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)

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

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 =




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.

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 =

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?

	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 [SO₂] = 0.350 M. Calculate K_c.

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

Kc =

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
2
3

Kc =

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. Original Concentration Quotient (OCQ)
Expression is same as for K_c except that in OCQ concentrations are in parentheses ( ), instead of brackets.

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

c. OCQ < K_c:

1. 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)

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?

K_c = 6.2 x 10 ^-14 at 2000 °C

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 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)
If K_c = 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. 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] = 0.80 M?

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
2
3
4

, ,,
[H2] =

[CO2] =

[H2O] =

[CO] =

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