Thermodynamics and Equilibrium for AP Chemistry

Practice problems for these concepts can be found at:

• Thermodynamics Review Questions for AP Chemistry
• Thermodynamics Free-Response Questions for AP Chemistry

Thus far, we have considered only situations under standard conditions. But how do we cope with nonstandard conditions? The change in Gibbs free energy under nonstandard conditions is:

ΔG = ΔG° + RT ln Q = ΔG° + 2.303 log Q

Q is the activity quotient, products over reactants. This equation allows the calculation of ΔG in those situations in which the concentrations or pressures are not 1.

Using the previous concept, calculate ΔG for the following at 500. K:

Note that Q, when at equilibrium, becomes K. This equation gives us a way to calculate the equilibrium constant, K, from a knowledge of the standard Gibbs free energy of the reaction and the temperature.

If the system is at equilibrium, then ΔG = 0 and the equation above becomes:

ΔG° = –RT ln K = –2.303 RT log K

For example, calculate ΔG° for:

2O³(g) 3O²(g) K_p = 4.17 ×10¹⁴

Note:° = 298 K

Answer:

Practice problems for these concepts can be found at:

• Thermodynamics Review Questions for AP Chemistry
• Thermodynamics Free-Response Questions for AP Chemistry