17
Chemical
Equilibrium 化學平衡
1
Chapter Goals
1. Basic Concepts
2. The Equilibrium Constant 平衡常數
3. Variation of Kc with the Form of the Balanced
Equation
4. The Reaction Quotient 反應商
5. Uses of the Equilibrium Constant, Kc
6. Disturbing a System at Equilibrium: Predictions
7. The Haber Process: A Commercial Application of
Equilibrium
8. Disturbing a System at Equilibrium: Calculations
9. Partial Pressures and the Equilibrium Constant
10. Relationship between Kp and Kc
11. Heterogeneous Equilibria
12. Relationship between Gorxn and the Equilibrium
Constant
13. Evaluation of Equilibrium Constants at Different
Temperatures
2
Basic Concepts基本概念
•Chemical reactions that can occur in either
direction are called reversible reaction可逆反應
•Reversible reactions do not go to completion
 Reactants are not completely converted to
products. (反應物不會完全轉成產物)
– They can occur in either direction
– Symbolically, this is represented as:
aA(g)+bB(g )  cC(g)+dD(g)
When A and B react to form C and D at the same rate at
which C and D react to form A and B, the system is
equilibrium (當A與B反應形成C和D的速率與C及D反應形成A
和B的速率相同時稱之為平衡)
3
Basic Concepts 基本概念
• Chemical equilibrium exists when two opposing
reactions occur simultaneously at the same rate.化
學平衡是指在可逆反應中，正逆反應速率相等，反應物和
生成物各組分濃度不再改變的狀態。
– A chemical equilibrium is a reversible reaction
that the forward reaction rate is equal to the
reverse reaction rate (化學平衡為可逆反應其正向反
應與反向反應的速率相同)
• Chemical equilibria are dynamic
equilibria (動態
平衡)
– Molecules are continually reacting, even though
the overall composition of the reaction mixture
does not change
4
Basic Concepts
• One example of a dynamic equilibrium can be
shown using radioactive 131I as a tracer in a
saturated PbI2 solution. (利用放射線碘131當作追蹤劑,
看放射線碘存在何處)
1. place solid PbI2* in a saturated PbI2 solution
H2O
*
PbI2(s)
Pb2+(aq)+2I-(aq)
2. Stir for a few minutes, then filter the solution
some of the radioactive iodine will go into solution
將固體PbI2置於水中,攪拌數分後,再經過過濾
一些放射線碘存於溶液中
5
Basic Concepts
• Graphically, this is a representation of the rates for
the forward and reverse reactions for this general
reaction
aA(g)+bB(g )  cC(g)+dD(g)
反應開始
Equilibrium is
established
達成平衡狀態
6
Basic Concepts
• One of the fundamental ideas of chemical
equilibrium is that equilibrium can be established
from either the forward or reverse direction
2SO2(g)+ O2(g )  2SO3(g)
達成平衡
0.02M
7
Basic Concepts
2SO2(g)+ O2(g )  2SO3(g)
開始莫耳數
反應改變莫耳數
反應後莫耳數
0.400mol 0.200mol
-0.056mol -0.028mol
0.344mol 0.172mol
2:
1:
0
+0.056mol
0.056mol
2
2SO2(g)+ O2(g )  2SO3(g)
開始莫耳數
反應改變莫耳數
反應後莫耳數
0
0
+0.424mol +0.212mol
0.424mol 0.212mol
2:
1:
0.500mol
-0.424mol
0.076mol
2
8
Basic Concepts
In 1.00 liter container 反應均為氣體,在固定體積
Initial conc.
Change due to rxn
Equilibrium conc’n
平衡濃度
Initial conc.
Change due to rxn
Equilibrium conc’n
2SO2(g)+ O2(g )  2SO3(g)
0.400M
-0.056M
0.344M
2:
0.200M
-0.028M
0.172M
1:
0
+0.056M
0.056M
2
0
0
+0.424M +0.212M
0.424M 0.212M
2:
1:
0.500M
-0.424M
0.076M
2
2SO2(g)+ O2(g )  2SO3(g)
9
The Equilibrium Constant
• For a simple one-step mechanism reversible
reaction such as:
A(g)+B(g )  C(g)+D(g)
• The rates of the forward and reverse
reactions can be represented as:
Forward rate (正反應速率): Ratef = kf[A][B]
Reverse rate (逆反應速率): Rater = kr[C][D]
10
The Equilibrium Constant
• When system is at equilibrium 當系統達成平衡
Ratef = Rater 正反應速率=逆反應速率
which represents the forward rate
kf[A][B] = kr[C][D]
which rearranges to
kf
[C][D]
=
kr
[A][B]
• Because the ratio of two constants is a constant we
can define a new constant as follows :
kf
[C][D]
kc =
= kc
kr
[A][B]
11
The Equilibrium Constant
• Similarly, for the general reaction:
aA(g)+bB(g )  cC(g)+dD(g)
we can define a constant: 平衡常數 Kc
[C]c[D]d
Kc=
[A]a[B]b
Products 產物
Reactants 反應物
This expression is valid for all reactions
12
The Equilibrium Constant
• Kc is the equilibrium constant平衡常數 .
• Kc is defined for a reversible reaction at a given
temperature as the product of the equilibrium
concentrations (in M) of the products, each raised
to a power equal to its stoichiometric coefficient in
the balanced equation, divided by the product of
the equilibrium concentrations (in M) of the
reactants, each raised to a power equal to its
stoichiometric coefficient in the balanced equation.
各物種的體積莫耳濃度均為平衡時的濃度。Kc的數值等於
方程式中各生成物濃度的係數次方相乘後，再除以各反應
物濃度的係數次方。定溫時無論反應的初濃度如何改變，
只要達到平衡時，其平衡常數均相等。
• 此常數的大小僅與物種、溫度有關，而與濃度、壓力的大
13
小無關。
The Equilibrium Constant
• Example 17-1: Write equilibrium constant
expressions for the following reactions at 500oC. All
reactants and products are gases at 500oC.
PCl5  PCl3 + Cl2
[PCl3][Cl2]
Kc=
[PCl5]
H2 + l2  2HI
Kc=
[HI]2
[I2] [I2]
4NH3 + 5O2  4NO + 6H2 O
[NO]4[H2O]6
Kc=
[NH3]4[O2]5
14
The Equilibrium Constant
Example 17-1: Calculation of Kc
Some nitrogen and hydrogen are placed in an empty 5.00liter container at 500oC. When equilibrium is established,
3.01mol of N2, 2.10 mol of H2, and 0.565 mol of NH3 are
present. Evaluate Kc for the following reaction at 500oC.
N2(g) + 3H2(g)  2NH3(g)
[N2]: 3.01mol/5L = 0.602 M
[H2]: 2.10mol/5L = 0.420 M
[NH3]: 0.565mol/5L = 0.113 M
(0.113)2
[NH3]2
Kc=
= 0.286
=
3
3
[N2][H2]
(0.602)(0.420)
15
The Equilibrium Constant
Example 17-2: One liter of equilibrium mixture from the
following system at a high temperature was found to
contain 0.172 mole of phosphorus trichloride, 0.086 mole
of chlorine, and 0.028 mole of phosphorus pentachloride.
Calculate Kc for the reaction.
One liter
Equil []’s M
PCl5  PCl3 + Cl2
0.028M 0.172M
[PCl3][Cl2]
Kc=
[PCl5]
(0.172)(0.086)
=
(0.028)
Kc= 0.53
0.086M
16
The Equilibrium Constant
Example 17-3: The decomposition of PCl5 was studied at
another temperature. One mole of PCl5 was introduced
into an evacuated 1.00 liter container. The system was
allowed to reach equilibrium at the new temperature. At
equilibrium 0.60 mole of PCl3 was present in the
container. Calculate the equilibrium constant at this
temperature.
Initial
Change
Equilibrium conc’n
PCl5  PCl3 + Cl2
1.00M
0
0
-0.60M
0.60M 0.60M
0.40M
0.60M 0.60M
[PCl3][Cl2]
K’c=
[PCl5]
(0.60)(0.60)
=
=0.90
(0.40)
At another temperature
17
The Equilibrium Constant
Example 17-4: At a given temperature 0.80 mole of N2 and
0.90 mole of H2 were placed in an evacuated 1.00-liter
container. At equilibrium 0.20 mole of NH3 was present.
Calculate Kc for the reaction.
N2: 0.8mole/1Liter = 0.8M H2: 0.9mol/1Liter = 0.9M
NH3: 0.2mol/1Liter = 0.2M
Initial
Change
Equilibrium conc’n
N2 + 3H2  2NH3
0.80M
0.90M
-0.10M -0.30M
0.70M
0.60M
[NH3]2
Kc=
[N2][H2]3
(0.20)2
=
(0.70)(0.60)3
0
0.20M
0.20M
=0.26
18
The Equilibrium Constant
Example 17-2: Calculation of Kc
We put 10.0 mol of N2O into a 2-L container at some
temperature, where it decomposes according to
2N2O (g)  2N2(g) + O2(g)
At equilibrium, 2.20 moles of N2O remain, Calculate the
value of Kc for the reaction
Initial [N2O]: 10.0mol/2L = 5.0 M
equili [N2O]: 2.20mol/2L = 1.1 M
Initial
Change
Equilibrium conc’n
2N2O (g)  2N2(g) + O2(g)
5.0M
-3.9M
1.1M
0
+3.9M
3.9M
(3.9)2(1.95)
[N2]2[O2]
Kc=
=
=
2
2
[N2O]
(1.1)
0
+1.95M
1.95M
24.5
19
Variation of Kc with the
Form of the Balanced Equation
• The value of Kc depends upon how the
balanced equation is written.
• From example 17-2 we have this reaction:
PCl5  PCl3 + Cl2
• This reaction has a
Kc=[PCl3][Cl2]/[PCl5]=0.53
20
Variation of Kc with the
Form of the Balanced Equation
Example 17-5: Calculate the equilibrium constant for the
reverse reaction by two methods, i.e, the equilibrium
constant for this reaction.
PCl3 + Cl2  PCl5
Equil. []’s 0.172M 0.086M 0.028 M
The concentrations are from Example 17-2.
[PCl5]
[PCl3][Cl2]
(0.028)
=
(0.172)(0.086)
K’c= 1.9
K’c =
21
Variation of Kc with the
Form of the Balanced Equation
K’c =
Kc =
[PCl5]
[PCl3][Cl2]
1
K’c
or
=
(0.028)
=1.9
(0.172)(0.086)
K’c=
1 = 1 =1.9
Kc
0.53
Large equilibrium constants indicate that
most of the reactants are converted to
products. (大的平衡常數表示大部分的反應物轉
成產物)
 Small equilibrium constants indicate that
only small amounts of products are formed.
(小平衡常數表示僅少數產物生成)
22

• 平衡狀態可由任一方向達成，其與反應物(A,B) ，
及生成物(C,D)之初濃度有關
---反應物之濃度大於平衡濃度
反應由反應物向生成物方向而達平衡
---生成物之初濃度大於平衡濃度
反應由生成物向反應物而達平衡
• Kc定溫下為常數，其值僅隨溫度改變而改變
• 不同之平衡狀態，平衡濃度值 ([A]、[B]、[C]、
[D])可以不同，但其比值恆等於Kc
• Kc值大小無法決定達成平衡之移動方向
---值大：平衡時，生成物較反應物多
---值小：平衡時，反應物較生成物
The Reaction Quotient
反應商數
• The mass action expression質量作用表示法 or reaction
quotient反應商數has the symbol Q.
– Q has the same form as Kc (Q即是Kc的另一表示形式)
• The major difference between Q and Kc is that the
concentrations used in Q are not necessarily
equilibrium values. (Q並不一定是達成平衡的濃度)
For this general reaction
aA(g)+bB(g )  cC(g)+dD(g)
[C]c[D]d
Q=
[A]a[B]d
Not necessarily
equilibrium
concentrations
24
The Reaction Quotient
• Why do we need another “equilibrium constant”
that does not use equilibrium concentrations?
• Q will help us predict how the equilibrium will
respond to an applied stress. Q值可用於預期反應受到
外力影響時 的反應方向
• To make this prediction we compare Q with Kc.
25
The Reaction Quotient
僅有產物
僅有反應物
When:
Q<Kc Forward reaction predominates until
equilibrium is established (反應向右)
Q=Kc The system is at equilibrium (達成平衡)
Q>Kc Reverse reaction predominates until
equilibrium is established (反應向左)
26
The Reaction Quotient
Example 17-6: The equilibrium constant for the following
reaction is 49 at 450oC. If 0.22 mole of I2, 0.22 mole of H2,
and 0.66 mole of HI were put into an evacuated 1.00-liter
container, would the system be at equilibrium? If not,
what must occur to establish equilibrium?
The concentrations given in the problem are not
necessarily equilibrium []’s. We can calculate Q
H2 + l2  2HI
0.22M
0.22M
0.66M
[HI]2
(0.66)2
Q=
=9.0
=
(0.22)(0.22)
[I2][H2]
Q=9.0 but Kc=49
Q<Kc
Forward reaction predominates until equilibrium is
established (反應會持續往右進行,直至達到平衡)
27
Uses of the Equilibrium
Constant,
K
c , is 3.00 for the
Example 17-7: The equilibrium constant, K
c
following reaction at a given temperature. If 1.00 mole
of SO2 and 1.00 mole of NO2 are put into an evacuated
2.00 L container and allowed to reach equilibrium, what
will be the concentration of each compound at
equilibrium?
SO2(g) + NO2(g)  SO3(g) +NO(g)
½M
½M
0
0
-x M
-x M
+x M
+x M
Equilibrium (0.5-x)M
(0.5-x)M
xM
xM
[SO3][NO]
(x)(x)
(x)2
Kc =
=3.0 =
=
[SO2][NO2]
(0.5-x)(0.5-x)
(0.5-x)2
x
0.865-1.73x=x
1.73=
0.5-x
x=0.316M=[SO3]=[NO]
[SO2]=[NO2]=0.5-0.316=0.184M
Initial
Change
28
Uses of the Equilibrium
Constant, Kc
Example 17-8: The equilibrium constant is 49 for the
following reaction at 450oC. If 1.00 mole of HI is put into
an evacuated 1.00-liter container and allowed to reach
equilibrium, what will be the equilibrium concentration of
each substance?
H2(g) + I2(g)  2HI(g)
0
0
+x M
+x M
Equilibrium
xM
xM
(1.0-2x)2
[HI]2
Kc=
= 49 =
(x)(x)
[H2][I2]
Initial
Change
7.0= 1.0-2x
x
1.0M
-2x M
1.0-2x M
7.0x=1.0-2x
x=0.11M=[H2]=[I2]
[HI]=1.0-(2x0.11)=0.78M
29
Disturbing a System at
Equilibrium: Predictions
•LeChatelier’s Principle - If a change of conditions
(stress) is applied to a system in equilibrium, the
system responds in the way that best tends to
reduce the stress in reaching a new state of
equilibrium
•勒沙特原理：一平衡系統中，加一影響此反應平衡之因素時，
反應會向抵銷此影響因素的方向進行
•Some possible stresses to a system at
equilibrium are:
1.Changes in concentration of reactants or products.
2.Changes in pressure or volume (for gaseous
reactions)
3.Changes in temperature
•增加反應物濃度或移除生成物時，平衡往生成物方向移動
•氣相反應中，增加壓力或減少反應體積，平衡則往莫耳數減少之方向移動
30
Disturbing a System at
Equlibrium: Predictions
• For convenience we may express the amount
of a gas in terms of its partial pressure rather
than its concentration. 以分壓表示
• To derive this relationship, we must solve the
ideal gas equation理想氣體方程式.
PV=nRT
P=(n/V)RT
because (n/V) has the units mol/L (濃度)
P=MRT
Thus at constant T the partial pressure of a gas is
directly proportional to its concentration
定溫下,一氣體的分壓與其濃度成正比
31
Disturbing a System at
Equlibrium: Predictions
Changes in Concentration of Reactants and/or Products
改變反應物或產物的濃度
• Also true for changes in pressure for reactions involving gases.
– Look at the following system at equilibrium at 450oC.
H2(g) + I2(g)  2HI(g)
[HI]2
Kc=
[H2][I2]
=49
If some H2 is added, Q<Kc (分母變大,分子不變)
This favors the forward reaction (反應往右移動)
Equilibrium will shift to the right or product side
If we remove some H2, Q>Kc (分母變小,分子不變)
This favors the reverse reaction (反應往左進行)
Equilibrium will shift to the left or reactant side
32
Disturbing a System at
Equlibrium: Predictions
Changes in Volume (體積改變)
(and pressure for reactions involving gases)
–Predict what will happen if the volume of this system
at equilibrium is changed by changing the pressure
at constant temperature:
[N2O4]
2NO2(g)  N2O4(g) Kc= [NO ]2
2
If the volume is decreased, which increased the pressure
(體積減少,壓力增大), (V, P, [NO2] and [N2O4])
Q= (2[N2O4])/(2[NO2])2 = (2/4)Kc= (1/2) Kc
Q<Kc
This favors product formation or the forward reaction (反應向右)
If the volume is increased, which decreased the pressure,
Q>Kc
This favors the reactants or the reverse reaction (反應向左)
33
Disturbing a System at
Equlibrium: Predictions
3 Changing the Reaction Temperature (改變溫度)
– Consider the following reaction at equilibrium:
2SO2(g) + O2(g)  2SO3(g) Horxn=-198kJ/mol
Is heat a reactant or product in this reaction?
Heat is a product of this reaction! (放熱反應當作產物)
2SO2(g) + O2(g)  2SO3(g) +198kJ/mol
Increasing the reaction temperature (增加溫度) stresses the
products
This favors the reactant or reverse reaction (反應向左)
Decreasing the reaction temperature stresses the reactants
This favors the product or forward reaction (反應向右)
34
Disturbing a System at
Equlibrium: Predictions
若為放熱反應,提高溫度反應向左
A+BC+D+ heat
若為吸熱反應,提高溫度反應向右
A+B+ heat C+D
Disturbing a System at
Equlibrium: Predictions
•Introduction of a Catalyst 加入催化劑
– Catalysts decrease the activation energy of both the
forward and reverse reaction equally (催化劑會同時降低正
反應及負反應的活化能)
–Catalysts do not affect the position of equilibrium. (
因此催化劑不會改變平衡狀態)
•The concentrations of the products and reactants will
be the same whether a catalyst is introduced or not
•Equilibrium will be established faster with a catalyst (加
入催化劑可加速反應達成平衡)
36
Disturbing a System at
Equlibrium: Predictions
Example 17-9: Given the reaction below at equilibrium in a
closed container at 500oC. How would the equilibrium be
influenced by the following?
2
[NH3]
Kc=
o
N2(g) + 3H2(g)  2NH3(g) H rxn=-92kJ/mol
[N2][H2]3
Factors
a. Increasing the reaction temperature
b. Decreasing the reaction temperature
c. Increasing the pressure by decreasing
the volume
d. Increasing the concentration of H2
e. Decreasing the concentration of NH3
f. Introduction a platinum catalyst
Effect on
reaction procedure
 Left
 Right
 Right
 Right
 Right
No effect
37
Disturbing a System at
Equlibrium: Predictions
Example 17-10: How will an increase in pressure (caused
by decreasing the volume) affect the equilibrium in
each of the following reactions?
Effect on
Factors
a. H2(g) + I2(g)  2HI(g)
b. 4NH3(g)+ 5O2(g)  4NO(g)+6H2O(g)
c. PCl3(g) + Cl2(g)  PCl5(g)
d. 2H2(g) + O2(g)  2H2O(g)
equilibrium
No effect
 Left
 Right
 Right
假設壓力增加兩倍,則濃度增加兩倍
a.
b.
(2)4[NO] 4x(2)6[H2O]6
(2)2[HI]2
Q=
= Kc Q =
(2)[H2]x(2)[I2]
(2)4[NH3]4x(2)5[O2]5
c.
d.
(2)[PCl5]
(2)2[H2O]2
Q=
= 0.5Kc Q = 2
(2)[PCl3]x(2)[Cl2]
(2) [H2]2x(2)[O2]
= 2Kc
= 0.5Kc
Disturbing a System at
Equlibrium: Predictions
Example 17-11: How will an increase in temperature affect
each of the following reactions?
Factors
a. 2NO2 (g)  2N2O4(g) Horxn<0
b. H2(g)+ Cl2(g)  2HCl(g)+92kJ
c. H2(g) + l2(g)  2HI(g) Horxn=25kJ
Effect on
equilibrium
 Left
 Left
 Right
39
The Haber Process: A Practical
Application of Equilibrium
• The Haber process is used for the commercial
production of ammonia哈柏製氨法:為商業化產氨的方式
– This is an enormous industrial process in the US
and many other countries.
– Ammonia is the starting material for fertilizer
production.
• Look at Example 17-9. What conditions did we
predict would be most favorable for the production
of ammonia?
40
The Haber Process: A Practical
Application of Equilibrium
N2(g) + 3H2(g)
Fe & metal oxide
2NH3(g) Horxn=-92kJ/mol
N2 is obtained from liquid air; H2 obtain from coal gas
This reactions is run at T=450oC and P of N2 =200 to 1000atm
G<0 which is favorable H<0 also favorable
S<0 which is unfavorable G=H-TS △G < 0反應自然發生
However the reaction kinetics are very slow at low T
Haber’s solution to this dilemma
1. Increase T to increase rate, but yield is decreased (反應向左)
2. Increase reaction pressure to  right
3. Use excess N2 to  right
4. Remove NH3 periodically to  right
The reaction system never reaches equilibrium because
NH3 is removed . This increase the reaction yield and helps
with the kinetics (由於不斷的移除產物氨,所以無法達成平衡)
41
The Haber Process: A Practical
Application of Equilibrium
This diagram illustrates the
commercial system
devised for the Haber
process.
42
Disturbing a System at
Equilibrium: Calculations
To help with the calculations, we must determine the
direction that the equilibrium will shift by comparing Q
with Kc.
Example 17-12: An equilibrium mixture from the following
reaction was found to contain 0.20 mol/L of A, 0.30 mol/L
of B, and 0.30 mol/L of C. What is the value of Kc for this
reaction?
Equilibrium
Kc=
A(g)  B(g) + C(g)
0.20M 0.3 M 0.3M
[B][C]
[A]
(0.3)(0.3)
=
=0.45
(0.2)
43
Disturbing a System at
Equilibrium: Calculations
• If the volume of the reaction vessel were suddenly
doubled while the temperature remained constant, what
would be the new equilibrium concentrations?
1. Calculate Q, after the volume has been doubled
體積加倍,濃度均減半
A(g)  B(g) + C(g)
0.10M 0.15M 0.15M
[B][C]
Q = [A]
(0.15)(0.15)
=0.22
=
(0.10)
[B][C]
Kc =
[A]
(0.3)(0.3)
= (0.2)
=0.45
Q<Kc
44
Disturbing a System at
Equilibrium: Calculations
• Since Q<Kc the reaction will shift to the right to reestablish
the equilibrium. (O<Kc, 反應向右以達成另一平衡)
2. Use algebra to represent the new concentrations
A(g)  B(g) + C(g)
0.1M 0.15M
0.15M
-x M +x M
+x M
Equilibrium
(0.1-x) M (0.15+x) M (0.15+x) M
2
[B][C]
(0.15+x)
(a+b)2
Kc=
=0.45
=
[A]
(0.1-x)
=a2+2ab+b2
0.045-0.45x=0.0225+0.30x+x2
x2+0.75x-0.0225=0
Initial
Change
45
Disturbing a System at
Equilibrium: Calculations
x2+0.75x-0.0225=0
x= -b 
x=
x=
-0.75 
ax2+bx+c=0
b2-4ac
2a
(0.075)2-4(1)(-0.0225)
2x1
-0.75  0.081
2
x= -0.78 and 0.03M
Since 0<x<0.10
 x=0.03M
[A]=0.10-x=0.07 M
[B]=[C]=0.15+x=0.18 M
46
Disturbing a System at
Equilibrium: Calculations
Example 17-13: Refer to example 17-12. If the initial
volume of the reaction vessel were halved, while the
temperature remains constant, what will the new
equilibrium concentrations be? Recall that the original
concentrations were: [A]=0.20 M, [B]=0.30 M, and
[C]=0.30 M.
Instantaneous
Q=
A(g)  B(g) + C(g) 體積減半,濃度增倍
0.40M 0.60M 0.60M
[B][C]
[A]
(0.6)(0.6)
=
(0.40)
=0.90
Q>Kc thus the equilibrium shifts to the left
or reactant side
47
Disturbing a System at
Equilibrium: Calculations
•Set up the algebraic expressions to determine
the equilibrium concentrations
反應向左
A(g)  B(g) + C(g)
0.40M 0.60M
0.60M
+x M
-x M
-x M
Equilibrium
(0.4+x) M (0.60-x) M (0.60-x) M
(0.60-x)2 0.18+0.45x=0.36-1.2x+x2
[B][C]
=0.45 =
Kc=
(0.4+x) x2-1.65x+0.18=0
[A]
Initial
Change
(-1.65)2-4(1)(0.18)
x=
2x1
1.65 1.42
x=
2
x= 1.5 and 0.12M
1.65 
Since 0<x<0.60
 x=0.12M
[A]=0.40+x=0.52 M
[B]=[C]=0.60-x=0.48 M
48
Disturbing a System at
Equilibrium: Calculations
Example 17-14: A 2.00 liter vessel in which the following
system is in equilibrium contains 1.20 moles of COCl2, 0.60
moles of CO and 0.20 mole of Cl2. Calculate the
equilibrium constant.
CO(g) + Cl2(g)  COCl2(g)
Equilibrium
0.6/2
0.2/2
1.2/2
Equilibrium
0.3M
0.1M
0.6M
(0.6)
[COCl2]
=
Kc =
=20
[CO][Cl2] (0.30)(0.10)
49
Disturbing a System at
Equilibrium: Calculations
An additional 0.80 mole of Cl2 is added to the vessel at the
same temperature. Calculate the molar concentrations
of CO, Cl2, and COCl2 when the new equilibrium is
established. 0.80 mole of Cl2 in 2-liter vessel  0.4M of Cl2
CO(g) + Cl2(g)  COCl2(g)
Orig. Equil.
0.3M
(Stress) Add
(0.6)
Qc= (0.30)(0.50)=4
0.1M
0.6M
+0.4M
Q<Kc
New Initial
0.30 M
0.50 M
0.6 M
-x M
反應向右
Change
-x M
+x M
Equilibrium
(0.3-x) M (0.50-x)M (0.60+x) M
(0.6+x)
[COCl2]
20x2-17x+2.4=0
=
Kc =
=20
(0.3-x)(0.5-x)
[CO][Cl2]
Since 0<x<0.30  x=0.18M
17  (17)2-4(20)(2.3)
[CO]=0.30-x=0.12 M
x=
2x20
[Cl2]=0.5-x=0.32
50
X=0.67 and 0.18
[COCl]=0.6+x=0.78
Partial Pressures and the
Equilibrium Constant
• For gas phase reactions the equilibrium
constants can be expressed in partial
pressures rather than concentrations. (氣態的
反應中,平衡常數可以分壓表示)
• For gases, the pressure is proportional to the
concentration.(氣體的壓力與濃度成正比)
• We can see this by looking at the ideal gas
law.
–
–
–
–
PV = nRT
P = nRT/V
n/V = M
P= MRT and M = P/RT
51
Partial Pressures and the
Equilibrium Constant
• Consider this system at equilibrium at 5000C.
2Cl2(g) + 2H2O (g)  4HCl(g) + O2(g)
[HCl]4[O2]
Kc=
[Cl2]2 [H2O]2
(PHCl)4(PO2)
Kp=
(PCl2)2 (PH2O)2
52
Partial Pressures and the
Equilibrium Constant
P= MRT and M = P/RT
Kc=
PHCl
RT
PCl2
RT
4
2
PO 2
RT
PH2O
RT
Kc =Kp
2
1
RT
(PHCl)4(PO2)
x
=
2
2
(PCl2) (PH2O)
1
RT
5
1
RT
4
so for this reaction
Kc=Kp(RT)-1 or Kp=Kc(RT)1
Must use R = 0.0821 L atm/mol K
53
Relationship Between Kp and
Kc
• From the previous slide we can see that the
relationship between Kp and Kc is:
Kp=Kc(RT)∆n or Kc=Kp(RT)-∆n
∆n= (# of moles of gaseous products) –
(# of moles of gaseous reactants)
∆n= (氣體生成物的莫耳數和) –
(氣體反應物的莫耳數和)
2Cl2(g) + 2H2O (g)  4HCl(g) + O2(g)
∆n= (4+1)-(2+2)=1
Kp=Kc(RT)1or Kc=Kp(RT)-1
54
Relationship Between Kp and
Kc
Example 17-15: Nitrosyl bromide, NOBr, is 34% dissociated
by the following reaction at 25oC, in a vessel in which the
total pressure is 0.25 atmosphere. What is the value of
Kp?
2NOBr(g)  2NO(g) + Br2(g)
0
0
x atm
-0.34x atm
+0.34x atm +0.17x atm
Equilibrium (x-0.34x) atm (0.34x) atm (0.17x)atm
Initial
Change
Ptot=PNOBr + PNO+ PBr2
0.25atm = (x-0.34x)atm + 0.34x atm + 0.17x atm
0.25atm = 1.17x atm, thus x=0.21atm
55
Relationship Between Kp and
Kc
Because NOBr is 34% dissociated,
It is 66% undissociated
PNOBr=(x-0.34x) = 0.66x
PNOBr=(0.66)(0.21atm) = 0.14atm
PNO= 0.34x = (0.34) x (0.21atm) = 0.071 atm
PBr2= 0.17x = (0.17) x (0.21atm) = 0.036 atm
(PNO)2(PBr2)
(0.071)2(0.036)
-3
=
Kp=
=9.3x10
2
2
(0.14)
(PNOBr)
• The numerical value of Kc for this reaction can be
determined from the relationship of Kp and Kc.
2NOBr(g)  2NO(g) + Br2(g)
Kp=Kc(RT)∆n or Kc=Kp(RT)-∆n ∆n=1
Kc= (9.3x10-3)[(0.0821)(298)]-1
= 3.8 x10-4
56
Relationship Between Kp and
Kc
Example 17-16: Kc is 49 for the following reaction at 450oC.
If 1.0 mole of H2 and 1.0 mole of I2 are allowed to reach
equilibrium in a 3.0-liter vessel, (a)How many moles of I2
remain unreacted at equilibrium?
H2(g) + I2 (g)  2HI(g)
1/3M
1/3M
Initial
-x M
Change
-x M
+2x M
Equilibrium (0.33-x)M (0.33-x)M +2x M
[HI]2
(2x)2
=49 =
Kc=
[H2][I2]
(0.33-x)2
7.0=
2x
0.33-x
9x=2.31 x=0.256M
[H2]=[I2]=0.33-0.256=0.074
[HI]=2x0.256=0.51M
?mol I2 = 3.0L x0.074 =0.21 mol
57
Relationship Between Kp and
Kc
(b) What are the equilibrium partial pressures of H2, I2
and HI?
(c) What is the total pressure in the reaction vessel?
PH2=PI2=MRT=(0.074mol/L)(0.0821Latm/molK)(723K)
=4.4 atm
PHI=MRT=(0.051mol/L)(0.0821Latm/molK)(723K)
=30 atm
Ptot=PH2+PI2+PHI = 4.4 + 4.4 + 30
=38.8 atm
58
The Equilibrium Constant
• 由於實驗只能測量熱能變化量，無法測量某個物質的熱含
量，所以必須有一個公定的基準來作參照。
– 平衡常數以活性(activity)表示而不是以濃度表示
• 理想混合物的活性為該濃度或分壓與標準濃度或標準壓力
的比值
– 純固體或純液體，其活性activity等於1
– 理想溶液與理想氣體的活性為與標準濃度或標準壓力的
比值,故無單位
• 前面章節所描述均為單一相(如均為氣體)的平衡,稱之為
Homogeneous equilibria (均相平衡)
• 若兩種相以上的反應達成平衡時,則稱之為
Heterogeneous equilibria (異相平衡)
59
Heterogeneous Equlibria
•Homogeneous equilibria 均相平衡: only single phase
•Heterogeneous equilibria 異相平衡 have more than
one phase present.
– For example, a gas and a solid or a liquid and a gas.
CaCO3(s)  CaO(s) + CO2 (g) at 500oC
•How does the equilibrium constant differ for
heterogeneous equilibria?
– Pure solids and liquids have activities of unity.
– Solvents in very dilute solutions have activities that are
essentially unity.
– The Kc and Kp for the reaction shown above are:
Kc = [CO2] Kp = PCO2
• 在化學反應中，純固體或純液體的濃度不包含在反應的平衡表示式。
這個應用只有針對純固體或純液體，它並不會應用到溶液或氣體，因
為它們的濃度可以改變。
60
• 純固體和純液體的濃度是常數,而壓力變化不會影響其濃度
Heterogeneous Equlibria
What are the forms of Kc and Kp?
SO2(g) + H2O(l)  H2SO3(aq) at 25oC
H2O(l) is the solvent
[H2SO3]
Kc=
[SO2]
Kp=
1
PSO2
水溶液不因壓力改變而有所變化
61
Heterogeneous Equlibria
• What are Kc and Kp for this reaction?
CaF2(s)  Ca2+(aq) + 2F1-(aq) at 25oC
Kc = [Ca2+][F-]2
Kp is not defined; no gas involved
3Fe(s) + 4H2O(g)  Fe3O4 (s) + 4H2(g) at 500oC
Kc=
[H2]4
[H2O]4
(PH2 )4
Kp=
(PH2O)4
62
Heterogeneous Equlibria
Example 17-15: Kc and Kp for Heterogeneous Equlibria
Write both Kc and Kp for the following reversible
reactions.
(a)2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)
(b)2NH3(g) + H2SO4(l)  (NH4)SO4(s)
(c) S(s) + H2SO3(aq)  H2S2O3(aq)
(PSO2 )2
[SO2]2
Kp=
a. Kc=
3
[O2]
(PO2)3
1
1
K
=
b. Kc=
p
[NH3]2
(PNH3)2
[H2S2O3]
Kp: undefined, no gases involved
c. Kc=
[H2SO3]
63
Spontaneity of Physical and
Chemical Changes (Ch15, p575)
• Spontaneous changes happen without any
continuing outside influences. 自發性的改變指不受任
何外力影響下所進行的
– A spontaneous change has a natural direction.
• For example the rusting of iron occurs
spontaneously. (例如鐵生琇)
– Have you ever seen rust turn into iron metal without
man made interference?
• The melting of ice at room temperature occurs
spontaneously. (又如冰在室溫中溶化)
– Will water spontaneously freeze at room temperature?
64
The Two Aspects of
Spontaneity 自發性反應的兩個概念
• An exothermic reaction does not ensure
spontaneity.(放熱反應並不能確保為自發性反應)
– For example, the freezing of water is exothermic
but spontaneous only below 0oC.
• An increase in disorder of the system also does not
insure spontaneity.(增加系統的亂度亦不能確保為自發
性反應)
• It is a proper combination of exothermicity and
disorder that determines spontaneity.(需放熱及亂度
的適當組合才能決定反應是否為自發性反應)
65
Entropy熵, S
• Entropy is a measure of the disorder or randomness
of a system.熵用以測定一個系統的亂度
• As with H, entropies have been measured and
tabulated in Appendix K as So298.熵可被測量如附表K
• When:
– S > 0 disorder increases (which favors
spontaneity).亂度增加,有利於自發性的反應
– S < 0 disorder decreases (does not favor
spontaneity). 亂度減少, 不利於自發性的反應
熱力學第二定律 (The Second Law of Thermodynamics)
若反應趨向自發性形成,則其亂度必增加
(熱力學第一定律即質量守恆定律)
66
Entropy, S
• From the Second Law of Thermodynamics,
for a spontaneous process to occur:
Suniverse = Ssystem + Ssurroundings >0
• In general for a substance in its three
states of matter:
Sgas > Sliquid > Ssolid
增加亂度
67
• Entropy increase (Ssysytem>0), When
–
–
–
–
–
–
Temperature increase
Volume increase
Mixing of substance
Increasing particle number
Molecular size and complexity
Ionic compounds with similar formulas but different
charges
Example
Without doing a calculation, predict whether the entropy
change will be positive or negative
a) C2H6(g) +7/2 O2(g) 3H2O(g) + 2 CO2(g)
b) 3C2H2(g)  C6H6(l)
c) C6H12O6(s) + 6 O2(g) 6 CO2(g) + 2 H2O(l)
a) S0>0
b) S0<0
c) S0>0
d) Hg(l)< Hg(s) <Hg(g)
d) Hg(l), Hg(s), Hg(g)
e) C2H6(g), CH4(g) , C3H8(g)
e) CH4(g)< C2H6(g)< C3H8(g)
f) CaS(s), CaO(s)
f) CaO(s)< CaS(s)
68
Entropy, S
• The Third Law of Thermodynamics states,
“The entropy of a pure, perfect, crystalline
solid at 0 K is zero.”熱力學第三定律闡明”在絕
對零度時純物質晶體之亂度為零”
• This law permits us to measure the absolute
values of the entropy for substances. (依此定
律用以測定物質亂度的絕對值)
– To get the actual value of S, cool a substance to
0 K, or as close as possible, then measure the
entropy increase as the substance heats from 0
to higher temperatures.
– Notice that Appendix K has values of S not S.(所
以附錄K所指的是亂度而非亂度的變化)
69
Entropy, S
• Entropy changes for reactions can be
determined similarly to H for reactions. 反應
中熵的改變可與H相似
0
0
Srxn
= nSproducts
- nS0reactants
70
Entropy, S
• Example 15-15: Calculate the entropy change for
the following reaction at 25oC. Use appendix K.
2NO2(g) N2O4(g)
0
0
S0rxn= nSproducts
- nSreactants
0
0
= SN2O4(g) - 2SNO2(g)
= (304.2 J/molK) – 2(240.0 J/molK)
= -175.8J/molK
• The negative sign of S indicates that the system is
more ordered. (S為負值表示系統較整齊)
• If the reaction is reversed the sign of S changes.
–For the reverse reaction So298= +0.1758 kJ/K
• The + sign indicates the system is more disordered (較不規律).
71
Entropy, S
• Example 15-16: Calculate So298 for the reaction
below. Use appendix K.
3NO(g) N2O(g) + NO2(g)
0
S0298= SN0 2O(g) + S0NO2(g) - 3SNO(g)
= (219.7 + 240.0) – 3(210.4) J/molK
= -172.4J/molK
• Changes in S are usually quite small
compared to E and H. (亂度的變化較小)
– Notice that S has units of only a fraction of a kJ
while E and H values are much larger numbers
of kJ.
72
The Second Law of
Thermodynamics
• The second law of thermodynamics states,
“In spontaneous changes the universe tends
towards a state of greater disorder.若反應趨向
自發性形成,則其亂度必增加”
• Spontaneous processes have two
requirements:
1.The free energy change of the system must be
negative. (系統的自由能必須小於零)
2.The entropy of universe must increase.(亂度必須增
加)
• Fundamentally, the system must be capable of doing
useful work on surroundings for a spontaneous process
to occur.
73
Free Energy Change自由能的改變
, G,and Spontaneity
•In the mid 1800’s J. Willard Gibbs determined the
relationship of enthalpy焓, H, and entropy, S, that
best describes the maximum useful energy
obtainable in the form of work from a process at
constant temperature and pressure.
–The relationship also describes the spontaneity of
a system.
•The relationship is a new state function, G,
the Gibbs Free Energy 自由能.
G=H-TS
(at constant T and P)
74
Free Energy Change自由能的改變
, G, and Spontaneity
• The change in the Gibbs Free Energy, G, is a
reliable indicator of spontaneity of a physical
process or chemical reaction.
– G does not tell us how quickly the process
occurs. (自由能的改變無法告知反應進行多快)
• Chemical kinetics, the subject of Chapter 16, indicates
the rate of a reaction.
• Sign conventions for G.
–G > 0
– G = 0
– G < 0
reaction is nonspontaneous (不自發性反應)
system is at equilibrium (系統達到平衡)
reaction is spontaneous (自發性反應)
75
Free Energy Change, G,
and Spontaneity
• Changes in free energy obey the same type of
relationship we have described for enthalpy, H,
and entropy, S, changes.
0
G0rxn= nGproducts
- nG0reactants
76
Free Energy Change, G,
and Spontaneity
• Example 15-17: Calculate Go298 for the reaction in
Example 15-8. Use appendix K.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
0
0
Grxn
= [3G0fCO2(g) + 4GfH2O (g) ]- [G0fC3H8 (g) + 5G0fO2 (g) ]
= [3(-394.4)+4(-237.3)] – [(-23.49)+5(0)] kJ/mol
= -2108.9 kJ/mol
Go298 < 0, so the reaction is spontaneous at standard
state conditions.
If the reaction is reversed:
Go298 > 0, and the reaction is nonspontaneous at
standard state conditions.
77
Relationship Between Gorxn and
the Equilibrium Constant
Gorxn is the standard free energy change 標準狀態下
的自由能變化
 Gorxn is defined for the complete conversion of all
reactants to all products. (被定義為所有反應物完全轉為產物)
Grxn is the free energy change at nonstandard
conditions (Grxn 則為非標準狀態下自由能的變化)
• For example, concentrations other than 1 M or pressures
other than 1 atm.(例如濃度非1M或壓力非1大氣壓)
Grxn is related to Gorxn by the following relationship.
∆Grxn=∆Gorxn + RT lnQ or
∆G=∆Go + 2.303 RT logQ
R= universal gas constant (8.314J/molK)
T= absolute temperature (絕對溫度)
Q= reaction quotient (反應商)
78
Relationship Between Gorxn and
the Equilibrium Constant
• 當達成平衡時, G=0 and Q=Kc.
• Then we can derive this relationship:
0=∆Go + RT lnK or
0=∆Go + 2.303 RT logK
Which rearranges to:
∆Go = -RT lnK or
∆Go = -2.303 RT log K
反應物及生成物均為氣體時, K為 Kp;
反應物及生成物均為溶液時, K為 Kc;
若為異相時,均可使用
79
Relationship Between Gorxn and
the Equilibrium Constant
• For the following generalized reaction, the
thermodynamic equilibrium constant is defined as
follows:
aA(g)+bB(g )  cC(g)+dD(g)
(aC)c(ad)d
K =
Where
a
b
(aa) (ab)
aa is the activity of A
ac is the activity of C
ab is the activity of B
ad is the activity of D
80
Relationship Between Gorxn and
the Equilibrium Constant
• The relationships among Gorxn, K, and the
spontaneity of a reaction are:
Gorxn
<0
K
>1
Spontaneity at unit concentration
Forward reaction spontaneous
=0
=1
System at equilibrium
>0
<1
Reverse reaction spontaneous
81
Relationship Between Gorxn and
the Equilibrium Constant
82
Relationship Between Gorxn and
the Equilibrium Constant
Example 17-17: Calculate the equilibrium constant, Kp,
for the following reaction at 25oC from
thermodynamic data in Appendix K.
N2O4(g)  2NO2(g)
• Note: this is a gas phase reaction.
1. Calculate Gorxn
Gorxn =2 GofNO2(g) - GofN2O4(g)
Gorxn =2 (51.30kJ) – (97.82kJ)
Gorxn =4.78 kJ/mol rxn
Gorxn =4.78x103 j/mol rxn
This reaction is nonspontaneous
83
Relationship Between Gorxn and
the Equilibrium Constant
2. Calculate K from Gorxn =-RT lnKp
lnKp =Gorxn/-RT
=(4.78x103J/mol)/-(8.314J/molK)(298K)
=-1.93
Kp = e-1.93 =0.145= (PNO2)2/(PN2O4)
84
Relationship Between Gorxn and
the Equilibrium Constant
• Kp for the reverse reaction at 25oC can be
calculated easily, it is the reciprocal of the above
reaction.
2NO2(g) N2O4(g)
Gorxn =-4.78 kJ/mol
K’p = 1/Kp =1/0.145
= 6.9
= (PN2O4)/(PNO2)2
85
Relationship Between Gorxn
and the Equilibrium Constant
Example 17-18: At 25oC and 1.00 atmosphere, Kp = 4.3 x 1013 for the decomposition of NO . Calculate Go
2
rxn at
25oC.

2 NO2g   2 NOg   O 2g 
G
o
rxn
  RT ln K p
G
o
rxn
 (8.314
G
o
rxn
 (2480 J mol )( 28.47)
G
o
rxn
 7.06 10
J
mol K
4 J
)( 298 K ) ln 4.3 10
mol rxn
 70.6 kJ mol rxn
86
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