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So back to equilibrium, If K > 1, then the products have higher entropy. However, if K < 1, then the reactants have higher entropy. The system is going to go towards whatever side that will increase entropy, but there will be a limit, just like with spontaneity. In our clothes example, you would eventually run out of clothes to throw on your floor unless you started borrowing someone else's. This would be our "equilibrium" point.The free energy change of a chemical process under standard state conditions,Go, can be determined four different ways:
Using Free Energies of Formation to Determine Standard State Free Energy Changes
If we know the standard free energy changes of formation,Gof, of each species in a change we can determine the standard state free energy change,Go, for the change using the following equation:
Using Enthalpy Changes and Entropy Changes to Determine Standard State Free Energy Changes
If we know the enthalpy change,Ho, and the entropy change,So, for a chemical process, we can determine the standard state free energy change,Go, for the process using the following equation:
In this equation T is the temperature on the Kelvin scale. In introductory courses we make the assumption thatHo andSo, do not change as the temperature changes.
Using Equilibrium Constants to Determine Standard State Free Energy Changes
If we know the equilibrium constant, Keq, for a chemical change (or if we can determine the equilibrium constant), we can calculate the standard state free energy change,Go, for the reaction using the equation:
In this equation
- R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1.
- T is the temperature on the Kelvin scale.
- Keq is the equilibrium constant at the temperature T.
Using Cell Potentials to Determine Standard State Free Energy Changes
If we know the standard state cell potential, Eo, for an electrochemical cell (or if we can determine the standard state cell potential), we can calculate the standard state free energy change,Go, for the cell reaction using the equation:
If you look up or calculate the value of the standard free energy of a reaction, you will end up with units of kJ mol-1, but if you look at the units on the right-hand side of the equation, they include J - NOT kJ.
You must convert your standard free energy value into joules by multiplying the kJ value by 1000.
ln K
ln K (that is a letter L, not a letter I) is the natural logarithm of the equilibrium constant K. For the purposes of A level chemistry (or its equivalents), it doesn't matter in the least if you don't know what this means, but you must be able to convert it into a value for K.
How you do this will depend on your calculator. Once you have calculated a value for ln K, you just press the ex button. On my calculator, that is the same button as the ln function, but you have to press the shift key and then the ln button. If yours is different and it isn't obvious, read the instruction book!
Using the equation to work out values of K
Example 1
Suppose you have a fairly big negative value of ΔG° = -60.0 kJ mol-1.
The first thing you have to do is remember to convert it into J by multiplying by 1000, giving -60000 J mol-1.
And let's suppose that we are interested in the equilibrium constant for the reaction at 100°C - which is 373 K.
ΔG° = -RT ln K
-60000 = -8.314 x 373 x ln K
This gives ln K =19.35
Using the ex function on your calculator gives a value for K = 2.53 x 108.
That is a huge value for an equilibrium constant, and means that at equilibrium the reaction has almost gone to completion. In the equilibrium constant expression, there must be lots of products at the top and hardly any reactants at the bottom.
Example 2
Now let's repeat the same exercise with a fairly big positive value of ΔG° = +60.0 kJ mol-1.
And we will keep the same temperature as before - 373 K.
ΔG° = -RT ln K
+60000 = -8.314 x 373 x ln K
This gives ln K = -19.35
Using the ex function on your calculator gives a value for K = 3.96 x 10-9.
That is a tiny value for an equilibrium constant, and there has been virtually no reaction at all at equilibrium. In the equilibrium constant expression, there must be hardly any products at the top and lots of reactants at the bottom.
More examples
It is up to you now to play around with your own examples until you are confident of the mechanics of getting an answer.
What happens if you change the value of ΔG°? Try the calculations again with values closer to zero, positive and negative. What happens if you change the temperature?