Consider the Generalized Reaction: Aa + Bb = Cc + Dd

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.

c d c d [C] [D] [C]eq [D]eq Q  a b K  a b [A] [B] [A]eq [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.

1 The relationship between  G and Q: calculating  G under non-standard conditions

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 = ______

G = ______

2. As Q increases, what happens to the value of G? Is this as expected?

o The relationship between  G and Keq and between  G and Q

3. If a reaction is at equilibrium,

Q = ______

G = ______

4. Insert the values from question 3 into the equation for G above. Rearrange the equation to produce o an expression for G . Rearrange this expression to solve for Keq.

Go = ______

eq = ______

5. Substitute your expression for Go into the original equation for G.

G = ______

If Q = Keq, what is the value of G? ______

If Q < Keq, what is the sign of G? ______

If Q > Keq, what is the sign of G? ______

2 o o Relationship between  H ,  S , T, and Keq (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:

o G = -RTlnKeq

o Equate these two expressions for G and solve for lnKeq.

This is the equation used in experiment 12H!

A plot of lnKeq 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 Keq depend on temperature for an endothermic or exothermic reaction?

3 7. Consider the reaction 2NO2 (g) → N2O4 (g)

a. Is this reaction spontaneous under standard conditions at 298 K? At 350 K? Answer these

questions in two ways: determine the sign of G and compare Q to Keq.

4 b. Is the reaction spontaneous at 350 K if the reaction mixture consists of 1.60 atm of NO2 and 0.40 atm of N2O4? Answer in two ways as in part a.

5 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.

o G 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.