Chemical Equilibria

you are watching: Chemical Equilibria here

Suppose you know that the following equilibrium is exothermic ( H
H2 (g) + Cl2 (g) 2 HCl (g) Keq = 4 x 1031 at 300 K

Since the process is exothermic, heat is released by the system to the surroundings, so one may include heat in the equilibrium as a product:

H2 (g) + Cl2 (g) 2 HCl (g) + heat

Suppose the temperature of the system is increased to 500 K. The stress is too much heat, so the equilibrium will shift to the left to remove the excess heat. In addition, the value of the equilibrium constant decreases to 4 x 1018 , indicating the equilibrium lies further to the left.

When writing the mass action expression of an equilibrium, only include the components in the equilibrium which are in the same phase. Never include solids. Let’s look at some equilibrium reactions involving more than one phase, and their mass action expressions.
CaCO3 (s) CaO(s) + CO2(g)

Keq = [CO2]

Notice that the solids were not included as part of the mass action expression.

HCl(aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g)

Notice that the equilibrium mass action expression may be expressed in two ways; Keq1 is expressed in terms of the concentrations of the aqueous species, while Keq2 is expressed in terms of the concentration of the gaseous species. These two constants will have different values. Which expression do you use? That depends on the information given in the problem. If the concentrations of the aqueous species are given, use the expression for Keq1 . If the concentration for CO2 is given, use the expression for Keq2 . Realistically, one uses the expression whose quantities are most easily measured experimentally. The main idea is not to mix phases, and that the solid is not included in either expression.

READ:  Keri-Anne Croce | Towson University

Quantitative Problems

Next, let’s consider some quantitative problems dealing with chemical equilibria. By examining the amounts of reactants and products at equilibrium, one may numerically see how an equilibrium favors either reactants or products.

Suppose we are given the following equilibrium at 500 K:
CO(g) + 2 H2 (g) CH3OH(g)

The equilibrium concentrations are: [CO] = 0.0911 M, [H2] = 0.0822 M, [CH3OH] = 0.00892 M, what is the value of the equilibrium constant? Does the equilibrium favor reactants or products?

First, we need to write the mass action expression:

Next, substitute the equilibrium concentrations into the mass action expression, and calculate for the equilibrium constant:

Since the value of the equilibrium constant is greater than one, (Keq >1), the equilibrium favors the products.

Another type of equilibrium problem deals with finding the equilibrium concentrations given the initial concentrations and the value of the equilibrium constant. Suppose you are given the following equilibrium:
CO(g) + H2 O(g) CO2 (g) + H2 (g) Keq = 23.2 at 600 K

If the initial amounts of CO and H2O were both 0.100 M, what will be the amounts of each reactant and product at equilibrium?

For this type of problem, it is convenient to set a table showing the initial conditions, the change that has to take place to establish equilibrium and the final equilibrium conditions. Let’s begin by showing the initial conditions:

READ:  Joel Moore | Towson University

Initially, 0.100 M CO and 0.100 M H2O are present. Equilibrium hasn’t been established yet, so the amounts of CO2 and H2 are assumed to be zero.

To establish equilibrium, some CO and H2O has to react, so we will call the amount of CO and H2O reacted x, and the same x amount of CO2 and H2 must form:

The amounts of reactants and products present at equilibrium will be the combination of the initial amounts and the change. Just add the quantities together:

Substitute the above algebraic quantities into the mass action expression:

Since the algebraic expression is a perfect square, begin solving for x by taking the square root of both sides of the equation:

Multiply both sides by the denominator, 0.100 – x:

The term, 0.100 – x, cancels out on the left hand side, and the term, 4.85 must be distributed through the term 0.100 – x on the right hand side:
x = 0.485 – 4.85x

Add 4.85x to both sides of the equation:
x + 4.85x = 0.485 – 4.85x + 4.85 x

Combining terms:
5.85x = 0.485

Solve for x by dividing both sides by 5.85:

Recall that x represents the equilibrium quantities of both H2 and CO2 . The equilibrium quantities of CO and H2O is given by:
0.100 – x = 0.100 – 0.0829 = 0.017 M = [CO] = [H2O]

READ:  20 Grammar Rules | Grammar

© Copyright, 2001, L. Ladon. Permission is granted to use and duplicate these materials for non-profit educational use, under the following conditions: No changes or modifications will be made without written permission from the author. Copyright registration marks and author acknowledgement must be retained intact.

View more information:

See more articles in category: Grammar

Leave a Reply

Your email address will not be published. Required fields are marked *

Back to top button