information: How far the reaction is from equilibrium and which
direction the reaction must proceed in order to reach equilibrium. The further a reaction is from
equilibrium, the more work it can do.
Go and Keq provide the same information: The concentrations of reactants and products present
at equilibrium.
The second law of thermodynamics explains how Hrxn and Srxn determine the direction in which
a reaction will proceed.
Thermodynamics explains the relationship between temperature and Keq
Thermodynamics provides a way to use calorimetry to measure equilibrium constants.
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