Chem 110 

Summer 2006 
Lecture Notes::Lec 7_7 June 


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Let's look at examples of each method of calculating deltaG.
1. Let's look at another example of the first method, using the equation defining free energy: deltaG° = deltaH° TdeltaS°.
Example: Find the free energy of the reaction of sulfur dioxide and oxygen to give sulfur trioxide under standard conditions (25°C and 1 atm)
From Appendix 4 of your text we find:

We need to calculate deltaH° and deltaS° separately.
deltaH° = Sum(ndeltaH°_{f}_{(prod)})  Sum(ndeltaH°_{f(rct)})
deltaH° = 2 deltaH°_{f}_{(SO3(g))}  {2 deltaH°_{f}_{(SO2(g))} + deltaH°_{f}_{(O2(g))}}
deltaH° = 2(396 kJ/mol)  {2 (297 kJ/mol) + 0 kJ/mol)}
deltaH° = 198 kJ/mol.
deltaS° = Sum(nS°_{(prod)})  Sum(nS°_{(rct)})
deltaS° = 2 S°_{(SO3(g))}  {2 S°_{(SO2(g))} + S°_{(O2(g))}}
deltaS° = 2 (257 J/molK)  {2 (248 J/molK) + 205 J/molK}
deltaS° = 187 J/molK
deltaG° = deltaH° TdeltaS°
deltaG° = 198 kJ/mol  (298 K)(187 J/molK)
deltaG° = 142 x 10^{3}J/mol = 142 kJ/mol
2. Hess's Law method (deltaG is a state function and path independent):
Example:
From tabulated reactions we have:

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

We can add these reactions to give:

3. Use deltaG°_{f}: This is just like using deltaH°_{f}, and is obviously the easiest. However tabulations of deltaG°_{f} are often unavailable, so the other methods must be used.
Example. Consider the complete combustion of methanol:
CH_{3}OH_{(g)} + 3/2 O_{2(g)} CO_{2(g)} + 2 H_{2}O_{(g)} From a table:
deltaG°_{f} (kJ/mol) CH_{3}OH_{(g)} 163 O_{2(g)} 0 CO_{2(g)} 394 H_{2}O_{(g)} 229 We can now find deltaG°_{f} from the formula:
deltaG° = Sum(ndeltaG°_{f (prod)})  Sum(ndeltaG°_{f ( rct)})
deltaG° = deltaG°_{f(CO2(g))} + 2 deltaG°_{f(H2O(g))}  deltaG°_{f(CH3OH(g))}
deltaG° = 394 kJ/mol + 2 (229 kJ/mol)  (163 kJ/mol) = 689 kJ/mol
We've been looking at the calculation of deltaG under standard conditions. How does it change with deviations from standard conditions? We will look first at the change in deltaG with changes in pressure for an ideal gas because of its theoretical simplicity.
First recall the formula for free energy, deltaG = deltaH TdeltaS.
S_{large V} > S_{small}
since the particles are more spread out and more difficult to describe (positional entropy).
Similarly, we expect the entropy to be affected by pressure (positional entropy):
It turns out (the derivation is beyond our scope) that the free energy for an ideal gas under nonstandard 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.
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, deltaG = 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!
In Chem 109 we noted that a chemical reaction reaches equilibrium when the rates of the forward and reverse reactions are equal. This occurs in a a reaction with, for example a large deltaH, because although the % of particles with sufficient energy to go backwards is very small, there are huge numbers of them compared to the number of reactant molecules which readily go forward.
We also noted that a reaction at equilibrium has no tendency to change (appears to be stagnant) and is stable at the macroscopic level, even though it is very active at the microscopic level. From our new perspective of thermodynamics it is at its lowest free energy, deltaG = 0.
Let's look at why this occurs from the perspective of energy. Consider a favorable reaction, i.e. deltaG = (). Again, lets look at the components of deltaG, and follow the theoretical progress of the reaction from 100% reactants to 100% products. Thus:
Looking at deltaH first we can plot the progress of the reaction (Note that this is NOT the same as a reaction progress or reaction coordinate diagram  in that case we follow a single set of molecules through the reaction looking at the instantaneous energy of the molecules. In this case we are looking at the deltaG of the entire reaction mixture.):
Note that the relationship is a simple linear plot beginning at a high enthalpy for the reactants and going to a lower enthalpy for the products. The negative change indicates a favorable deltaH contribution to the deltaG of the reaction.
How about S? Let's assume a favorable deltaS for this reaction:
Note that S will generally not be linear, rather there will be a maximum value. This is expected since the entropy of a mixture of particles will be greater than the same number of identical particles.
As result of substituting the data of these two curves into deltaG = deltaH TdeltaS we will get a plot something like the one below, with a minimum value of deltaG occurring between the reactants and the products:
Finally, let's do a calculation of deltaG for a reaction which is not at equilibrium from values of the equilibrium constant, and the mass action expression, Q:
Example: Calculate deltaG given K and Q for the reaction of glucose and ATP to give glucose6phosphate and ADP. (This is the first reaction of Glycolysis, the major pathway in sugar metabolism, catalyzed by the enzyme hexokinase. It is frequently held at nonequilibrium values by keeping the enzyme largely turned off.)
 K' = 5000 [The prime indicates that the standard conditions for the reaction have been modified to pH = 7.0 instead of pH = 0) and 37°C (rather than 25°C).]
 Q = 0.04
First we can find deltaG°':
deltaG°' = RT ln K = (8.314 J/molK)(310 K) ln 5000
deltaG°' =  2.195 x 10^{4}J/mol We can now find deltaG:
deltaG = deltaG°' + RT ln Q
deltaG =  2.195 x 10^{4}J/mol + (8.314 J/molK)(310 K) ln 0.04
deltaG =  3.025 x 10^{4}J/mol =  3 x 10^{4}J/mol
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© R A Paselk
Last modified 7 June2006