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.

If Q < Kor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.

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 understandard 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!)______

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 Goand solve for lnK.

-RTlnK = Ho – TSo

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!

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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