EQUILIBRIUM WORKSHEET 2
Consider the following equilibrium:
N2O4(g)
2NO2(g)
H° = 58.0 kJ
In what direction will the equilibrium shift when each of the following changes is made
to a system at equilibrium: (a) add N2O4; (b) remove NO2; (c) increase the total pressure
by adding N2(g); (d) increase the volume; (e) decrease the temperature?
SOLUTION Le Châtelier's principle can be used to determine the effects of each of
these changes.
(a) The system will adjust, so as to decrease the concentration of the added N2O4; the
equilibrium consequently shifts to the right, in the direction of products.
(b) The system will adjust to this change by shifting to the side that produces more NO2;
thus, the equilibrium shifts to the right.
(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the
reaction. The partial pressures of NO2 and N2O4 are unchanged, and there is no shift in
the position of the equilibrium.
(d) The system will shift in the direction that occupies a larger volume (more gas
molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect
observed in Figure 15.13.)
(e) The reaction is endothermic; therefore, we can imagine heat as a reagent on the
reactant side of the equation. Decreasing the temperature will shift the equilibrium in the
direction that produces heat, and so the equilibrium shifts to the left, toward the formation
of more N2O4. Note that only this last change also affects the value of the equilibrium
constant, K.
For the reaction
PCl5(g)
PCl3(g) + Cl2(g)
H° = 87.9 kJ
in what direction will the equilibrium shift when (a) Cl2(g) is added; (b) the temperature
is increased; (c) the volume of the reaction system is decreased; (d) PCl5(g) is added?
Answers: (a) left; (b) right; (c) left; (d) right
determine the standard enthalpy change for the reaction
N2(g) + 3H2(g)
2NH3(g)
Determine how the equilibrium constant for this reaction should change with
temperature.
SOLUTION Recall that the standard enthalpy change for a reaction is given by the sum
of the standard molar enthalpies of formation of the products, each multiplied by its
coefficient in the balanced chemical equation, less the same quantities for the reactants.
At 25°C, Hf° for NH3(g) is -46.19 kJ/mol. The Hf° values for H2(g) and N2(g) are zero
by definition, because the enthalpies of formation of the elements in their normal states at
25°C are defined as zero (Section 5.7). Because 2 mol of NH3 is formed, the total
enthalpy change is
(2 mol)(-46.19 kJ/mol) - 0 = 292.38 kJ
The reaction in the forward direction is exothermic. An increase in temperature causes
the reaction to shift in the reverse direction--in our example, in the direction of less NH3
and more N2 and H2. This effect is seen in the values for Kp presented in Table 15.3.
Notice that Kp changes very markedly with change in temperature and that it is larger at
lower temperatures. This is a matter of great practical importance. To form ammonia at a
reasonable rate requires higher temperatures. Yet at higher temperatures, the equilibrium
constant is smaller and so the percentage conversion to ammonia is smaller. To
compensate for this, higher pressures are needed because high pressure favors ammonia
formation.
determine the enthalpy change for the reaction
2POCl3(g)
2PCl3(g) + O2(g)
Use this result to determine how the equilibrium constant for the reaction should change
with temperature. Answer: H° = 508 kJ; the equilibrium constant for the reaction will
increase with increasing temperature
At 448°C the equilibrium constant, Kc, for the reaction
H2(g) + I2(g)
2HI(g)
is 51. Predict how the reaction will proceed to reach equilibrium at 448°C if we start with
2.0 × 10-2 mol of HI, 1.0 × 10-2 mol of H2, and 3.0 × 10-2 mol of I2 in a 2.0-L container.
SOLUTION The initial concentrations are
[HI] = 2.0 × 10-2 mol/2.0 L = 1.0 × 10-2 M
[H2] = 1.0 × 10-2 mol/2.0 L = 5.0 × 10-3 M
[I2] = 3.0 × 10-2 mol/2.0 L = 1.5 × 10-2 M
The reaction quotient is
Because Q < Kc, [HI] will need to increase and [H2] and [I2] decrease to reach
equilibrium; the reaction will proceed from left to right.
At 1000 K the value of Kc for the reaction 2SO3(g)
2SO2(g) + O2(g) is 4.08 × 10-3.
Calculate the value for Q and predict the direction in which the reaction will proceed
toward equilibrium if the initial concentrations of reactants are [SO3] = 2 × 10-3 M; [SO2]
= 5 × 10-3 M; [O2] = 3 × 10-2 M. Answer: Q = 0.2; the reaction will proceed from right to
left, forming more SO3.
For the Haber process, N2(g) + 3H2(g)
2NH3(g), Kp = 1.45 × 10-5 at 500°C. In an
equilibrium mixture of the three gases at 500°C the partial pressure of H2 is 0.928 atm
and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibrium
mixture?
SOLUTION Because the mixture is in equilibrium, we need not worry about initial
concentrations. We tabulate the equilibrium pressures as follows:
N2(g) + 3H2(g)
2NH3(g)
Equilibrium pressure (atm):
0.4320.928
x
Because we do not know the equilibrium pressure of NH3, we represent it with a variable,
x. At equilibrium the pressures must satisfy the equilibrium expression:
We now rearrange the equation to solve for x:
x2 5 (1.45 3 10-5)(0.432)(0.928)3 5 5.01 3 10-6
We can always double-check our answer by using it to recalculate the value of the
equilibrium constant:
At 500 K the reaction PCl5(g)
PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium
mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm.
What is the partial pressure of Cl2 in the equilibrium mixture? Answer: 1.22 atm
A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of
the equilibrium constant, Kc, for the reaction
H2(g) + I2(g)
2HI(g)
at 448°C is 50.5. What are the concentrations of H2, I2, and HI in the flask at equilibrium?
SOLUTION In this case, unlike Sample Exercise 15.10, we are not given any of the
equilibrium concentrations. We must develop some relationships that relate the initial
concentrations to those at equilibrium. The procedure is similar in many regards to that
outlined in Sample Exercise 15.7.
First, we construct a table in which we tabulate the initial concentrations: [H2] = 1.000 M,
[I2] = 2.000 M, and [HI] = 0.
Second, we use the stoichiometry of the chemical equation to determine the changes in
concentrations that occur as the reaction proceeds to equilibrium. Clearly, the
concentration of HI cannot be zero at equilibrium. The concentrations of H2 and I2 will
decrease as equilibrium is established, and that of HI will increase. Let's represent the
change in concentration of H2 by the variable x. The balanced chemical equation tells us
the relationship between the changes in the concentrations of the three gases: For each x
moles of H2 per liter that react, x moles of I2 per liter are also consumed, and 2x moles of
HI per liter are produced.
Third, we use the initial concentrations and the concentration changes to express the
equilibrium concentrations. With all of our entries we now have the following table:
H2(g) + I2(g)
2HI(g)
Fourth, we substitute the equilibrium concentrations into the equilibrium expression and
solve for the single unknown, x:
Expanding this expression leads to a quadratic equation in x:
4x2 = 50.5(x2 - 3.000x + 2.000)
46.5x2 - 151.5x + 101.0 = 0
Solving the quadratic equation (Appendix A.3) leads to two solutions for x:
When we substitute the first of these solutions, x = 2.323, into the expressions for the
equilibrium concentrations, we find "negative" concentrations of H2 and I2. A negative
concentration is not chemically meaningful, so we reject this solution. We use the other
solution, x = 0.935, to find the equilibrium concentrations:
[H2] = 1.000 -x = 0.065 M
[I2] = 2.000 -x = 1.065 M
[HI] = 2x = 1.870 M
Finally, we can double-check our solution by putting these numbers into the equilibrium
expression:
Whenever we use the quadratic equation to solve an equilibrium problem, one of the
solutions will not be chemically meaningful and is rejected.
For the equilibrium PCl5(g)
PCl3(g) + Cl2(g), the equilibrium constant, Kp, has the
value 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial
pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this
temperature? Answer: PPCI5 = 0.97 atm; PPCI3 = PCI2 = 0.693 atm
A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium
at 472°C. The equilibrium mixture of gases was analyzed and found to contain 0.1207 M
H2, 0.0402 M N2, and 0.00272 M NH3. From these data calculate the equilibrium
constant, Kc, for
SOLUTION
Nitryl chloride, NO2Cl, is in equilibrium with NO2 and Cl2:
At equilibrium the concentrations of the substances are [NO2Cl] = 0.00106 M, [NO2] =
0.0108 M, and [Cl2] = 0.00538 M. From these data calculate the equilibrium constant, Kc.
Answer: 0.558
We often don't know the equilibrium concentrations of all chemical species in an
equilibrium. However, if we know the equilibrium concentration of at least one species,
we can generally use the stoichiometry of the reaction to deduce the equilibrium
concentrations of the other species in the chemical equation. We will use the following
procedure to do this:
1. Tabulate the known initial and equilibrium concentrations of all species involved in the
equilibrium.
2. For those species for which both the initial and equilibrium concentrations are known,
calculate the change in concentration that occurs as the system reaches equilibrium.
3. Use the stoichiometry of the reaction to calculate the changes in concentration for all
the other species in the equilibrium.
4. From the initial concentrations and the changes in concentration, calculate the
equilibrium concentrations. These are used to evaluate the equilibrium constant.
We illustrate the procedure in Sample Exercise 15.7.
A mixture of 5.000 × 10-3 mol of H2 and 1.000 × 10-2 mol of I2 is placed in a 5.000-L
container at 448°C and allowed to come to equilibrium. Analysis of the equilibrium
mixture shows that the concentration of HI is 1.87 × 10-3 M. Calculate Kc at 448°C for the
reaction
SOLUTION First, we tabulate the initial and equilibrium concentrations of all the
species in the equilibrium. We also provide space in our table for listing the changes in
concentrations. As shown, it is convenient to use the chemical equation as the heading for
the table.
In this exercise the initial concentrations of H2 and I2 must be calculated:
Thus, the first entries in the table are:
H2(g) + I2(g)
2HI(g)
Second, we calculate the change in concentration of HI, using the initial and equilibrium
values. The change is the difference between the equilibrium and initial values, 1.87 × 103
M.
Third, we use the stoichiometry of the reaction to calculate the changes in the other
species. The balanced chemical equation indicates that for each 2 mol of HI formed, 1
mol of H2 must be consumed. Thus, the amount of H2 consumed is
The same line of reasoning gives us the amount of I2 consumed, which is also 0.935 × 103
M.
Fourth, we calculate the equilibrium concentrations, using the initial concentrations and
the changes. The equilibrium concentration of H2 is the initial concentration minus that
consumed:
[H2] = 1.000 × 10-3 M - 0.935 × 10-3 M = 0.065 × 10-3 M
Likewise, the equilibrium concentration of I2 is
[I2] = 2.000 × 10-3 M - 0.935 × 10-3 M = 1.065 × 10-3 M
The completed table now looks like this:
H2(g) + I2(g)
2HI(g)
Finally, now that we know the equilibrium concentration of each reactant and product,
we can use the equilibrium expression to calculate the equilibrium constant:
Sulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g)
2SO2(g) + O2(g). Initially the vessel is charged at 1000 K with SO3(g) a concentration of
6.09 × 10-3 M. At equilibrium the SO3 concentration is 2.44 × 10-3 M. Calculate the value
of Kc at 1000 K. Answer: 4.08 × 10-3
Using the value of Kc obtained in Sample Exercise 15.6, calculate Kp for
N2(g) + 3H2(g)
2NH3(g)
at 472°C.
SOLUTION The relationship between Kc and Kp is given by Equation 15.16. There are 2
mol of gaseous products (2NH3) and 4 mol of gaseous reactants (1N2 + 3H2). Therefore,
n = 2 - 4 = -2. (Remember that functions are always based on products minus
reactants.) The temperature, T, is 273 + 472 = 745 K. The value for the ideal-gas
constant, R, is 0.0821 L-atm/mol-K. The value of Kc from Sample Exercise 15.6 is 0.105.
We therefore have
Practice Exercise
For the equilibrium 2SO3(g)
2SO2(g) + O2(g) at a temperature of 1000 K, Kc has the
-3
value 4.08 × 10 . Calculate the value for Kp. Answer: 0.335
Write the equilibrium expressions for Kc and Kp for each of the following reactions:
(a) CO2(g) + H2(g)
(b) SnO2(s) + 2CO(g)
CO(g) + H2O(l)
Sn(s) + 2CO2(g)
SOLUTION
(a) The equilibrium expressions are
Because H2O appears in the reaction as a pure liquid, its concentration does not appear in
either equilibrium expression.
(b) The equilibrium expressions are
Because SnO2 and Sn are both pure solids, their concentrations do not appear in either
equilibrium expression.
Write the equilibrium expressions for Kc and Kp for the reaction 3Fe(s) + 4H2O(g)
Fe3O4(s) + 4H2(g). Answer: Kc = [H2]4/[H2O]4; Kp=P4H2/P4H2O