1 17 Chemical Equilibrium 化學平衡. 2 Chapter Goals 1.Basic Concepts 2.The Equilibrium Constant 平衡常數 3.Variation of K c with the Form of the Balanced Equation

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1717Chemical Chemical

Equilibrium Equilibrium 化學化學平衡平衡

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Chapter GoalsChapter Goals1. Basic Concepts2. 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: Predictions7. The Haber Process: A Commercial Application

of Equilibrium8. Disturbing a System at Equilibrium:

Calculations9. Partial Pressures and the Equilibrium Constant10.Relationship between Kp and Kc

11.Heterogeneous Equilibria12.Relationship between Go

rxn and the Equilibrium Constant

13.Evaluation of Equilibrium Constants at Different Temperatures

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Basic ConceptsBasic Concepts 基本概念基本概念•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 )

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Basic ConceptsBasic Concepts 基本概念基本概念• 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

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Basic ConceptsBasic 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 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

H2O

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

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Basic ConceptsBasic Concepts• Graphically, this is a representation of the

rates for the forward and reverse reactions for this general reaction


Equilibrium isestablished 達成平衡狀態

反應開始

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Basic ConceptsBasic 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)

0.02M

達成平衡

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Basic ConceptsBasic Concepts

2SO2(g)+ O2(g ) 2SO3(g)

0.400mol開始莫耳數 0.200mol 0-0.056mol反應改變莫耳數 -0.028mol +0.056mol0.344mol反應後莫耳數 0.172mol 0.056mol

2SO2(g)+ O2(g ) 2SO3(g)

0開始莫耳數 0 0.500mol+0.424mol 反應改變莫耳數 +0.212mol -0.424mol

0.424mol反應後莫耳數 0.212mol 0.076mol

2: 1: 2: 1: 2 2

2: 1: 2: 1: 2 2

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Basic ConceptsBasic Concepts

2SO2(g)+ O2(g ) 2SO3(g)

0.400MInitial conc. 0.200M 0-0.056MChange due to

rxn-0.028M +0.056M

0.344MEquilibrium conc’n平衡濃度

0.172M 0.056M

2SO2(g)+ O2(g ) 2SO3(g)

0Initial conc. 0 0.500M+0.424MChange due to

rxn+0.212M -0.424M

0.424MEquilibrium conc’n

0.212M 0.076M

2: 1: 2: 1: 2 2

2: 1: 2: 1: 2 2

In 1.00 liter container反應均為氣體 , 在固定體積

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The Equilibrium ConstantThe Equilibrium Constant• For a simple one-step mechanism

reversible reaction such as:

• 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]

A(g)+B(g ) C(g)+D(g)

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The Equilibrium ConstantThe 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

kr

kc= kc =[C][D][A][B]

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The Equilibrium ConstantThe Equilibrium Constant• Similarly, for the general reaction:

we can define a constant: 平衡常數 Kc

Kc =[C]c[D]d

[A]a[B]b


Products 產物Reactants 反應物

This expression is valid for all reactions

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The Equilibrium ConstantThe 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、壓力的大小無關。

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The Equilibrium ConstantThe Equilibrium Constant• Example 17-1: Write equilibrium constant

expressions for the following reactions at 500oC. All reactants and products are gases at 500oC.

Kc =[PCl3][Cl2]

[PCl5]

PCl5 PCl3 + Cl2 H2 + l2 2HI

Kc =[HI]2

[I2][I2]

4NH3 + 5O2 4NO + 6H2

O Kc =[NO]4[H2O]6

[NH3]4[O2]5

The Equilibrium ConstantThe Equilibrium Constant

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Example 17-1: Calculation of Kc Some nitrogen and hydrogen are placed in an

empty 5.00-liter 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.

Example 17-1: Calculation of Kc Some nitrogen and hydrogen are placed in an

empty 5.00-liter 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)

Kc [N2][H2]3

[NH3]2

=

[N2]: 3.01mol/5L = 0.602 M[H2]: 2.10mol/5L = 0.420 M[NH3]: 0.565mol/5L = 0.113 M

=(0.602)(0.420)3

(0.113)2

=0.286

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The Equilibrium ConstantThe Equilibrium ConstantExample 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.PCl5 PCl3 + Cl2

0.028MEquil []’s M 0.172M 0.086M

Kc =[PCl3][Cl2]

[PCl5]

=(0.172)(0.086)

(0.028)Kc=0.53

One liter

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The Equilibrium ConstantThe Equilibrium ConstantExample 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.PCl5 PCl3 + Cl2

1.00MInitial 0 0

K’c =[PCl3][Cl2]

[PCl5]

-0.60MChange 0.60M 0.60M0.40MEquilibrium

conc’n0.60M 0.60M

=(0.60)(0.60)

(0.40)At another temperature=0.90

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The Equilibrium ConstantThe Equilibrium ConstantExample 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 + 3H2 2NH3 0.80MInitial 0.90M 0

-0.10MChange -0.30M 0.20M0.70MEquilibrium

conc’n0.60M 0.20M

=(0.20)2

(0.70)(0.60)3 =0.26

Kc [N2][H2]3

[NH3]2

=

N2: 0.8mole/1Liter = 0.8M H2: 0.9mol/1Liter = 0.9MNH3: 0.2mol/1Liter = 0.2M

The Equilibrium ConstantThe Equilibrium Constant

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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

At equilibrium, 2.20 moles of N2O remain, Calculate the value of Kc for the reaction

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

At equilibrium, 2.20 moles of N2O remain, Calculate the value of Kc for the reaction

2N2O (g) 2N2(g) + O2(g)

Kc [N2O]2

[N2]2[O2]=

Initial [N2O]: 10.0mol/2L = 5.0 M

= (1.1)2

(3.9)2(1.95)=24.5

equili [N2O]: 2.20mol/2L = 1.1 M2N2O (g) 2N2(g) + O2(g) 5.0MInitial 0 0

-3.9MChange +3.9M +1.95M1.1MEquilibrium

conc’n3.9M 1.95M

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Variation of KVariation of Kcc with the with the Form of the Balanced Form of the Balanced

EquationEquation• The value of Kc depends upon how the

balanced equation is written.• From example 17-2 we have this reaction:

• This reaction has a Kc=[PCl3][Cl2]/[PCl5]=0.53

PCl5 PCl3 + Cl2

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EquationEquationExample 17-5: Calculate the equilibrium constant

for the reverse reaction by two methods, i.e, the equilibrium constant for this reaction.

Equil. []’s 0.172M 0.086M 0.028 MThe concentrations are from Example

17-2.

PCl3 + Cl2 PCl5

K’c =[PCl3][Cl2][PCl5]

=(0.172)(0.086)

(0.028)

K’c=1.9

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EquationEquation

Large equilibriumLarge equilibrium constants constants indicate indicate that that most of the reactantsmost of the reactants are are converted to products. (converted to products. ( 大的平衡常數表大的平衡常數表示大部分的反應物轉成產物示大部分的反應物轉成產物 ))

Small equilibrium constantsSmall equilibrium constants indicate indicate that only that only small amounts of products small amounts of products are formed.are formed. (( 小平衡常數表示僅少數產物生小平衡常數表示僅少數產物生成成 ))

K’c = [PCl3][Cl2][PCl5] =1.9=

(0.172)(0.086)(0.028)

Kc = K’c

1 K’c = Kc

1or =0.53

1 =1.9

•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

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The Reaction Quotient The Reaction Quotient 反應商數反應商數

• 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 )

Q =[C]c[D]d

[A]a[B]d

For this general reactionaA(g)+bB(g ) cC(g)+dD(g)

Not necessarily equilibrium concentrations

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The Reaction QuotientThe 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.

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The Reaction QuotientThe Reaction Quotient

When:Q<Kc Forward reaction predominates until equilibrium is established ( 反應向右 )Q=Kc The system is at equilibrium (equilibrium ( 達成平衡達成平衡 ))

Q>Kc Reverse reaction predominates until equilibrium is established ( 反應向左 )

僅有反應物僅有產物

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The Reaction QuotientThe Reaction QuotientExample 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

Q = [HI]2

[I2][H2]=

(0.22)(0.22)(0.66)2

=9.0

Q=9.0 but Kc=49Q<Kc

Forward reaction predominates until equilibrium is established ( 反應會持續往右進行 ,直至達到平衡 )

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Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc

Example 17-7: The equilibrium constant, Kc, is 3.00 for the 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 0Initial-x M -x M +x M +x MChange(0.5-x)M (0.5-x)M xM xMEquilibrium

Kc = [SO2][NO2][SO3][NO]

=(0.5-x)(0.5-x)

(x)(x)=3.0 =

(0.5-x)2

(x)2

1.73=0.5-x

x 0.865-1.73x=x x=0.316M=[SO3]=[NO][SO2]=[NO2]=0.5-0.316=0.184M

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Uses of the Equilibrium Uses of the Equilibrium Constant, KConstant, Kcc

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 1.0MInitial+x M +x M -2x MChange

x M x M 1.0-2x MEquilibrium

Kc = [H2][I2][HI]2

= (x)(x)(1.0-2x)2

= 49

7.0=x

1.0-2x 7.0x=1.0-2x x=0.11M=[H2]=[I2][HI]=1.0-(2x0.11)=0.78M

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Disturbing a System at Disturbing a System at Equilibrium: PredictionsEquilibrium: 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

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Disturbing a System at Disturbing a System at Equlibrium: PredictionsEqulibrium: 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=nRTP=(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

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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)

=49= [H2][I2][HI]2

Kc

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

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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:

2NO2(g) N2O4(g) = [NO2]2

[N2O4]Kc

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>KcThis favors the reactants or the reverse reaction ( 反應向左 )

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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 )

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

reactantsThis favors the product or forward reaction ( 反應向右 )

2SO2(g) + O2(g) 2SO3(g) +198kJ/mol


若為放熱反應 , 提高溫度反應向左

若為吸熱反應 , 提高溫度反應向右

A+BC+D+ heat

A+B+ heat C+D

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•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 )

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Example 17-9: Given the reaction below at equilibrium in a closed container at 500oC. How would the equilibrium be influenced by the following?

N2(g) + 3H2(g) 2NH3(g) Horxn=-92kJ/mol

FactorsEffect on

reaction procedurea. Increasing the reaction temperatureb. Decreasing the reaction temperaturec. Increasing the pressure by decreasing the volumed. Increasing the concentration of H2

e. Decreasing the concentration of NH3

f. Introduction a platinum catalyst

Left Right Right

Right RightNo effect

Kc [N2][H2]3

[NH3]2

=


Example 17-10: How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?

FactorsEffect on

equilibriuma. 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)

No effect

Left

Right

Right

假設壓力增加兩倍 , 則濃度增加兩倍

Q (2)2[HI]2

(2)[H2]x(2)[I2]=

a. = Kc Q

(2)4[NH3]4x(2)5[O2]5

(2)4[NO] 4x(2)6[H2O]6

=b.

= 2Kc

Q (2)[PCl5]

(2)[PCl3]x(2)[Cl2]=

c. = 0.5KcQ

(2)2[H2O]2

(2)2[H2]2x(2)[O2]=

d. = 0.5Kc

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Example 17-11: How will an increase in increase in temperature temperature affect each of the following reactions?

FactorsEffect on

equilibriuma. 2NO2 (g) 2N2O4(g) Ho

rxn<0

b. H2(g)+ Cl2(g) 2HCl(g)+92kJ

c. H2(g) + l2(g) 2HI(g) Horxn=25kJ

Left

Left

Right

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The Haber Process: A The Haber Process: A Practical Application of Practical Application of

EquilibriumEquilibrium• 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?

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EquilibriumEquilibriumN2(g) + 3H2(g) 2NH3(g) Ho

rxn=-92kJ/mol

Fe & metal oxide

N2 is obtained from liquid air; H2 obtain from coal gasThis 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-TS △ G < 0 反應自然發生

However the reaction kinetics are very slow at low THaber’s solution to this dilemma

1. Increase T to increase rate, but yield is decreased ( 反應向左 )2. Increase reaction pressure to right3. Use excess N2 to right4. 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 )

42


EquilibriumEquilibriumThis diagram illustrates the commercial system devised for the Haber process.

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Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: CalculationsTo 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?

A(g) B(g) + C(g)

0.20M0.3 M 0.3MEquilibrium

= (0.2)

(0.3)(0.3)=

[B][C][A]

Kc =0.45

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Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: 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

A(g) B(g) + C(g) 0.10M0.15M0.15M

= (0.10)(0.15)(0.15)

=[B][C][A]Q =0.22

體積加倍 , 濃度均減半

= (0.2)(0.3)(0.3)

=[B][C][A]Kc =0.45

Q<Kc

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Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: 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 0.15MInitial-x M +x M +x MChange

(0.1-x) M(0.15+x) M (0.15+x) MEquilibrium

= (0.1-x)(0.15+x)2

=0.45=[B][C][A]

Kc

0.045-0.45x=0.0225+0.30x+x2

x2+0.75x-0.0225=0

(a+b)2 =a2+2ab+b2

46

Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations

x2+0.75x-0.0225=0 ax2+bx+c=0

x= -b b2-4ac 2a

x= -0.75 (0.075)2-4(1)(-0.0225)

2x1

x= -0.75 0.081

2x= -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

47

Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: CalculationsExample 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.A(g) B(g) + C(g)

0.40M0.60M0.60MInstantaneous

= (0.40)(0.6)(0.6)

=[B][C][A]

Q =0.90

體積減半 , 濃度增倍

Q>Kc thus the equilibrium shifts to the left or reactant side

48

Disturbing a System at Disturbing a System at Equilibrium: CalculationsEquilibrium: Calculations•Set up the algebraic expressions to determine the equilibrium concentrationsA(g) B(g) + C(g)

0.40M 0.60M 0.60MInitial+x M -x M -x MChange

(0.4+x) M(0.60-x) M(0.60-x) MEquilibrium

= (0.4+x)(0.60-x)2

=0.45=[B][C]

[A]Kc

0.18+0.45x=0.36-1.2x+x2

x2-1.65x+0.18=0

x= 1.65 (-1.65)2-4(1)(0.18)

2x1

x= 1.65 1.42

2x= 1.5 and 0.12M

Since 0<x<0.60 x=0.12M [A]=0.40+x=0.52 M[B]=[C]=0.60-x=0.48 M

反應向左

49


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)

0.6/2 0.2/2 1.2/2Equilibrium

= (0.30)(0.10)

(0.6)=

[COCl2][CO][Cl2]

Kc =20

0.3M 0.1M 0.6MEquilibrium

50


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.

CO(g) + Cl2(g) COCl2(g)0.3M 0.1M 0.6MOrig. Equil.

+0.4M(Stress) Add

0.30 M 0.50 M 0.6 MNew Initial Q<Kc 反應向右-x M -x M +x MChange

(0.3-x) M(0.50-x)M(0.60+x) MEquilibrium

= (0.3-x)(0.5-x)(0.6+x)

=[COCl2]

[CO][Cl2]Kc =20

0.80 mole of Cl2 in 2-liter vessel 0.4M of Cl2

= (0.30)(0.50)(0.6)

Qc=4

20x2-17x+2.4=0

x= 17 (17)2-4(20)(2.3)

2x20X=0.67 and 0.18

Since 0<x<0.30 x=0.18M [CO]=0.30-x=0.12 M[Cl2]=0.5-x=0.32[COCl]=0.6+x=0.78

51

Partial Pressures and the Partial Pressures and the Equilibrium ConstantEquilibrium 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

52


• Consider this system at equilibrium at 5000C.2Cl2(g) + 2H2O (g) 4HCl(g) + O2(g)

=[HCl]4[O2][Cl2]2 [H2O]2Kc =

(PHCl)4(PO2)

(PCl2)2 (PH2O)2Kp

53


=Kc

PHCl 4

RT PO2

RT PCl2

2

RT PH2O

2

RT

=(PHCl)4(PO2

)(PCl2)

2 (PH2O)2 x

1 5

RT

1 4

RT

=KpKc 1

RTso for this reaction

Kc=Kp(RT)-1 or Kp=Kc(RT)1

Must use R = 0.0821 L atm/mol K

P= MRT and M = P/RT

54

Relationship Between KRelationship Between Kpp and Kand Kcc

• 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 )

Kp=Kc(RT)1or Kc=Kp(RT)-1

2Cl2(g) + 2H2O (g) 4HCl(g) + O2(g)

∆n= (4+1)-(2+2)=1

55


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)x atm 0 0Initial

-0.34x atm +0.34x atm+0.17x atmChange(x-0.34x) atm(0.34x) atm (0.17x)atmEquilibrium

Ptot=PNOBr + PNO+ PBr2

0.25atm = (x-0.34x)atm + 0.34x atm + 0.17x atm0.25atm = 1.17x atm, thus x=0.21atm

56


Because NOBr is 34% dissociated,It is 66% undissociatedPNOBr=(x-0.34x) = 0.66xPNOBr=(0.66)(0.21atm) = 0.14atmPNO= 0.34x = (0.34) x (0.21atm) = 0.071 atmPBr2

= 0.17x = (0.17) x (0.21atm) = 0.036 atm=

(PNO)2(PBr2)

(PNOBr)2Kp =(0.071)2(0.036)(0.14)2 =9.3x10-3

• The numerical value of Kc for this reaction can be determined from the relationship of Kp and Kc.

Kp=Kc(RT)∆n or Kc=Kp(RT)-∆n ∆n=1Kc= (9.3x10-3)[(0.0821)(298)]-1

= 3.8 x10-4

2NOBr(g) 2NO(g) + Br2(g)

57

Relationship Between KRelationship Between Kpp and Kand KccExample 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 1/3MInitial-x M -x M +2x MChange

(0.33-x)M(0.33-x)M+2x MEquilibrium

Kc=[H2][I2][HI]2

= (0.33-x)2

(2x)2

=49

7.0=0.33-x

2x 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

58


(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 atmPHI=MRT=(0.051mol/L)(0.0821Latm/molK)(723K) =30 atm

Ptot=PH2+PI2+PHI = 4.4 + 4.4 + 30

=38.8 atm

59

The Equilibrium ConstantThe 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 )

60

Heterogeneous EqulibriaHeterogeneous 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.

•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:

CaCO3(s) CaO(s) + CO2 (g) at 500oC

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。• 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

61

Heterogeneous EqulibriaHeterogeneous Equlibria What are the forms of Kc and Kp? SO2(g) + H2O(l) H2SO3(aq) at 25oC

Kc= [SO2][H2SO3]

Kp= PSO2

1

H2O(l) is the solvent

水溶液不因壓力改變而有所變化

62

Heterogeneous EqulibriaHeterogeneous 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= [H2O]4

[H2]4

Kp=(PH2O)4

(PH2 )4

Heterogeneous EqulibriaHeterogeneous Equlibria

63

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)

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)

a. Kc= [O2]3

[SO2]2

Kp= (PO2)3

(PSO2 )2

b. Kc= [NH3]21 Kp=(PNH3

)2

1

c. Kc= [H2SO3][H2S2O3]Kp: undefined, no gases involved

64

Spontaneity of Physical Spontaneity of Physical and Chemical Changesand Chemical Changes

(Ch15, p575)(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?

65

The Two Aspects of The Two Aspects of Spontaneity Spontaneity 自發性反應的兩個概念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 )

66

EntropyEntropy熵熵 , S, 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 So

298.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 )

67

Entropy, SEntropy, S• From the Second Law of Thermodynamics, for

a spontaneous process to occur:

• In general for a substance in its three states of matter:

Sgas > Sliquid > Ssolid

Suniverse = Ssystem + Ssurroundings >0

增加亂度

• 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

68

Example Without doing a calculation, predict whether the entropy change will be positive or negativea)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)

e) C2H6(g), CH4(g) , C3H8(g)

f) CaS(s), CaO(s)

d) Hg(l)< Hg(s) <Hg(g)

e) CH4(g)< C2H6(g)< C3H8(g) f) CaO(s)< CaS(s)

69

Entropy, SEntropy, 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 )

70

Entropy, SEntropy, 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

Srxn= nSproducts - nSreactants0 0 0

71

Entropy, SEntropy, S• Example 15-15: Calculate the entropy change for

the following reaction at 25oC. Use appendix K.

2NO2(g) N2O4(g)

Srxn= nSproducts - nSreactants0 0 0

= SN2O4(g) - 2SNO2(g)0 0

= (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 reactionSo

298= +0.1758 kJ/K •The + sign indicates the system is more disordered

( 較不規律 ).

72

Entropy, SEntropy, S• Example 15-16: Calculate So

298 for the reaction below. Use appendix K.

3NO(g) N2O(g) + NO2(g)

= (219.7 + 240.0) – 3(210.4) J/molK

= -172.4J/molK

S298= SN2O(g) + SNO2(g) - 3SNO(g)0 0 0 0

• 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.

73

The Second Law of The Second Law of ThermodynamicsThermodynamics

•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.

74

Free Energy ChangeFree Energy Change 自由能自由能的改變的改變 , , G,and SpontaneityG,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-TS (at constant T and P)

75

Free Energy ChangeFree Energy Change 自由能自由能的改變的改變 , , G, and G, and SpontaneitySpontaneity

• 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 reaction is nonspontaneous ( 不自發性反應 )

G = 0 system is at equilibrium ( 系統達到平衡 ) G < 0 reaction is spontaneous ( 自發性反應 )

76

Free Energy Change, Free Energy Change, G, G, and Spontaneityand Spontaneity

• Changes in free energy obey the same type of relationship we have described for enthalpy, H, and entropy, S, changes.

Grxn= nGproducts - nGreactants0 0 0

77

Free Energy Change, Free Energy Change, G, G, and Spontaneityand Spontaneity

• Example 15-17: Calculate Go298 for the

reaction in Example 15-8. Use appendix K.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

Grxn= [3GfCO2(g) + 4GfH2O (g) ]- [GfC3H8 (g) + 5GfO2 (g) ]0 0 00 0

= [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.

78

Relationship Between Relationship Between GGoorxnrxn

and the Equilibrium and the Equilibrium Constant Constant

Gorxn is the standard free energy change 標準狀態下的

自由能變化Go

rxn 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 Go

rxn by the following relationship.∆Grxn=∆Go

rxn + RT lnQ or

∆G=∆Go + 2.303 RT logQR= universal gas constant (8.314J/molK)T= absolute temperature (絕對溫度 )Q= reaction quotient ( 反應商 )

79


and the Equilibrium and the Equilibrium ConstantConstant

• 當達成平衡時, G=0 and Q=Kc. • Then we can derive this relationship:

0=∆Go + RT lnK or0=∆Go + 2.303 RT logKWhich 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

80



• For the following generalized reaction, the thermodynamic equilibrium constant is defined as follows:

K =(aC)c(ad)d

(aa)a(ab)b


Where

aa is the activity of A ab is the activity of B ac is the activity of C ad is the activity of D

81



• The relationships among Gorxn, K, and

the spontaneity of a reaction are:

Gorxn K Spontaneity at unit concentration

< 0 > 1 Forward reaction spontaneous

= 0 = 1 System at equilibrium

> 0 < 1 Reverse reaction spontaneous

82



83



Example 17-17: Calculate the equilibrium constant, Kp, for the following reaction at 25oC from thermodynamic data in Appendix K.

• Note: this is a gas phase reaction.N2O4(g)

2NO2(g)

1. Calculate Gorxn

Gorxn =2 Go

fNO2(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

84



2. Calculate K from Gorxn =-RT lnKp

lnKp =Gorxn/-RT

=(4.78x103J/mol)/-(8.314J/molK)(298K) =-1.93Kp = e-1.93

=0.145= (PNO2)2/(PN2O4

)

85



• 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

86

Relationship Between Relationship Between GGoo

rxnrxn and the Equilibrium and the Equilibrium ConstantConstant

Example 17-18: At 25oC and 1.00 atmosphere, Kp = 4.3 x 10-13 for the decomposition of NO2. Calculate Go

rxn at 25oC.

g2gg2 O NO 2 NO 2

rxn molkJ

rxn molJ4o

rxn

molJo

rxn

13-K mol

Jorxn

porxn

6.70 1006.7G

)47.28)(2480(G

104.3ln )K 298)(314.8(G

Kln RT G

Documents

1 17 Chemical Equilibrium 化學平衡. 2 Chapter Goals 1.Basic Concepts 2.The Equilibrium Constant 平衡常數 3.Variation of K c with the Form of the Balanced Equation