CHAPTER 13: CHEMICAL EQUILIBRIUM
Questions 29,33,39,41,43,45,47,51 ANSWERS
2
PNO
 PO2
= 6.3 × 10 13
Kp =
33.
Solids and liquids do not appear in equilibrium expressions. Only gases and dissolved solutes
appear in equilibrium expressions.
a. K =
2
PNO
2
=
(6.5  10 5 ) 2 (4.5  10 5 )
29.
(0.55) 2
PH 2O
[H 2O]
; Kp = 2
2
PNH3  PCO 2
[ NH3 ] [CO2 ]
c. K = [O2]3; Kp = PO3 2
39.
H2O(g) + Cl2O(g)  2 HOCl(g)
3
b. K = [N2][Br2]3; Kp = PN 2  PBr
2
d. K =
PH 2 O
[H 2O]
; Kp =
PH 2
[H 2 ]
[HOCl]2
= 0.0900
[H 2 O][Cl2 O]
K=
Use the reaction quotient Q to determine which way the reaction shifts to reach equilibrium. For
the reaction quotient, initial concentrations given in a problem are used to calculate the value for
Q. If Q < K, then the reaction shifts right to reach equilibrium. If Q > K, then the reaction shifts
left to reach equilibrium. If Q = K, then the reaction does not shift in either direction because the
reaction is already at equilibrium.
2
a.
 1.0 mol 


2
1.0 L 
[HOCl]0


 1.0 × 102
Q 
[H 2 O]0 [Cl 2 O]0
 0.10 mol  0.10 mol 



 1.0 L  1.0 L 
Q > K, so the reaction shifts left to produce more reactants to reach equilibrium.
2
b.
 0.084 mol 


2.0 L 

 0.090 = K; at equilibrium
Q 
 0.98 mol   0.080 mol 

 

 2.0 L   2.0 L 
c.
 0.25 mol 


3.0 L 

 110
Q 
 0.56 mol   0.0010 mol 

 

 3.0 L   3.0 L 
2
Q > K, so the reaction shifts to the left to reach equilibrium.
41.
CaCO3(s)
⇌ CaO(s) + CO2(g)
Kp = PCO 2 = 1.04
a. Q = PCO2 ; we only need the partial pressure of CO2 to determine Q because solids do not
appear in equilibrium expressions (or Q expressions). At this temperature, all CO2 will be in
the gas phase. Q = 2.55, so Q > Kp; the reaction will shift to the left to reach equilibrium; the
mass of CaO will decrease.
b. Q = 1.04 = Kp, so the reaction is at equilibrium; mass of CaO will not change.
c. Q = 1.04 = Kp, so the reaction is at equilibrium; mass of CaO will not change.
d. Q = 0.211 < Kp; the reaction will shift to the right to reach equilibrium; mass of CaO will
increase.
43.
K=
[ H 2 ]2 [ O 2 ]
[H 2 O]2
, 2.4  10 3 
Moles of O2 = 2.0 L 
45.
(1.9  10 2 ) 2 [O 2 ]
(0.11) 2
, [O2] = 0.080 M
0.080 mol O 2
= 0.16 mol O2
L
SO2(g) + NO2(g) ⇌ SO3(g) + NO(g) K =
[SO 3 ][ NO]
[SO 2 ][ NO2 ]
To determine K, we must calculate the equilibrium concentrations. The initial concentrations are:
2.00 mol
[SO3]0 = [NO]0 = 0; [SO2]0 = [NO2]0 =
= 2.00 M
1.00 L
Next, we determine the change required to reach equilibrium. At equilibrium, [NO] = 1.30
mol/1.00 L = 1.30 M. Because there was zero NO present initially, 1.30 M of SO2 and 1.30 M
NO2 must have reacted to produce 1.30 M NO as well as 1.30 M SO3, all required by the balanced
reaction. The equilibrium concentration for each substance is the sum of the initial concentration
plus the change in concentration necessary to reach equilibrium. The equi-librium concentrations
are:
[SO3] = [NO] = 0 + 1.30 M = 1.30 M; [SO2] = [NO2] = 2.00 M − 1.30 M = 0.70 M
We now use these equilibrium concentrations to calculate K:
K=
47.
[SO3 ][ NO]
(1.30)(1.30)
= 3.4

[SO 2 ][ NO2 ] (0.70)(0.70)
When solving equilibrium problems, a common method to summarize all the information in the
problem is to set up a table. We commonly call this table an ICE table because it summarizes
initial concentrations, changes that must occur to reach equilibrium, and equilibrium
concentrations (the sum of the initial and change columns). For the change column, we will
generally use the variable x, which will be defined as the amount of reactant (or product) that
must react to reach equilibrium. In this problem, the reaction must shift right to reach equilibrium
because there are no products present initially. Therefore, x is defined as the amount of reactant
SO3 that reacts to reach equilibrium, and we use the coefficients in the balanced equation to relate
the net change in SO3 to the net change in SO2 and O2. The general ICE table for this problem is:
2 SO3(g)
Initial
Change
Equil.
⇌
2 SO2(g)
+
O2(g)
K=
[SO 2 ]2 [O 2 ]
[SO 3 ]2
12.0 mol/3.0 L
0
0
Let x mol/L of SO3 react to reach equilibrium.
x

+x
+x/2
4.0  x
x
x/2
From the problem, we are told that the equilibrium SO2 concentration is 3.0 mol/3.0 L =
1.0 M ([SO2]e = 1.0 M). From the ICE table setup, [SO2]e = x, so x = 1.0. Solving for the other
equilibrium concentrations: [SO3]e = 4.0  x = 4.0  1.0 = 3.0 M; [O2] = x/2 = 1.0/2 = 0.50 M.
K=
[SO 2 ]2 [O 2 ] (1.0) 2 (0.50)
=
= 0.056
[SO 3 ]2
(3.0) 2
Alternate method: Fractions in the change column can be avoided (if you want) be defining x
differently. If we were to let 2x mol/L of SO3 react to reach equilibrium, then the ICE table setup
is:
[SO 2 ]2 [O 2 ]
2 SO3(g) ⇌ 2 SO2(g)
+ O2(g)
K=
[SO 3 ]2
Initial
4.0 M
0
0
Let 2x mol/L of SO3 react to reach equilibrium.
Change
2x

+2x
+x
Equil.
4.0  2x
2x
x
Solving: 2x = [SO2]e = 1.0 M, x = 0.50 M; [SO3]e = 4.0  2(0.50) = 3.0 M; [O2]e = x
= 0.50 M
These are exactly the same equilibrium concentrations as solved for previously, thus K will be the
same (as it must be). The moral of the story is to define x in a manner that is most comfortable
for you. Your final answer is independent of how you define x initially.
51.
Q = 1.00, which is less than K. The reaction shifts to the right to reach equilibrium. Summarizing
the equilibrium problem in a table:
SO2(g)
Initial
Change
Equil.
+
NO2(g)
⇌
SO3(g)
0.800 M
0.800 M
0.800 M
x mol/L of SO2 reacts to reach equilibrium
−x
−x
→
+x
0.800 − x
0.800 − x
0.800 + x
+
NO(g)
K = 3.75
0.800 M
+x
0.800 + x
Plug the equilibrium concentrations into the equilibrium constant expression:
K=
[SO 3 ][ NO]
(0.800  x) 2
, 3.75 
; take the square root of both sides and solve
[SO 2 ][ NO2 ]
(0.800  x) 2 for x:
0.800  x
= 1.94, 0.800 + x = 1.55 − (1.94)x, (2.94)x = 0.75, x = 0.26 M
0.800  x
The equilibrium concentrations are:
[SO3] = [NO] = 0.800 + x = 0.800 + 0.26 = 1.06 M; [SO2] = [NO2] = 0.800 − x = 0.54 M