4 Bohr's Theory of the Hydrogen Atom
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CH4(g) + Cl2(g)
CH3Cl(g) + HCl(g)
i
0.00100
0.00100
0
0
c
–x
–x
+x
+x
e
0.00100 – x
0.00100 – x
x
x
We fill in the bottom row (the equilibrium
concentrations) by adding the top- and
middle-row entries in each column. We then
use the equilibrium expression, in this case
[CH3Cl][HCl]
[CH4][Cl2]
to solve for x using the value of Kc and the
equilibrium concentrations from the bottom
row of the completed table.
Kc =
4.50 × 103 =
(x)(x)
x2
=
(0.00100 – x)(0.00100 – x)
(0.00100 – x) 2
Taking the square root of both sides of this equation gives
x2
(0.00100 – x) 2
x
67.08 =
0.00100 – x
4.50 × 103 =
and solving for x gives
0.06708 – 67.08x = x
0.06708 = 68.08x
x = 9.85 × 10–4
CHECK
We used x to represent the concentrations of both products at
equilibrium, and the amount by which each reactant concentration
decreases. Therefore, the equilibrium concentrations are
[CH4] = [Cl2] = (0.00100 – 9.85 × 10–4) M = 1.5 × 10–5 M
and
[CH3Cl] = [HCl] = 9.85 × 10–4 M
We can verify our results by plugging the equilibrium
concentrations back into the equilibrium expression. We should
get a number very close to the original equilibrium constant.
CH4(g) + Cl2(g)
CH3Cl(g) + HCl(g)
i
0.00100
0.00100
c
–9.7 × 10–4
–9.7 × 10–4
+9.7 × 10–4
+9.7 × 10–4
e
3.0 × 10–5
3.0 × 10–5
9.7 × 10–4
9.7 × 10–4
0
0
The equilibrium concentration of HCl is equal to that of
CH3Cl, 9.7 × 10–4 M. Further, the amount by which each reactant
concentration decreased is also 9.7 × 10–4 M. Therefore, the
equilibrium concentrations are [CH4] = [Cl2] = 3.0 × 10–5 M;
and [CH3Cl] = [HCl] = 9.7 × 10–4 M. Plugging these equilibrium
concentrations into the equilibrium expression gives
[CH3Cl][HCl]
(9.7 × 10–4) 2
Kc =
=
= 1.0 × 103.
[CH4][Cl2]
(3.0 × 10–5) 2
Therefore, at 2000°C, Kc for this reaction is 1.0 × 103.
Kc =
[CH3Cl][HCl]
(9.85 × 10–4) 2
=
= 4.3 × 103
[CH4][Cl2]
(1.5 × 10–5) 2
In this case, the small difference between the result and the
original Kc value (4.5 × 103) is due to rounding.
(See Visualizing Chemistry questions VC15.1–VC15.4 on page 705.)
What's the point?
Equilibrium or ice tables can be used to solve a variety of
equilibrium problems. We begin by entering the known
concentrations and then use the stoichiometry of the reaction to
complete the table. When the bottom row of the ice table is
complete, we can use the equilibrium expression to calculate either
equilibrium concentrations, or the value of the equilibrium constant.
679
680
CHAPTER 15 Chemical Equilibrium
Worked Example 15.11
Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,
H2(g) + I2(g)
2HI(g)
is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and I2?
Strategy Construct an equilibrium table to determine the equilibrium concentration of each species in terms of an unknown (x);
solve for x, and use it to calculate the equilibrium molar concentrations.
Setup Insert the starting concentrations that we know into the equilibrium table.
H2(g)+I2(g)
2HI(g)
0.2400.240 0
Initial concentration (M):
Change in concentration (M):
Equilibrium concentration (M):
Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the
reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium.
According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x),
and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in
concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations.
H2(g)+I2(g)
2HI(g)
0.2400.240 0
Initial concentration (M):
Change in concentration (M):−x−x+2x
0.240 − x0.240 − x
Equilibrium concentration (M):
2x
Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.
Kc =
[HI]2
[H2][I2]
(2x) 2
(2x) 2
=
(0.240 − x) (0.240 − x) (0.240 − x) 2
2x
√54.3 =
0.240 − x
54.3 =
x = 0.189
Using the calculated value of x, we can determine the equilibrium concentration of each species as follows:
[H2] = (0.240 − x) M = 0.051 M
[I2] = (0.240 − x) M = 0.051 M
[HI] = 2x = 0.378 M
Think About It
Always check your answer by inserting the calculated concentrations into the equilibrium expression.
(0.378) 2
[HI]2
=
= 54.9 ≈ Kc
[H2][I2] (0.051) 2
The small difference between the calculated Kc and the one given in the problem statement is due to rounding.
Practice Problem A T T E M PT Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations
are [H2] = [I2] = 0 M and [HI] = 0.525 M.
Practice Problem B U I L D Determine the initial concentration of HI if the initial concentrations of H2 and I2 are both 0.10 M
and their equilibrium concentrations are both 0.043 M at 430°C.
SECTION 15.5 Calculating Equilibrium Concentrations
681
Practice Problem C O N C E P T UA L I Z E Consider the reaction A(g) + B(g)
C(g). The diagram shown on the right
depicts the starting condition for a system. Without knowing the value of Kc determine which of the following diagrams [(i)–(iv)]
could represent the system at equilibrium. Select all that apply.
AA
==
BB
==
CC
==
(i)(i)
(ii)(ii)
(iii)
(iii)
(iv)
(iv)
Worked Example 15.12
For the same reaction and temperature as in Worked Example 15.11, calculate the equilibrium concentrations of all three species
if the starting concentrations are as follows: [H2] = 0.00623 M, [I2] = 0.00414 M, and [HI] = 0.0424 M.
Strategy Using the initial concentrations, calculate the reaction quotient, Qc, and compare it to the value of Kc (given in the
problem statement of Worked Example 15.11) to determine which direction the reaction will proceed to establish equilibrium.
Then, construct an equilibrium table to determine the equilibrium concentrations.
Setup
(0.0424) 2
[HI]2
=
= 69.7
[H2][I2] (0.00623) (0.00414)
Therefore, Qc > Kc, so the system will have to proceed to the left (reverse) to reach equilibrium. The equilibrium table is
H2(g)+I2(g)
2HI(g)
0.006230.004140.0424
Initial concentration (M):
Change in concentration (M):
Equilibrium concentration (M):
Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and
the concentrations of H2 and I2 will increase. Therefore, the table should be filled in as follows:
H2(g)+I2(g)
2HI(g)
0.006230.004140.0424
Initial concentration (M):
Change in concentration (M):+x+x−2x
Equilibrium concentration (M):
0.00623 + x0.00414 + x0.0424 − 2x
Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.
Kc =
54.3 =
[HI]2
[H2][I2]
(0.0424 − 2x) 2
(0.00623 + x)(0.00414 + x)
It isn’t possible to solve this equation the way we did in Worked Example 15.11 (by taking the square root of both sides) because
the concentrations of H2 and I2 are unequal. Instead, we have to carry out the multiplications,
54.3(2.58 × 10−5 + 1.04 × 10−2x + x2) = 1.80 × 10−3 − 1.70 × 10−1x + 4x2
Collecting terms we get
50.3x2 + 0.735x − 4.00 × 10−4 = 0
(Continued on next page)
682
CHAPTER 15 Chemical Equilibrium
This is a quadratic equation of the form ax2 + bx + c = 0. The solution for the quadratic equation (see Appendix 1) is
x=
−b ± √b2 − 4ac
2a
Here we have a = 50.3, b = 0.735, and c = −4.00 × 10−4, so
−0.735 ± √ (0.735) 2 − 4(50.3) (−4.00 × 10−4
2(50.3)
x=
x = 5.25 × 10−4 or
x = −0.0151
−4
Only the first of these values, 5.25 × 10 , makes sense because concentration cannot be a negative number. Using the calculated
value of x, we can determine the equilibrium concentration of each species as follows:
[H2] = (0.00623 + x) M = 0.00676 M
[I2] = (0.00414 + x) M = 0.00467 M
[HI] = (0.0424 − 2x) M = 0.0414 M
Think About It
Checking this result gives
Kc =
(0.0414) 2
[HI]2
=
= 54.3
[H2][I2] (0.00676) (0.00467)
Practice Problem A T T E M P T Calculate the equilibrium concentrations of H2, I2, and HI at 430°C if the initial concentrations
are [H2] = [I2] = 0.378 M and [HI] = 0 M.
Practice Problem B U I L D At 1280°C the equilibrium constant Kc for the reaction
Br2(g)
−3
−2
is 1.1 × 10 . If the initial concentrations are [Br2] = 6.3 × 10
these two species at equilibrium.
2Br(g)
M and [Br] = 1.2 × 10−2 M, calculate the concentrations of
Practice Problem C O N C E P T UA L I Z E For the reaction of hydrogen and iodine gases to form hydrogen iodide, indicate
which of these diagrams [(i)–(iv)] represents a starting condition from which equilibrium can be established.
I2 =
H2 =
(i)
(ii)
HI =
(iii)
(iv)
When the magnitude of K (either Kc or KP) is very small, the solution to an equilibrium problem can be simplified—making it unnecessary to use the quadratic equation.
Worked Example 15.13 illustrates this.
Worked Example 15.13
At elevated temperatures, iodine molecules break apart to give iodine atoms according to the equation
I2(g)
2I(g)
−12
Kc for this reaction at 205°C is 3.39 × 10 . Determine the concentration of atomic iodine when a 1.00-L vessel originally
charged with 0.00155 mol of molecular iodine at this temperature is allowed to reach equilibrium.
Strategy Construct an equilibrium table and use initial concentrations and the value of Kc to determine the changes in
concentration and the equilibrium concentrations.
SECTION 15.5 Calculating Equilibrium Concentrations
683
Setup The initial concentration of I2(g) is 0.00155 M and the original concentration of I(g) is zero. Kc = 3.39 × 10−12.
Solution We expect the reactant concentration to decrease by some unknown amount x. The stoichiometry of the reaction
indicates that the product concentration will increase by twice that amount, 2x.
I2(g)
2I(g)
0.00155
Initial concentration (M):
0
Change in concentration (M):−x+2x
0.00155 − x
Equilibrium concentration (M):
2x
Inserting the equilibrium concentrations into the equilibrium expression for the reaction gives
3.39 × 10−12 =
(2x) 2
[I]2
=
[I2] (0.00155 − x)
We could solve for x using the quadratic equation, as we did in Worked Example 15.12. However, because the magnitude of Kc is
very small, we expect this equilibrium to lie far to the left. The means that very little of the molecular iodine will break apart to
give atomic iodine; and the value of x [amount of I2(g) that reacts] will be very small. In fact, the value of x will be negligible
compared to the original concentration of I2(g). Therefore, 0.00155 − x ≈ 0.00155, and the solution simplifies to
3.39 × 10−12 =
(2x) 2
4x2
=
0.00155 0.00155
3.39 × 10−12 (0.00155)
= 1.31 × 10−15 = x2
4
x = √1.31 × 10−15 = 3.62 × 10−8 M
According to our ice table, the equilibrium concentration of atomic iodine is 2x; therefore, [I(g)] = 2 × 3.62 × 10−8 = 7.24 × 10−8 M.
Think About It
Having solved for x by this method, we can see that it is indeed insignificant compared to the original concentration of iodine. To an
appropriate number of significant figures, 0.00155 − 3.62 × 10−8 = 0.00155.
Practice Problem A T T E M P T Aqueous hydrocyanic acid (HCN) ionizes according to the equation
HCN(aq)
−10
At 25°C, Kc for this reaction is 4.9 × 10
aqueous HCN.
H+(aq) + CN−(aq)
. Determine the equilibrium concentrations of all species in a 0.100-M solution of
Practice Problem B U I L D Consider a weak acid, HA, that ionizes according to the equation
HA(aq)
H+(aq) + A−(aq)
At 25°C, a 0.145-M solution of aqueous HA is found to have a hydrogen ion concentration of 2.2 × 10−5 M. Determine the Kc for
this reaction at 25°C.
Practice Problem C ONCEPTUALIZE Each of the diagrams [(i)–(iv)] shows a system before
(i) and after equilibrium is established.
Indicate which diagram best represents a system in which you can neglect the x in the solution, as you did in Worked Example 15.13.
(i)
(i)
(ii)
(iii)
(iii)
(ii)
(iv)
684
CHAPTER 15 Chemical Equilibrium
Here is a summary of the use of initial reactant concentrations to determine
equilibrium concentrations.
Student Annotation: If Q < K, the reaction will
occur as written. If Q > K, the reverse reaction
will occur.
Student Hot Spot
Student data indicate you may struggle with
determining K values. Access the SmartBook to
view additional Learning Resources on this topic.
1.Construct an equilibrium table, and fill in the initial concentrations (including
any that are zero).
2. Use initial concentrations to calculate the reaction quotient, Q, and compare Q to
K to determine the direction in which the reaction will proceed.
3.Define x as the amount of a particular species consumed, and use the stoichiometry of the reaction to define (in terms of x) the amount of other species consumed
or produced.
4. For each species in the equilibrium, add the change in concentration to the initial
concentration to get the equilibrium concentration.
5. Use the equilibrium concentrations and the equilibrium expression to solve for x.
6.Using the calculated value of x, determine the concentrations of all species at
equilibrium.
7. Check your work by plugging the calculated equilibrium concentrations into the
equilibrium expression. The result should be very close to the Kc stated in the
problem.
The same procedure applies to KP.
Worked Example 15.14 shows how to solve an equilibrium problem using partial
pressures.
Worked Example 15.14
A mixture of 5.75 atm of H2 and 5.75 atm of I2 is contained in a 1.0-L vessel at 430°C. The equilibrium constant (KP) for the
reaction
H2(g) + I2(g)
2HI(g)
at this temperature is 54.3. Determine the equilibrium partial pressures of H2, I2, and HI.
Strategy Construct an equilibrium table to determine the equilibrium partial pressures.
Setup The equilibrium table is
H2(g)+I2(g)
2HI(g)
5.755.75 0
Initial partial pressure (atm):
Change in partial pressure (atm):−x−x+2x
Equilibrium partial pressure (atm):
5.75 − x5.75 − x
Solution Setting the equilibrium expression equal to KP,
54.3 =
(2x) 2
(5.75 − x) 2
Taking the square root of both sides of the equation gives
√54.3 =
2x
5.75 − x
7.369 =
2x
5.75 − x
7.369(5.75 − x) = 2x
42.37 − 7.369x = 2x
42.37 = 9.369x
x = 4.52
The equilibrium partial pressures are PH2 = PI2 = 5.75 − 4.52 = 1.23 atm, and PHI = 9.04 atm.
2x
SECTION 15.5 Calculating Equilibrium Concentrations
685
Think About It
Plugging the calculated partial pressures into the equilibrium expression gives
(PHI ) 2
(9.04) 2
=
= 54.0
(PH2 ) (PI2 )
(1.23) 2
The small difference between this result and the equilibrium constant given in the problem statement is due to rounding.
Practice Problem A T T E M P T Determine the equilibrium partial pressures of H2, I2, and HI if we begin the experiment with
1.75 atm each of H2 and I2 at 430°C.
A = pressures of H , I , and HI (at 430°C) if we begin the experiment
Practice Problem B U I L D Determine the equilibrium partial
2 2
= PHI = 2.5 atm.
with the following conditions: PH2 = 0.25 atm, PI2 = 0.050 B
atm,
C = A(g) + B(g)
Practice Problem C ONCEPTUALIZE Consider the reaction
C(s) + D(s). The diagram shown on the
right represents a system at equilibrium. Each of the following
D =diagrams [(i)–(iv)] is missing spheres of a particular color. Indicate
how many spheres of the missing color must be included for each diagram to represent a system at equilibrium.
A=
B=
C=
D=
(i)
(ii)
(iii)
(iv)
Section 15.5 Review
Using Equilibrium Expressions to Solve Problems
Use the following information to answer questions (i)
15.5.1 and 15.5.2:(ii)
Kc for the reaction(iii)
A+B
2C
is 1.7 × 10−2 at 250°C.
(iv)
15.5.1 What will the equilibrium concentrations of A, B, and C be at this temperature
if [A]i = [B]i = 0.750 M ([C]i = 0)?
(a) 6.1 × 10−3 M, 6.1 × 10−3 M, 0.092 M (d)0.70 M, 0.70 M, 0.092 M
(b)0.046 M, 0.046 M, 0.092 M
(e)0.087 M, 0.087 M, 0.66 M
(c)0.70 M, 0.70 M, 0.046 M
15.5.2 What will the equilibrium concentrations of A, B, and C be at this temperature
if [C]i = 0.875 M ([A]i = [B]i = 0)?
(a)0.41 M, 0.41 M, 0.82 M
(d)0.43 M, 0.43 M, 0.44 M
(b)0.41 M, 0.41 M, 0.054 M
(e)0.43 M, 0.43 M, 0.43 M
(c)0.43 M, 0.43 M, 0.0074 M
15.5.3If Kc = 3 for the reaction X + 2Y
Z at a certain temperature, then
for each of the mixtures of X, Y, and Z shown here, in what direction
must the reaction proceed to achieve equilibrium?
X=
Y=
Z=
(a) Right, right
(b) Neither, left
(c) Neither, right
(d) Left, neither
(e) Right, neither
686
CHAPTER 15 Chemical Equilibrium
15.6 LE CHÂTELIER’S PRINCIPLE: FACTORS
THAT AFFECT EQUILIBRIUM
One of the interesting and useful features of chemical equilibria is that they can be
manipulated in specific ways to maximize production of a desired substance. Consider,
for example, the industrial production of ammonia from its constituent elements by
the Haber process.
N2(g) + 3H2(g)
2NH3(g)
More than 100 million tons of ammonia is produced annually by this reaction, with
most of the resulting ammonia being used for fertilizers to enhance crop production.
Clearly it would be in the best interest of industry to maximize the yield of NH3. In
this section, we will learn about the various ways in which an equilibrium can be
manipulated to accomplish this goal.
Le Châtelier’s principle states that when we change the conditions of a system
at equilibrium, the system will respond by shifting in the direction that minimizes the
effect of the change. We may change the conditions of a system at equilibrium by any
of the following means.
∙ The addition of a reactant or product
∙ The removal of a reactant or product
∙A change in volume of the system, resulting in a change in concentration or
partial pressure of the reactants and products
∙ A change in temperature
“Shifting” refers to the occurrence of either the forward or reverse reaction such that
the effect of the change is partially offset as the system reestablishes equilibrium. An
equilibrium that shifts to the right is one in which more products are produced by the
forward reaction. An equilibrium that shifts to the left is one in which more reactants
are produced by the reverse reaction. Using Le Châtelier’s principle, we can predict
the direction in which an equilibrium will shift, given the specific change that is made.
Addition or Removal of a Substance
Again using the Haber process as an example,
N2(g) + 3H2(g)
2NH3(g)
consider a system at 700 K, in which the equilibrium concentrations are as follows:
[N2] = 2.05 M [H2] = 1.56 M [NH3] = 1.52 M
Student Annotation: Remember that at
equilibrium, the reaction quotient, Qc , is equal
to the equilibrium constant, Kc.
Using these concentrations in the reaction quotient expression, we can calculate the
value of Kc for the reaction at this temperature as follows:
Qc =
[NH3]2
[N2][H2]3
=
(1.52) 2
(2.05)(1.56) 3
= 0.297 = Kc
If we were to change the conditions of this system by adding more N2, increasing its
concentration from 2.05 M to 3.51 M, the system would no longer be at equilibrium. To
see that this is true, use the new concentration of nitrogen in the reaction quotient expression. The new calculated value of Qc (0.173) is no longer equal to the value of Kc (0.297).
Qc =
[NH3]2
[N2][H2]3
=
(1.52) 2
(3.51)(1.56) 3
= 0.173 ≠ Kc
For this system to reestablish equilibrium, the net reaction will have to shift in such
a way that Qc is again equal to Kc, which is constant at a given temperature. Recall
SECTION 15.6 Le Châtelier’s Principle: Factors That Affect Equilibrium
Figure 15.6 Adding more of a reactant
4.0
3.5
Concentration (M )
687
to a system at equilibrium causes the equilibrium position to shift toward product.
The system responds to the addition of N2
by consuming some of the added N2 (and
some of the other reactant, H2) to produce
more NH3.
N2
3.0
2.5
2.0
NH3
1.5
H2
1.0
0.5
0.0
Time
[NH3]2
[N2][H2]
3
=
(1.52)2
3
(2.05)(1.56)
= 0.297
Original equilibrium mixture
[NH3]2
3
[N2][H2]
=
(1.52)2
3
(3.51)(1.56)
= 0.173
[NH3]2
3
[N2][H2]
Immediately after addition of N2
=
(1.64)2
(3.45)(1.38)3
= 0.297
After equilibrium
has been reestablished
from Section 15.4 that when Q is less than K, the reaction proceeds to the right to
achieve equilibrium. Likewise, an equilibrium that is changed in such a way that Q
becomes less than K will shift to the right to reestablish equilibrium. This means that
the forward reaction, the consumption of N2 and H2 to produce NH3, will occur. The
result will be a net decrease in the concentrations of N2 and H2 (thus making the
denominator of the reaction quotient smaller), and a net increase in the concentration of
NH3 (thus making the numerator larger). When the concentrations of all species are such
that Qc is again equal to Kc, the system will have established a new equilibrium position,
meaning that it will have shifted in one direction or the other, resulting in a new equilibrium concentration for each species. Figure 15.6 shows how the concentrations of N2, H2,
and NH3 change when N2 is added to the original equilibrium mixture.
Conversely, if we were to remove N2 from the original equilibrium mixture, the
lower concentration in the denominator of the reaction quotient would result in Qc
being greater than Kc. In this case the reaction will shift to the left. That is, the reverse
reaction will take place, thereby increasing the concentrations of N2 and H2 and
decreasing the concentration of NH3 until Qc is once again equal to Kc.
The addition or removal of NH3 will cause a shift in the equilibrium, too. The
addition of NH3 will cause a shift to the left; the removal of NH3 will cause a shift
to the right. Figure 15.7(a) shows the additions and removals that cause this equilibrium to shift to the right. Figure 15.7(b) shows those that cause it to shift to the left.
In essence, a system at equilibrium will respond to the addition of a species by
consuming some of that species, and it will respond to the removal of a species by
producing more of that species. It is important to remember that the addition or
removal of a species from an equilibrium mixture does not change the value of the
equilibrium constant, K. Rather, it changes temporarily the value of the reaction quotient, Q. Furthermore, to cause a shift in the equilibrium, the species added or removed
must be one that appears in the reaction quotient expression. In the case of a heterogeneous equilibrium, altering the amount of a solid or liquid species does not change
the position of the equilibrium because doing so does not change the value of Q.
Student Annotation: While reestablishing
equilibrium causes a decrease in the N2
concentration, the final concentration will still be
higher than that in the original equilibrium
mixture. The system responds to the the added
reactant by consuming part of it.
Student Hot Spot
Student data indicate you may struggle with
how concentration changes affect equilibria.
Access the SmartBook to view additional
Learning Resources on this topic.
Worked Example 15.15 explores the effects of changing conditions for a system at
equilibrium.
addition
addition
addition
N2(g) + 3H2(g)
2NH3(g)
removal
(a)
N2(g) + 3H2(g)
removal removal
(b)
2NH3(g)
Figure 15.7 (a) Addition of a reactant or
removal of a product will cause an equilibrium to shift to the right. (b) Addition of a
product or removal of a reactant will cause
an equilibrium to shift to the left.
688
CHAPTER 15 Chemical Equilibrium
Worked Example 15.15
Hydrogen sulfide (H2S) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce
elemental sulfur.
2H2S(g) + O2(g)
2S(s) + 2H2O(g)
For each of the following scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither:
(a) addition of O2(g), (b) removal of H2S(g), (c) removal of H2O(g), and (d) addition of S(s).
Strategy Use Le Châtelier’s principle to predict the direction of shift for each case. Remember that the position of the equilibrium
is only changed by the addition or removal of a species that appears in the reaction quotient expression.
Setup Begin by writing the reaction quotient expression,
[H2O]2
Qc =
[H2S]2[O2]
Because sulfur is a solid, it does not appear in the expression. Changes in the concentration of any of the other species will cause a
change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression for Qc will shift the
equilibrium to the right.
addition
addition
2H2S(g) + O2(g)
2S(s) + 2H2O(g)
removal
Removal of a reactant or addition of a product that appears in the expression for Qc will shift the equilibrium to the left.
addition
2H2S(g) +
removal
2S(s) + 2H2O(g)
O2(g)
removal
Solution
(a) Shift to the right (b) Shift to the left (c) Shift to the right (d) No change
Think About It
In each case, analyze the effect the change will have on the value of Qc. In part (a), for example, O2 is added, so its concentration
increases. Looking at the reaction quotient expression, we can see that a larger concentration of oxygen corresponds to a larger overall
denominator—giving the overall fraction a smaller value. Thus, Q will temporarily be smaller than K and the reaction will have to shift to
the right, consuming some of the added O2 (along with some of the H2S in the mixture) to reestablish equilibrium.
Practice Problem A T T E M P T For each change indicated, determine whether the equilibrium
PCl3(g) + Cl2(g)
PCl5(g)
will shift to the right, shift to the left, or neither: (a) addition of PCl3(g), (b) removal of PCl3(g), (c) removal of PCl5(g), and
(d) removal of Cl2(g).
Practice Problem B U I L D What can be added to the equilibrium that will (a) shift it to the left, (b) shift it to the right, (c) not
A=
shift it in either direction? A =
−
+
A
=
AgCl(s) + 2NH3(aq)
Ag(NH3)2(aq) + B
Cl=
(aq)
B=
B
=
C
=
A
=
Practice Problem C ONCEPTUALIZE Consider the reaction A(g) + B(g)
C(s) + D(s). The first of the diagrams
C=
on the right represents a system at equilibrium where A = , B = , C = , and D = . The second diagram on the right
D = after more A has been added. Which of the following diagrams [(i)–(iv)] could represent the
represents the system immediately
D=
B=
C=
system after equilibrium has been reestablished? Select all that apply.
C=
D=
D=
A=
B=
C=
(i)
(ii)
(iii)
(i)
(i)
(i)
(i)
(iv)
(ii)
(ii)
(ii)
D=
(ii)
(iii)
(iii)
(iii)
(iii)
(iv)
(iv)
(iv)
(iv)
SECTION 15.6 Le Châtelier’s Principle: Factors That Affect Equilibrium
689
Figure 15.8 The effect of a volume
d ecrease (pressure increase) on the
N2O4(g)
2NO2(g) equilibrium. When
volume is decreased, the equilibrium is
driven toward the side with the smallest
number of moles of gas.
Changes in Volume and Pressure
If we were to start with a gaseous system at equilibrium in a cylinder with a movable
piston, we could change the volume of the system, thereby changing the concentrations of the reactants and products.
Consider again the equilibrium between N2O4 and NO2.
N2O4(g)
2NO2(g)
At 25°C the equilibrium constant for this reaction is 4.63 × 10−3. Suppose we have an
equilibrium mixture of 0.643 M N2O4 and 0.0547 M NO2 in a cylinder fitted with a movable piston. If we push down on the piston, the equilibrium will be disturbed and will
shift in the direction that minimizes the effect of this disturbance. Consider what happens
to the concentrations of both species if we decrease the volume of the cylinder by half.
Both concentrations are initially doubled: [N2O4] = 1.286 M and [NO2] = 0.1094 M. If
we plug the new concentrations into the reaction quotient expression, we get
Qc =
[NO2]2eq
[N2O2]eq
=
(0.1094) 2
= 9.31 × 10−3
1.286
which is not equal to Kc, so the system is no longer at equilibrium. Because Qc is
greater than Kc, the equilibrium will have to shift to the left for equilibrium to be
reestablished (Figure 15.8).
In general, a decrease in volume of a reaction vessel will cause a shift in the
equilibrium in the direction that minimizes the total number of moles of gas. Conversely, an increase in volume will cause a shift in the direction that maximizes the
total number of moles of gas.
Worked Example 15.16 shows how to predict the equilibrium shift that will be caused
by a volume change.
Worked Example 15.16
For each reaction, predict in what direction the equilibrium will shift when the volume of the reaction vessel is decreased.
(a) PCl5(g)
PCl3(g) + Cl2(g)
(b) 2PbS(s) + 3O2(g)
(c) H2(g) + I2(g)
2HI(g)
2PbO(s) + 2SO2(g)
Strategy Determine which direction minimized the number of moles of gas in the
reaction. Count only moles of gas.
Setup We have (a) 1 mole of gas on the reactant side and 2 moles of gas on the
product side, (b) 3 moles of gas on the reactant side and 2 moles of gas on the product
side, and (c) 2 moles of gas on each side.
Student Annotation: It is a common error to
count all the species in a reaction to determine
which side has fewer moles. To determine what
direction shift a volume change will cause, it is
only the number of moles of gas that matters.
Solution (a) Shift to the left (b) Shift to the right (c) No shift
(Continued on next page)