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.

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General Facts

Determining Kc

What Kc can tell you

LeChatelier's Principle

Determining Kp

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I. General Facts

aA_(g) + bB_(g) <------> cC_(g) + dD_(g)

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

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

• K_c 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

• 2SO_2(g) + O_2(g) <-----> 2SO_3(g)
  K_c = [SO₃]² /[SO₂]² [O₂]

• SO_2(g) + 1/2O_2(g) <-----> SO_3(g)
  K_c = [SO₃] / [SO₂] [O₂]^1/2

• 4NH_3(g) + 5O_2(g) <-----> 4NO_(g) + 6H₂O_(g)
  K_c = [NO]⁴ [H₂O]⁶ / [NH₃]⁴ [O₂]⁵

• Pb(NO₃)_2(s) <-----> PbO_(g) + 2NO_2(g) + 1/2O_2(g)
  K_c = [PbO] [NO₂]² [O₂]^1/2

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Determining K_c:

The value of K_c 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:
H_2(g) + CO_2(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

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

ANSWER: K_c = 1.6

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 K_c for the reaction PCl_3(g) + Cl_2(g) <-----> PCl_5(g)

K_c = [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 =




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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 K_c for the reaction. NOTE: Make sure to change moles to Molarity by dividing the number of moles by volume of the container (5.00 L)

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 =

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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 SO_3(g) at 832 °C. What is the K_c for the reaction

2SO_3(g) <-----> O_2(g) + 2SO_2(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)
1	2SO3	0.350	-0.186	0.164
2	O2	0	+0.093	0.093
3	2SO2	0	+0.186	0.186

To solve for K_c 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.

K_c = [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 SO_3(g) and at equilibrium [SO₂] = 0.350 M. Calculate K_c.

	species	original conc. (M)	change	equilibrium conc (M)
1	2 SO3	0.500	-0.350	0.150
2	O2	0	+0.175	0.175
3	2 SO2	0	+0.350	0.350

Kc =

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

a. Consider the reaction

H_2(g) + I_2(g) <-----> 2HI_(g).

Suppose we start with 0.2000 mol H_2(g) and 0.1000 mol I_2(g) in a one-liter flask. When equilibrium is reached 48.0% of the H_2(g) will have been consumed. Calculate K_c.

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)
1	H2	0.2000	-0.0960	0.1040
2	I2	0.1000	-0.0960	0.0040
3	2HI	0	+0.192	0.192

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

b. At another temperature, we start with 0.450 mol H_2(g) and 0.350 mol I_2(g) in a one-liter flask. When equilibrium has been reached we find that 30.0% of the I_2(g) has reacted. Calculate K_c.

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	H2	0.450	-0.105	0.345
2	I2	0.350	-0.105	0.245
3	2 HI	0	+0.210	0.210

Kc =

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What K_c can tell you:

1. whether a reaction is likely to be feasible

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

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

2. the direction of the reaction

a. Reacion Quotient (Q)
Expression is same as for K_c except that in Q concentrations are taken when the system is not at equilibrium.

b. Q > K_c:
1. the numerator is larger than denomenator and so the denominator must become larger to reach equilbirium, therefore more reactants are produced
   
2. reaction proceeds <----

c. Q < K_c:

1. the denomenator is larger than numerator and at equilibrium the denominator must become smaller, hence the reactants are used up
   
2. reaction goes ----->

Examples

a. Given the equation N_2(g) + O_2(g) ---> 2NO_(g)       K_c = 6.2 x 10 ^-14 at 2000 °C

if we start with 0.0520 mol of N_2(g), 0.0124 mol O_2(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 Q is larger than the K_c, 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 N_2(g), and 0.060 mol O_2(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 K_c and [conc] of all species but one
1. For the reaction of N₂O_(g) + 1/2O_2(g) <-----> 2NO_(g)        K_c = 1.7 x 10 ^-13 at 25 °C

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

K_c = [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. N_2(g) + 3H_2(g) <-----> 2NH_3(g) K_c 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. H_2(g) + I_2(g) <-----> 2HI_(g)     K_c = 50.2 at 445 °C.

If we start with 0.100 M H_2(g), 0.100 M I_2(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)
1	H2	0.100	-x	0.100-x
2	I2	0.100	-x	0.100-x
3	2HI	0.050	+2x	0.050+2x

Set up the problem as before:

K_c = [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. H_2(g) + CO_2(g) <-----> H₂O_(g) + CO_(g)    K_c = 1.6 at 990 °C

What are equilibrium concentrations of all species, if we start with 0.250 M H_2(g), 0.250 M CO_2(g) and 0.100 M H₂O_(g)?

	species	original conc. (M)	change	equilibrium conc (M)
1	H2	0.250	-x	0.250 - x
2	CO2	0.250	-x	0.250 - x
3	H2O	0.100	+x	0.100 + x
4	CO	0	+x	x

[H2] =

[CO2] =

[H2O] =

[CO] =

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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:
   a. adding species shifts direction away from the species added.
   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 K_c to become larger
b. decrease in temperature causes K_c to become smaller

2. If forward reaction is exothermic:

a. increase in temperature causes K_c to become smaller
b. decrease in temperature causes K_c 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, 2SO_2(g) + O_2(g) <-----> 2SO_3(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: COCl_2(g) <-----> CO_(g) + Cl_2(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

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III. K_p

A. General Reaction:

aA_(g) + bB_(g) <-----> cC_(g) + dD_(g)

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

"P" is equilibrium partial pressure in atmospheres

B. K_p vs K_c

1. Concentration in M are used to determine K_c . Partial pressures in atmospheres are used to determine K_p .

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: K_p = ( K_c)( 0.0821T)^*Δng

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

a. Determine K_p for the reaction 2SO_2(g) + O_2(g) <-----> 2SO_3(g) if K_c = 56 at 900 K

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

K_p = 56(0.0821 x 900)^-1 = 56 / (0.0821 x 900) = 0.76

b. For the following reaction and conditions, 2NaHCO_3(s) <-----> Na₂CO_3(s) + CO_2(g) + H₂O_(g) at 100°C     (K_p = 0.23)
K_c is determined to be = ?

Kc =

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