Chemical Equilibrium in the Gas Phase

The concept of equilibrium is an extremely important one to master. It helps to explain the behavior of chemical reactions and buffer systems, and can be used to drive reactions that normally don't occur, or occur to only a small extent.

The following notes describe the characteristics of gaseous systems at equilibrium, but equilibrium in aqueous solutions and the liquid phase behave in essentially the same manner. There are a variety of example problems that are worked out, plus practice problems that you can do. You should work these practice problems out on paper, then type in your answer(s) for a particular problem and click on "Check Answer" to see if it is correct. A PowerPoint review can be downloaded to help you summarize all of the kinds of problems are in this competency.

I. General Facts

aA(g) + bB(g) <------> cC(g) + dD(g)

Kc = [C]c [D]d / [A]a [B]b

• Kc has a constant value at a given temperature, independent of original concentration, volume of container, or pressure.

• Kc depends upon the way the equation was written--given in terms of the forward ( ----> ) reaction.

• For reactions involving liquids and solids, as well as gases, the liquids and solids are not included in the equilibrium expression.
Examples

• 2SO2(g) + O2(g) <-----> 2SO3(g)

Kc = [SO3]2 /[SO2]2 [O2]

• SO2(g) + 1/2O2(g) <-----> SO3(g)

Kc = [SO3] / [SO2] [O2]1/2

• 4NH3(g) + 5O2(g) <-----> 4NO(g) + 6H2O(g)

Kc = [NO]4 [H2O]6 / [NH3]4 [O2]5

• Pb(NO3)2(s) <-----> PbO(g) + 2NO2(g) + 1/2O2(g)

Kc = [PbO] [NO2]2 [O2]1/2
Determining Kc:
The value of Kc can be determined in three different ways, depending on what data you have:

1. Given the concentration of products and reactants at equilibrium

a. Given the reaction:

H2(g) + CO2(g) <-----> H2O(g) + CO(g) at 99°C

with the following equilibrium concentrations:

[H2] = 0.61 M, [CO2] = 1.6 M, [H2O] = 1.1 M, [CO] = 1.4 M

Kc = [H2O][CO] / [H2][CO2] = [1.1M][1.4M] / [0.61M][1.6M]

b. At a certain temperature, the equilibrium mixture of PCl5 , PCl3 , and Cl2 has the following concentrations:

[PCl3] = 0.035 M, [PCl5] = 0.017 M, [Cl2] = 0.074 M

Calculate Kc for the reaction PCl3(g) + Cl2(g) <-----> PCl5(g)

Kc = [PCl5] / [PCl3][Cl2] = [ ? M] / [ ? M][ ? M]

NOTE: all answers to the interactive problems in this document that are less than one, need to have a zero in front of the decimal point.

 Kc =

c. Solid ammonium chloride decomposes to ammonia gas and hydrogen chloride gas. At a certain temperature the equilibrium mixture contained in a 5.00 L flask is analyzed and found to contain 0.243 mol of ammonia, 1.12 g of ammonium chloride, and 0.683 mol of hydrogen chloride gas. Calculate Kc for the reaction.

K c = [NH3][HCl] = [ ? M][ ? M] ( remember, solids are not included in gaseous equilibrium so the ammonium chloride is not a part of the equilibrium expression)

 Kc =

2. Given the initial concentration of one species and the equilibrium concentration of another.

a. Before equilibrium is reached, a one-liter flask contains only 0.350 mol of SO3(g) at 832 °C. What is the Kc for the reaction

2SO3(g) <-----> O2(g) + 2SO2(g)

if at equilibrium 0.093 mol of oxygen is present?

Make a table similar to this with the following column headings:  species, original conc. (M), change, and equilibrium conc (M). Then fill in all the data that is given in the problem, such as starting concentrations, any equilibrium concentrations, and the reacting species from the balanced equation. Include the coefficient as part of each species so you do not overlook molar rations when determining the change for each one. This helps immensely as you calculate the changes and equilibrium concentrations for all species that you need for the equilibrium expression.

species original conc. (M) change equilibrium conc (M)
12SO3 0.350-0.186 0.164
2O2 0 +0.093 0.093
32SO2 0 +0.1860.186

To solve for Kc you should first write the equilibrium expression, then substitute in the values for the equilibrium concentrations for each species. Be sure to pay attention to all exponents.

Kc = [O2][SO2]2 / [SO3]2 = (0.093)(0.186)2 / (0.164)2 = 0.12

b. At another temperature, we start with 0.500 M of pure SO3(g) and [SO2] = 0.350 M. Calculate Kc.

species original conc. (M) change equilibrium conc (M)
1
2
3
 Kc =

3. Given the initial concentration of one species and how much of that species was used.

a. Consider the reaction

H2(g) + I2(g) <-----> 2HI(g).

Suppose we start with 0.2000 mol H2(g) and 0.1000 mol I2(g) in a one-liter flask. When equilibrium is reached 48.0% of the H2(g) will have been consumed. Calculate Kc.

Solution

If 48% of the H2 has been used up, then (0.2000 mol)(0.48) or 0.0960 mol of H2 has been converted into HI. Since the mole ratio of I2 to H2 is 1:1, then this is also the amount of I2 that has been used up. This amount then, is the change in concentration for both the H2 and the I2 and is shown as a negative change. The change in concentration of the HI is therefore a +0.192 mol since there are 2 moles of HI produced for every mole of H2 used up.

species original conc. (M) change equilibrium conc (M)
1H2 0.2000 -0.0960 0.1040
2I2 0.1000 -0.0960 0.0040
32HI 0 +0.192 0.192

Kc = [HI]2/[H2][I2] = (0.192)2 / (0.1040)(0.0040) = 89

b. At another temperature, we start with 0.450 mol H2(g) and 0.350 mol I2(g) in a one-liter flask. When equilibrium has been reached we find that 30.0% of the I2(g) has reacted. Calculate Kc.

This time, 30.0% of 0.350 mol, or 0.105 mol of I2 has been converted into HI. This is the change in concentration, then, for both the H2 and the I2.

species original conc. (M) change equilibrium conc (M)
1
2
3

 Kc =

What Kc can tell you:
1. whether a reaction is likely to be feasible

a. a very small Kc means that a reaction is not likely to happen.

b. a very large Kc, the reaction is feasible.

2. the direction of the reaction

a. Original Concentration Quotient (OCQ)

Expression is same as for Kc except that in OCQ concentrations are in parentheses ( ), instead of brackets.

b. OCQ > Kc:
1. the denominator must become larger to reach equilbirium, therefore more reactants are produced
2. reaction proceeds <----

c. OCQ < Kc:

1. at equilibrium the denominator must become smaller, hence the reactants are used up
2. reaction goes ----->

Examples

a. Given the equation N2(g) + O2(g) ---> 2NO(g)       Kc = 6.2 x 10 -14 at 2000 °C

if we start with 0.0520 mol of N2(g), 0.0124 mol O2(g) and 0.0020 mol NO(g) in a five-liter (5.0 L) flask which direction will be favored?

You need to find the concentration in mol/liter, so each amount of gas should be divided by 5.0 L.

OCQ = (NO)2 / (N2)(O2) = (0.0020/5.0)2 / (0.052/5.0)(0.0124/5.0) = 0.0065

Since the value of the OCQ is larger than the Kc, the reaction will go in the direction that will increase the denominator, therefore it will go to the <---

b. At the same temperature, in which direction will the reaction proceed if we start with 0.00031 mol NO(g), 0.020 mol N2(g), and 0.060 mol O2(g) in a 2.0 L flask?

The reaction will go to the (1) --> or (2) <-- ?
 The reaction will go

3. The concentration of species at equilibrium.

a. Given Kc and [conc] of all species but one

1. For the reaction of N2O(g) + 1/2O2(g) <-----> 2NO(g)        Kc = 1.7 x 10 -13 at 25 °C

What is [NO] if [N2O] = 0.0035 M and [O2] 0.0027 M?

Kc = [NO]2 / [N2O][O2]1/2 = 1.7 x 10-13 = [NO]2 / (0.0035)(0.0027)1/2

[NO]2 = (1.7 x 10-13)(0.0035)(0.0027)1/2

Therefore, [NO] = 5.6 x 10-9 M

2. N2(g) + 3H2(g) <-----> 2NH3(g) Kc at 25° C = 5.2 x 10-5.

How many grams of ammonia are in a 10.0 L flask if [N2] = 2.00 M and [H2] = 0.80M?

First, calculate the equilibrium concentration of the ammonia: (don't forget to use the concentration in mol/L)

 [NH3] =

Then use this concentration in moles/liter and calculate the mass of the ammonia you would have in the 10.0 L container:

 grams of ammonia =

b. Given only the orginal conc of the species

1. H2(g) + I2(g) <-----> 2HI(g)     Kc = 50.2 at 445 °C.

If we start with 0.100 M H2(g), 0.100 M I2(g) and 0.050 M HI(g) what are equilibrium concentrations of each?

First, set up a table with original concentrations, change, and equilibrium concrations. This time, however, since the amount of change is not known, use "x" to represent the change as shown in the table below.

species original conc. (M) change equilibrium conc (M)
1H2 0.100 -x 0.100-x
2I2 0.100 -x 0.100-x
32HI 0.050 +2x 0.050+2x

Set up the problem as before:

Kc = [HI]2 / [H2][I2] = (0.050+2x)2 / (0.100-x)2 = 50.2

Take the square root of each side: (0.050+2x / (0.100-x) = 7.09
Now, combine terms, solve for "x", and then substitute the value for "x" and get the equilibrium concentration for each species.
0.050 + 2x = 0.709 - 7.09x
9.09x = 0.659
x = 0.0725

[H2] = 0.100 - 0.072 = 0.028 M
[I2] = 0.100 - 0.072 = 0.028 M
[HI] = 0.050 + 2(0.072) = 0.194 M

2. H2(g) + CO2(g) <-----> H2O(g) + CO(g)    Kc = 1.6 at 990 °C

What are equilibrium concentrations of all species, if we start with 0.250 M H2(g), 0.250 M CO2(g) and 0.100 M H2O(g)?

species original conc. (M) change equilibrium conc (M)
1

2

3

4

 [H2] = [CO2] = [H2O] = [CO] =

II. Effects of Changes on an Equilibrium System
LeChatelier's Principle

LeChatelier's Principle states that whenever a system at equilibrium is subjected to a stress, then the equilibrium will shift in a direction so as to releive that stress.

These stresses and their effects are summarized below:

A. Adding or subtracting a product or reactant

1. if species added is a solid or liquid, there is no effect on a gaseous equilibrium.

2. if species added or subtracted is a gas then:
b. subtracting a species shifts direction towards the species removed.

B. Change in volume

1. if volume is decreased, the reaction proceeds towards the side with least moles of gas.

2. if volume is increased, the reaction proceeds towards side with most moles of gas.

3. if in the balanced equation there are the same number of moles of gas on both sides, a volume change will not affect the equilibrium.

C. Change in pressure

1. increase in pressure shifts equilibrium in direction of decrease in the number of moles of gas

2. decrease in pressure shifts equilibrium in direction of increase in number of moles of gas.

3. if in the balanced equation there are the same number of moles of gas on both sides, a pressure change will not affect the equilibrium.

D. Change in temperature:

1. If forward reaction is endothermic:
a. increase in temperature causes Kc to become larger
b. decrease in temperature causes Kc to become smaller

2. If forward reaction is exothermic:

a. increase in temperature causes Kc to become smaller
b. decrease in temperature causes Kc to become larger

Examples: use (1) if rxn goes to the right, (2) to the left, and (3) if rxn is not affected,

1. Consider the reaction, 2SO2(g) + O2(g) <-----> 2SO3(g)   (DH < 0)

What is the effect of each change on the above equilibrium system?

Note: there are 3 moles of gas on the left and 2 moles on the right.

 1. Adding more O2(g) 2. Removing SO3(g) 3. Increasing the volume of the container 4. Pressure increase 5. Temperature increase

2. Now, consider the reaction: COCl2(g) <-----> CO(g) + Cl2(g)    (DH > 0)

Note: there is one mole of gas on the left and 2 moles on the right.

 1. Decrease CO 2. Increase Cl2 3. Decrease volume of the container 4. Decrease pressure on the container 5. Increase temperature

III. Kp
A. General Reaction:

aA(g) + bB(g) <-----> cC(g) + dD(g)

Kp = (PC)c(PD)d / (PA)a(PB)b

"P" is equilibrium partial pressure in atmospheres

B. Kp vs Kc

1. Concentration in M are used to determine Kc . Partial pressures in atmospheres are used to determine Kp .

2. Both are independent of:
a. starting amounts of reactants and products
b. volume of the container
c. total pressure

3. Both vary with temperature.

C. Related by: Kp = ( Kc)( 0.0821T)*Δng

*Δng = change in number of moles of gas (product - reactant)

a. Determine Kp for the reaction 2SO2(g) + O2(g) <-----> 2SO3(g) if Kc = 56 at 900 K

Δng is -1 because the number of moles of gas goes from 3 moles originally to 2 moles of product.

Kp = 56(0.0821 x 900)-1 = 56 / (0.0821 x 900) = 0.76

b. For the following reaction and conditions, 2NaHCO3(s) <-----> Na2CO3(s) + CO2(g) + H2O(g) at 100°C     (Kp = 0.23)

Kc is determined to be = ?
 Kc =

Review of equilibrium, Kc, OCQ, LeChatelier's Principle and Kp:
The equilibrium review consists of a PowerPoint presentation that has been saved to be viewed through a web browser. There are two versions, one that requires Netscape 4.6 or higher and Internet Explorer 5.0. The other can be used with any version of either browser.

Equilibrium Review: Netscape 4.6 and IE 5.0
Equilibrium Review: for any version