20.8: Calculations of Free Energy and Keq
What are these formations called when they point down?
Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and redeposits on the rock as the carbon dioxide is dissipated into the environment.
Equilibrium Constant and
At equilibrium the for a reversible reaction is equal to zero. relates the concentrations of all substances in the reaction at equilibrium. Therefore we can write (through a more advanced treatment of thermodynamics) the following equation:
The variable is the ideal gas constant (8.314 J/K • mol), is the Kelvin temperature, and is the natural logarithm of the equilibrium constant.
When is large, the products of the reaction are favored and the negative sign in the equation means that the is negative. When is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative and so the sign of is positive. The Table below summarizes the relationship of to :
Description | |||
>1 | positive | negative | Products are favored at equilibrium. |
1 | 0 | 0 | Reactants and products are equally favored. |
<1 | negative | positive | Reactants are favored at equilibrium. |
Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.
Sample Problem: Gibbs Free Energy and the Equilibrium Constant
The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at 25°C.
The actual concentrations of each gas would be difficult to measure, and so the _{ } for the reaction can more easily calculated from the , which is equal to 173.4 kJ/mol.
Step 1: List the known values and plan the problem.
Known
Unknown
In order to make the units agree, the value of will need to be converted to J/mol (173,400 J/mol). To solve for , the inverse of the natural logarithm, , will be used.
Step 2: Solve .
Step 3: Think about your result.
The large positive free energy change leads to a _{ } value that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.
Sample Problem: Free Energy from
The solubility product constant of lead(II) iodide is 1.4 × 10 ^{ -8 } at 25°C. Calculate for the dissociation of lead(II) iodide in water.
Step 1: List the known values and plan the problem .
Known
Unknown
The equation relating to _{ } can be solved directly.
Step 2: Solve.
Step 3: Think about your result.
The large, positive indicates that the solid lead(II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.
Summary
- The relationship between and _{ } is described.
- Calculations involving these two parameters are shown.
Practice
Questions
Read the material at the link below and answer the following questions:
http://www.chem1.com/acad/webtext/thermeq/TE5.html
- What is the difference between and ?
- At equilibrium, why does the equation between free energy and equilibrium constant reduce to ?
- What other equilibrium units could we use?
Review
Questions
- When _{ } is large, what will be the sign of ?
- When _{ } is small, are reactants or products favored?
- What does stand for?
Image Attributions
Description
Learning Objectives
- Describe the relationship between and .
- Perform calculations involving these two parameters.
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Date Created:
Jun 25, 2013Last Modified:
Jan 12, 2015Vocabulary
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