1.1 What is Chemistry?

http://www.ck12.org Chapter 20. Entropy and Free Energy

FIGURE 20.4

and the desired products would be less favored. However, the industrial manufacture of ammonia uses this reaction
at temperatures of 400-450°C. Why are elevated temperatures used? Knowing whether reactants or products are
energetically favored is one important factor for optimizing production of a desired product, but it is not the only
relevant variable. As it turns out, this reaction is quite slow, so even if the products are heavily favored at equilibrium,
it takes a long time to reach that equilibrium. Increasing the temperature speeds up the reaction, and this advantage
makes up for the slightly less favorable equilibrium constant.

Example 20.5

The equilibrium constant for the following reaction is 1.6× 10 −^10 at 25°C:

AgCl(s)Ag+(aq) + Cl−(aq)

Calculate∆G°.

Answer:

Simply plug the relevant values into the following equation:

∆G◦=−RT ln Keq

∆G◦=−( 8 .314 J/K·mol)(298 K)ln( 1. 6 × 10 −^10 )

∆G◦=− 55 ,884 J/mol

Note that the answer will be in J/mol (not kJ/mol) if we cancel the units correctly. Rounding the correct number of
significant figures and converting to the usual units for free energy values,∆G° = 56 kJ/mol for this reaction. This is
a very large number, indicating that the reactants are highly favored, as would be expected for a relatively insoluble
ionic compound.

One type of equilibrium that we will be studying extensively in the following chapter is the ionization of an acid in
water:

HA(aq)+H 2 O(l)A−(aq)+H 3 O+(aq)

where HA is a generic acid, and A−is its conjugate base.