1
Chapter 15 - Homework Solutions
15.3)
The following diagram represents an equilibrium mixture for a reaction,
A + X
º AX
(Copyright by Prentice Hall)
Red = A
Blue = X
Red-Blue = AX
If the volume is 1 L, is K greater or smaller than 1?
Since the question gives us volume but no other info we will assume this is Kc.
This means we need molar concentrations. While we specifically have particles
this is related to moles and mole ratios and particle ratios are the same. So we can
use the number of particles in place of moles to determine molarities.
[A] = 3/1L = 3 M
[X] = 2/1L = 2 M
[AX]
8
8
Kc = ----------- = ----------- = ----- = 1.33
[A] [X]
(3) (2)
6
[AX] = 8/1L = 8 M
2
15.6)
The reaction A2 + B2 º 2 AB has an equilibrium constant Kc = 1.5 . The
diagrams below represent reaction mixtures containing reactants and products.
(Copyright by Prentice Hall)
Red = A2
Blue = B2
Red-Blue = AB
[AB]2
Kc = ------------- = 1.5
[A2] [B2]
Express the contents in the boxes in terms of concentrations (molarities as in
15.3 above). Need to calculate the reaction quotient, Qc, and compare to Kc.
Remember, products in numerator, reactants in denominator
Q=K
at equilibrium
Q>K
not at equil.: numerator too large, denominator too small
Too much product, not enough reactant to be at equilibrium
- reaction must proceed from right to left in the reverse
direction toward reactant to reach equil. (use reactant &
produce product)
Q<K
not at equil.: numerator too small, denominator too large Not enough product, too much reactant, to be at
equilibrium - reaction must proceed from left to right in
the forward direction toward product to reach equil. (use
reactant & produce product)
3
15.6) (cont.)
a) & b)
Which reaction(s) are at equilibrium and if not what direction do the
proceed to reach equilibirum?
Mixture 1: 1 A2, 1 B2, 6 AB
[AB]2
(6)2
Qc = ------------- = ---------- = 36
[A2] [B2]
(1) (1)
Qc > Kc
36 > 1.5
The mixture is not at equilibrium. Since Q is larger than K there is too much
product and the reaction proceeds to the left (reverse direction) toward
reactants.
Mixture 2: 3 A2, 2 B2, 3 AB
[AB]2
(3)2
Qc = ------------- = ---------- = 1.5
(3) (2)
[A2] [B2]
Qc = Kc
1.5 = 1.5
The mixture is at equilibrium.
Mixture 3: 3 A2, 3 B2, 2 AB
[AB]2
(2)2
Qc = ------------- = ---------- = 0.44
[A2] [B2]
(3) (3)
Qc < Kc
0.44 < 1.5
The mixture is not at equilibrium. Since Q is smaller than K there is too
much reactant and the reaction proceeds to the right (forward direction)
toward products.
4
15.7)
The reaction A2(g) + B(g) º A(g) + AB(g) has an equilibrium constant Kp =
2 . The diagram below shows a mixture containing A atoms, A2 molecules and AB
molecules.
(Copyright by Prentice Hall)
Red = A atoms
Blue = B atoms
Red-Blue = AB molec.
How many B atoms should be shown in the diagram if the system is at equilibrium?
You are given Kp but since )n=0 (change in moles of gas) Kp = Kc. It is easier to
work the problem using Kc. As in 15.3, we can use the # particles in place of # of
moles for molarity. Thus we have [A] = 2 M, [AB] = 4 M, [A2] = 2 M. We want to
determine [B]. Solve Kc expression for [B].
Kp = Kc.
since )n = 0 ()n = mol gaseous products - mol gaseous
reactants)
[A] [AB]
Kc = ------------- = 2
[A2] [B]
[A] [AB]
(2) (4)
[B] = ------------- = --------- = 2 M
[A2] Kc
(2) (2)
Therefore, 2 B’s (blue spheres) should be added to the diagram.
5
15.9)
The reaction A2(g) + B(g) º A(g) + AB(g) is at equilibrium. The diagrams
below show mixtures containing A atoms, B atoms, A2 molecules and AB
molecules at 300 K (diagram a) and 500 K (diagram b).
(Copyright by Prentice Hall)
Red = A atoms
Blue = B atoms
Is the reaction exothermic or endothermic?
For temperature changes you treat heat as a reactant or product. Also, changes in
temperature are the only changes that cause a numerical change in the
equilibrium constant, K.
Remember:
)H < 0 (-);
exothermic;
heat is a product
- inc T (add heat, a product), shift left (K dec)
- dec T (remove heat, a product), shift right (K inc)
Reactants
º Products + Heat
)H > 0 (+);
endothermic;
heat is a reactant
- inc T (add heat, a reactant), shift right (K inc)
- dec T (remove heat, a reactant), shift left (K dec)
Heat + Reactants
º Products
* continued on next page *
6
15.9) (cont.)
)H < 0
)H > 0
T 8, 7, K 9
T 8, 6, K 8
T 9, 6, K 8
T 9, 7, K 9
T & K move in
opposite direction
T & K move in
same direction
Product favored by
LOW T
Product favored by
HIGH T
When it says product is favored by low T it means that low temperature produces
more product. When it says product is favored by high T it means that high
temperature produces more product. In neither case does it mean that we always
want product. There are times we don’t want product (for instance, if going from
reactant to product means an explosion we don’t want product to form).
Diagram (a), 300 K:
[A] = 3 M, [AB] = 5 M, [A2] = 1 M, [B] = 1 M
[A] [AB]
(3) (5)
Kc = ------------- = ---------- = 15
(1) (1)
[A2] [B]
Diagram (b), 500 K:
[A] = 1 M, [AB] = 3 M, [A2] = 3 M, [B] = 3 M
[A] [AB]
(1) (3)
Kc = ------------- = ---------- = 0.333
[A2] [B]
(3) (3)
Since Kc decreases as T increases (move in opposite directions) this means the
reaction is exothermic (shifts to left, reverse direction, as temp. inc.).
7
The following are some general comments about equilibrium and its relationship to
rates, rate constants and concentrations of reactants and products.
8
15.12)
9
15.13)
10
15.16)
11
15.18)
This problem relates Kp and Kc. They are not necessarily numerically equal. A
relationship can be derived between the two, assuming ideal gas behavior. The
following is the derivation for the balanced equation given , but holds in general.
a A (g) + b B (g)
º x X (g) + y Y (g)
[X]x [Y]y
Kc = ------------[A]a [B]b
(PX)x (PY)y
Kp = --------------(PA)a (PB)b
Assuming IGL behavior:
PA = (nA/V) RT = (MA) RT
PA = [A] RT
Substitute this for each partial pressure in Kp
([X] RT)x ([Y] RT)y
[X]x [Y]y (RT)x (RT)y
Kp = -------------------------= --------------------------([A] RT)a ([B]) RT)b
[A]a [B]b (RT)a (RT)b
= Kc (RT){(x +y) - (a + b)}
Kp = Kc (RT))n
)n = {(x +y) - (a + b)}
mol gas mol gas
product reactant
(Change in moles of gas).
R = 0.08206 LCatm/molCK,
Calculate Kc at 303 K for SO2 (g) + Cl2 (g)
T in Kelvin
º SO2Cl2 (g) given Kp =34.5
n = 1 & (1 +1) = &1
Kc = Kp (RT)&)n
= (34.5) {(0.0821 LCatm/molCK)(303 K)}& (& 1) = (34.5) (24.86418)
= 857.81 = 858
12
15.21)
º SO3 (g)
SO2 (g) + 1/2 O2 (g)
Kp = 1.85
(1)
at 1000 K
PSO3
Kp1 = ----------------(PSO2) (PO2)1/2
When you multiply a rxn by some factor the K for the new rxn is equal to original
K raised to the power of the factor the rxn was multiplied by (the multiplier).
Proof
aA + bB
W
xX + yY
(1)
[X]x [Y]y
K1 = ------------[A]a [B]b
Multiply by n
(aCn) A + (bCn) B
W
(xCn) X + (yCn) Y
(2)
[X]x [Y]y n
[X]xn [Y]yn
K2 = --------------= (-------------)
= K1n
an
bn
a
b
[A] [B]
[A] [B]
Multiply by -1 reverses rxn.
-aA + - bB
W
-xX + -yY
want (+) coeff - like in math switch reactants & products to other sides
xX + yY
W
aA + bB
Krev for rev. rx is 1/Kfor or (Kfor)&1 (reciprocal of Kfor for the forward rx)
Krev = (Kfor)&1
13
15.21) (cont.)
a) Calculate K for the reverse rxn, Krev
SO3 (g)
Krev
º SO2 (g) + 1/2 O2 (g)
(2)
(PSO2) (PO2)1/2
1
= ----------------- = ------ = (Kfor)&1
PSO3
Kfor
Krev = 1/(1.85) = 0.54054 = 0.541
b) Calculate K for the following reaction:
2 SO2 (g) + O2 (g)
º 2 SO3 (g)
(3)
This is equation (1) multiplied by a factor of 2.
Kp1
(PSO3)2
= ---------------(PSO2)2 (PO2)
Kp3 = (Kp1)2 = (1.85)2 = 3.4225 = 3.42
c) Calculate Kc3 (for reaction (3) in part b).
Need )n: )n = 2 & (2 +1) = &1
Kp = Kc (RT))n
Kc = Kp (RT)&)n
= (3.4225) {(0.0821 LCatm/molCK)(1000 K)}& (& 1) = (3.4225) (82.1)
= 280.987
= 281
14
15.23)
Find K for a reaction that is the sum of two reactions.
A (aq) + B (aq)
W
K1 = 1.9 x 10-4
C (aq)
C (aq) + D (aq) W E (aq) + A (aq)
----------------------------------------------------B (aq) + D (aq)
W
E (aq)
K2 = 8.5 x 102
K3
If two rxn’s (eqn’s) can be added to give a 3rd rxn (eqn) then K3 is the product of
the K1 and K2.
Rx1 + Rx2 = Rx3
K1 * K2 = K3
[C]
[E] [A]
[E]
---------- * ---------- = ---------[A] [B] [C] [D]
[B] [D]
([C] and [A] cancel out)
For the reactions given we have the following:
CoO (s) + H2 (g)
W
Co (s) + H2O (g)
Kc1 = 67
CoO (s) + CO (aq)
W
Co (s) + CO2 (g)
Kc2 = 490
15
15.23) (cont.)
We need to turn eqn 2 around (reverse it) so the CO2 is a reactant and the CO is a
product. This means Kc2, rev = Kc2!1. We add the eqns and cancel things that appear
on both sides.
CoO (s) + H2 (g)
W
Co (s) + H2O (g)
Co (s) + CO2 (g) W CoO (s) + CO (aq)
---------------------------------------------------------H2 (g) + CO2 (g)
W
K3 = Kc1 * Kc2, rev
= (67) (1/490)
= 0.13673
= 0.14
CO (aq) + H2O (g)
Kc1 = 67
Kc2, rev = 1/490
K3
16
15.25)
2 Hg2O (s)
º 4 Hg (R) + O2 (g)
What would Kc look like if you included everything (including pure solid and liq)?
[Hg]4 [O2]
Kc = -------------[Hg2O]2
However, we normally exclude pure solids and liquids. Their molar conc. are
essentially constant so their conc. are incorporated into the equilibrium constant.
Kc [Hg2O]2
K = --------------= [O2]
[Hg]4
’
c
Kc = [O2]
a) The same thing is done for Kp (ignore solids and liquids).
Kp = PO
2
b)
As noted above, the molar concentrations for solids and liquids are constant.
Their concentration is a ratio of moles to volume occupied (which is related
to their density). This doesn’t change whether you have a small or large
amounts.
NOTE:
PURE SOLIDS and LIQUIDS do NOT appear in the expression for K or Q.
(Aqueous and gaseous substances should be in K or Q.)
17
15.27)
2 HI (g)
º H2 (g) + I2 (g)
At 425°C,
[HI] = 3.53 x 10&3 M
[H2] = 4.79 x10&4 M
[I2] = 4.79 x 10&4 M
[H2] [I2]
(4.79 x10&4) (4.79 x10&4)
Kc = ------------= -------------------------------= 0.01841 = 0.0184
(3.53 x 10&3)2
[HI]2
18
15.30)
15.32)
Given initial conc. and asked for K
Set up an Equilibrium Table (ICE table)
This table should be done in Molarity (M) or Pressure (atm)
If you use moles in the table you must convert the moles to conc. (M) or
pressure (atm) before substituting into the expression for K.
* See Next Page *
19
15.32) (cont.)
20
15.34)
21
15.36)
22
15.38)
N2 (g) + 3 H2 (g)
º 2 NH3 (g)
(PNH )2
Kp = --------------= 4.51 x 10!5
3
(PN ) (PH )
3
2
at 450 °C
2
(Qp has the same form)
a)
PNH = 98 atm, PN = 45 atm,
3
2
PH = 55 atm
2
(PNH )2
(98)2
Qp = --------------= -------------= 1.282 x 10!3 = 1.3 x 10!3
3
3
(PN ) (PH )
(45) (55)
3
2
Qp > Kp
2
too much product, not enough reactant; the reaction shifts left
toward reactants (reverse direction) to achieve equilibrium.
b)
PNH = 57 atm,
PN = 143 atm,
3
2
PH = 0 atm (no H2)
2
(PNH )2
(57)2
Qp = --------------= -------------= 4
(PN ) (PH )3
(143) (0)3
3
2
Qp > Kp
2
too much product, not enough reactant; the reaction shifts left
toward reactants (reverse direction) to achieve equilibrium. This
one should be obvious. There is no H2 (a reactant) initially. The
only way to get H2 and reach equil. is for the reaction to move
left. As NH3 decomposes H2 forms. There must be some H2
present to be at equil.
c)
PNH = 13 atm,
PN = 27 atm,
3
2
PH = 82 atm
2
(PNH )2
(13)2
Qp = --------------= -------------= 1.135 x 10!5 = 1.1 x 10!5
3
3
(PN ) (PH )
(27) (82)
3
2
Qp < Kp
2
not enough product, too much reactant; the reaction must shift to
right toward products (forward direction) to achieve
equilibrium.
23
15.39)
SO2Cl2 (g)
º SO2 (g) + Cl2 (g)
[SO2] [Cl2]
Kc = ---------------- = 0.078
[SO2Cl2]
at 100°C
Given Kc and molarities of SO2Cl2 and SO2. Want partial pressure of Cl2.
[SO2] = 0.052M
[SO2Cl2] = 0.108 M
Easiest thing to do is to find conc. of Cl2, [Cl2], and then its partial pressure.
[SO2Cl2] Kc
(0.108) (0.078)
[Cl2] = ---------------- = --------------------- = 0.162 M
[SO2] [Cl2]
(0.052)
Use Ideal Gas Law, PV = nRT
PCl = (n/V) RT = (M) RT = [Cl2] RT = (0.162 M)(0.08206 atm/MCK)(373K)
2
= 4.958 atm = 5.0 atm
15.41)
24
15.41) (cont.)
b)
H2 (g) + I2 (g)
º 2 HI (g)
Kc = 55.3
at 700 K
An equilibrium mixture in a 2.00-L flask has 0.056 g H2 and 4.36 g I2. What
is the mass of HI in the flask?
0.056 g H2
1 mol H2
[H2] = --------------- × ------------------ = 0.013889 M = 0.014 M
2.00 L
2.01588 g H2
4.36 g I2
1 mol I2
[I2] = ------------- × -------------------- = 0.008589 M = 0.00859 M
2.00 L
253.80894 g I2
[HI]2
Kc = -----------[H2] [I2]
[HI] = {Kc [H2] [I2]}1/2
[HI] = {(55.3) (0.013889) (0.008589)}1/2
[HI] = 0.081223 = 0.081 M
0.081223 mol HI
127.9124 g HI
? g HI = 2.00 L × ----------------------- × -------------------1 L soln
1 mol HI
= 20.779
= 21 g HI
25
15.44)
Calculate the initial conc. of Br2 and Cl2 and use equil. (ICE) table to find equil. conc.
[Br2] = [Cl2] = 0.25 mol/1.0 L = 0.25 M
ICE Table - In MOLARITY (or atm when Kp involved)
Br2 (g) + Cl2 (g) W 2 BrCl (g)
Initial
0.25 M
0.25 M
0
!x
!x
+ 2x
Change
-----------------------------------------------------Equil
(0.25 ! x)
(0.25 ! x)
2x
[BrCl]2
Kc = --------------- = 7.0
[Br2] [Cl2]
(2x)2
(2x)2
Kc = --------------------------- = --------------= 7.0
(0.25 ! x)2
(0.25 ! x) (0.25 ! x)
This eqn is a perfect square - we can take the square root of both sides.
(2x)
--------------- = (7.0)1/2 = 2.645
(0.25 ! x)
2.645 (0.25 ! x) = 2x
0.6614 ! (2.645)x = 2x
0.6614 = (4.645)x
x = 0.1423 = 0.14 M
[Br2] = (0.25 ! x) = (0.25 ! 0.1423) = 0.1076 = 0.11 M
[Cl2] = (0.25 ! x) = (0.25 ! 0.1423) = 0.1076 = 0.11 M
[BrCl] = 2x = 2 (0.1423) = 0.2847 = 0.28 M
Check:
Kc = (0.2847)2/(0.1076)2 = 7.0
26
15.46)
Solid NH4HS is placed in a closed container and decomposes until equilibrium is
reached. What are the concentrations of NH3 and H2S at equilibrium?
NH4HS (s) W NH3 (g) + H2S (g)
initial
C
0M
0M
change
-x
+x
+x
----------------------------------------------------------------------equil
constant
+x
+x
Kc = [NH3] C [H2S] = 1.2 x 10-4
x2 = 1.2 x 10-4
NH4HS (s) doesn’t appear in K
(pure solids and liquids do not
appear in K)
x = (1.2 x 10-4)1/2 = 1.0954 x 10-2 M
[NH3] = [H2S] = 1.1 x 10-2 M
15.47)
a)
Solid CaSO4 is placed in a closed container and decomposes until equilibrium
is reached. What are the concentrations of Ca2+ and SO42& at equilibrium?
CaSO4 (s) W Ca2+ (aq) + SO42& (aq)
initial
C
0M
0M
change
-x
+x
+x
------------------------------------------------------------------------equil
constant
+x
+x
Kc = [Ca2+] C [SO42&] = 2.4 x 10-5
x2 = 2.4 x 10-5
CaSO4 (s) doesn’t appear in K
(pure solids and liquids do not
appear in K)
x = (2.4 x 10-5)1/2 = 4.8989 x 10-3 M
[Ca2+] = [SO42&] = 4.9 x 10-3 M
27
15.47) (cont.)
b) What is the minimum mass of CaSO4(s) needed to reach equil. in 3.0 L of soln.?
A saturated soln of CaSO4(aq) is 4.9 x 10-3 M (this is the dissolved amount of
CaSO4, which is [Ca2+] = [SO42&])
4.9 x 10-3 mol
136.14 g CaSO4
? g CaSO4 = 3.0 L soln × ------------------- × ---------------------- = 2.0 g CaSO4
1 L soln
1 mol CaSO4
15.49)
I2 (g) + Br2 (g)
W
2 IBr (g)
Kc = 280
at 150°C
The reaction is started with 0.500 mol IBr in a 1.00-L flask. Since no I2 or Br2 are
present initially the reaction proceeds in the reverse direction (right to left) toward
the reactants. Need to set up an ICE table. Since the reaction goes in the reverse
direction the negative x is on the right-hand side of the table and the plus x’s go on
the left-hand side.
[IBr] = 0.500 mol/1.00 L = 0.500 M
I2
+
Br2
W
2 IBr
initial
0.000
0.000
0.500
change
+x
+x
& 2x
------------------------------------------------------------equil
x
x
0.500 & 2x
[IBr]2
(0.500 & 2x)2
(0.500 & 2x)2
Kc = ------------- = ------------------- = -----------------= 280
[I2] [Br2]
(x) (x)
x2
Even though the reaction is going right to left (reverse direction), you still write Kc
for the balanced equation as written. That is the equation the given Kc refers to.
* continued on next page *
28
15.49) (cont.)
(0.500 & 2x)2
-----------------= 280 perfect square (take square root of both sides)
x2
(0.500 & 2x)
----------------- = (280)1/2 = 16.733
x
0.500 & 2x = (16.733) (x)
0.500 = (16.733) x + 2x
0.500 = (18.733) x
x = 0.02669 M
[I2] = [Br2] = 0.027 M
[IBr] = 0.500 & 2 (0.02669) = 0.44661 M = 0.447 M