Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Fall 2003
Lecture Notes::Lec 10_17 September	© R. Paselk 2003
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DG Calculations, cont.

Last time we started our look at the various methods of calculating free energy with the "formula" method. Let's now look at teh next two methods.

2. Hess's Law method (DG is a state function and path independent):

Example:

From tabulated reactions we have:

1) 2 CH4(g) + 3 O2(g) Æ 2 CO(g) + 4 H2O(g) DG° = -1088 kJ/mol 2) CH4(g) + 2 O2(g) Æ CO2(g) + 2 H2O(g) DG° = -801 kJ/mol

Reversing reaction (1) and multiplying reaction (2) by 2 to cancel methane:

2 CO(g) + 4 H2O(g) Æ 2 CH4(g) + 3 O2(g) DG° = +1088 kJ/mol 2 x (CH4(g) + 2 O2(g) Æ CO2(g) + 2 H2O(g)) DG° = -1602 kJ/mol

We can add these reactions to give:

2 CO(g) + O2(g) Æ 2 CO2(g)  DG° = -514 kJ/mol

3. Use DG°_f: This is just like using DH°_f, and is obviously the easiest. However tabulations of DG°_f are often unavailable, so the other methods must be used.

Example. Consider the complete combustion of methanol:

CH₃OH_(g) + 3/2 O_2(g) Æ CO_2(g) + 2 H₂O_(g)

From a table:

DG°f (kJ/mol) CH3OH(g) -163 O2(g) 0 CO2(g) -394 H2O(g) -229

We can now find DG°_f from the formula:

DG° = Sn_pDG°_f(prod) - Sn_rDG°_f(rct)

DG° = DG°_f(CO2(g)) + 2 DG°_f(H2O(g)) - DG°_f(CH3OH(g))

DG° = -394 kJ/mol + 2 (-229 kJ/mol) - (-163 kJ/mol) = -689 kJ/mol

 

The Pressure Dependence of DG

We've been looking at the calculation of DG under standard conditions. How does it change with deviations from standard conditions? We will look first at the change in DG with changes in pressure for an ideal gas because of its theoretical simplicity.

First recall the formula for free energy, DG = DH -TDS.

• For an ideal gas DH does not change with pressure. (Remember that the particles in an ideal gas are completely independent of one another, thus there is no heat gain or loss in the system when pressure changed since the interactions of the gas particles doe not change [no bond making or breaking, no attraction or repulsion changes].)
• What about DS? In this case we do expect a change with pressure. After all the entropy obviously changes with DV. For and ideal gas:

It turns out (the derivation is beyond our scope) that the free energy for an ideal gas under non-standard pressure is related to standard conditions by:

So how do we apply this to a reaction? Consider the synthesis of ammonia from nitrogen and hydrogen:

Recall that, in general:

and

Substituting:

Collecting terms:

Consolidating and taking advantage of log relationships:

Thus the expression for the reaction is more complex than for a simple gas, with the necessity to take differences in moles into account as well as pressure.

DG and Equilibria

You may have noticed in our treatment of pressure that the final expression included the expression for the Mass Action Quotient for ammonia synthesis:

So we can rewrite the general expression as:

This is a very general expression, which works for gases with pressures in the Mass Action Quotient, or any chemical system with moles (or activities) in the Mass Action Quotient.

Now suppose we look at a system at equilibrium, where Q = K_eq. Then our expression for free energy becomes:

Of course for a reaction at equilibrium, DG = 0, so we see that:

So we see that the standard free energy is directly related to the equilibrium constant for a chemical reaction. We can also see that the amount of energy available from a reaction (or which must be put in to make a reaction go) is related to how far the system is from equilibrium!

© R A Paselk

Last modified 17 September 2003