Untitled - Kelly Walsh High School

260 CHEMISTRY FOR THE UTTERLY CONFUSED
Don’t Forget!
The Gibbs free energy for the following process is 25.2 kJ/mol. Determine the
equilibrium constant for this process at 25°C:
I 2(g) K 2 I(g)
The following equation relates the Gibbs free energy to an equilibrium constant:
∆G RT ln K
We will need to rearrange this equation:
G
ln K
RT
We can now enter the given values into this relationship. We must not forget a
kJ/J conversion and a C/K conversion.
ln K
(25.2 kJ/mol)
(8.314 J/mol·K) (25 273)K a103 J
1kJ b
ln K 10.1712 (unrounded)
K e 10.1712 3.8256 10 5 3.82 10 5
A system, at 298 K, contains the following gases NO, O 2, and NO 2. These gases
are part of the following reaction:
2 NO(g) O 2(g) l 2 NO 2(g)
The concentrations of the gases are: NO 2.00 M, O 2 0.500 M, and NO 2
1.00 M. Determine G for the system.
Even though the system is at standard temperature, the system is not standard
because the concentrations are not all one. For this reason, we must use the
relationship:
G G° RT ln Q
Anytime one or more of the following conditions are not met, the system is not
standard, and this equation must be used. The conditions are 298 K, 1 M (for
any reactant or product), and 1 atm (for any reactant or product that is a gas).