Chemical Equilibrium

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Chemical Equilibrium

Dr. Kendrew K. W. MakDepartment of Chemistry

The Chinese University of Hong Kong

Textbook: John Green, Sadru Damji, Chemistry for the International Baccalaureate Diploma Programme, 2nd Ed.

IJSO Training (Phase 3)

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Reversible（可逆） and Irreversible（不可逆）

Reactions go to completion:products are much more energetically favorable than the reactants.

Reactions occur easily: low activation energy ( 活化能 )

Reactions that do not occur: the activation energy is too high, or the reaction is not energetically favorable

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Equilibrium Reaction（平衡反應）

For some chemical systems, the energies of the reactants and products are comparable so that the reactions are reversible – they can occur in either direction.

When a reaction attained the state of chemical equilibrium ( 化學平衡 ), the concentrations of reactants and products remain constant.

Example: The Haber Process (哈柏法 )

N2 (g)

nitrogen

+ 3 H2 (g)

hydrogen

heatpressurecatalyst

2 NH3

ammonia

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Equilibrium Reaction

Chemical equilibrium is a state of dynamic equilibrium ( 動態平衡 ) that occurs in a closed system when the forward and reverse reactions of a reversible reaction ( 可逆反應 ) occur at the same rate.

A + B C + D

Forward rate ( 正向速率 ) = kf[A][B]Backward rate ( 逆向速率 ) = kb[C][D]

At equilibrium, kf[A][B] = kb[C][D]At equilibrium, kf[A][B] = kb[C][D]

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Equilibrium Reaction

At equilibrium, all of the species involved (reactants and products) are present at constant concentrations.

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Dynamic Equilibrium（動態平衡）

When equilibrium is reached, it appears that a chemical reaction has stopped.

In reality, both forward and reverse reactions are still occurring, but there is a balance between transformation of reactants into products, and transformation of products into reactants.

The rate of the forward reaction equals to the rate of the reverse reaction.The rate of the forward reaction equals to the rate of the reverse reaction.

Equilibrium reactions are represented with a double arrow between reactants and products, showing their reversible and dynamic nature.

CH3COOH (aq) CH3COO-(aq) + H+(aq)

acetic acid(醋酸)

acetate ion(醋酸根離子)

hydrogen ion(氫離子)

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Dynamic Equilibrium

For a specific reaction, the equilibrium state will be the same, no matter the equilibrium is approached from which side.

N2O4 (g) 2 NO2 (g)

Initial concentration 1 mol dm-3 0 mol dm-3

Equilibrium concentration 0.4 mol dm-3 1.2 mol dm-3

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Catalyst and Equilibrium

If a catalyst is present, the same equilibrium state will be attained, but it will be attained more quickly.

If a catalyst is present, the same equilibrium state will be attained, but it will be attained more quickly.

A catalyst speeds up both the forward and the backward reactions.

The overall effect is to produce exactly the same concentrations at equilibrium, whether or not a catalyst is in the reaction mixture.

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The Equilibrium Constant

A quotient of equilibrium concentrations of reactant and product substances that has a constant value for a given reaction at a given temperature is called an equilibrium constant ( 平衡常數 ).

Example 1: N2(g) + O2(g) 2 NO(g)

Equilibrium constant = K =[NO]2

[N2][O2]

a A + b B + ... c C + d D + ...

ba

dc

BA

DC

][][

][][κconstant mEquilibriu

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Example 2: 1/8 S8(s) + O2(g) SO2(g)

K' =[SO2(g)]

[S8(s)]1/8 [O2]K =

[SO2(g)]

[O2]

Which one is correct?

Since sulphur ( 硫 ) is a solid substances, and for any solid the number of molecules per unit volume remains the same at any given temperature. Therefore the concentration of sulphur is not changed as the reaction proceeds.

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Example 3: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

K =[NH4

+] [OH-]

[NH3]

Because the molar concentration ( 摩爾濃度 ) of water is effectively constant for reactions involving dilute solutions, the concentration of water is not included in the equilibrium constant expression.

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Equilibrium Constant Expressions for Related Reactions

N2(g) + 3 H2(g) 2 NH3(g) K1 =[N2][H2]

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[NH3]2

= 2.4 x 107

1/2 N2(g) + 3/2 H2(g) NH3(g) K2 =[N2]1/2[H2]

3/2

[NH3]

= (2.4 x 107)1/2 = 4.9 x 103

N2(g) + 3 H2(g)2 NH3(g) K3 =[N2][H2]

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[NH3]2

2.4 x 107

1= = 4.2 x 10-8

Whenever the stoichiometric coefficients ( 化學計量系數 ) of a balanced equation are multiplied by some factor, the equilibrium constant ( 平衡常數 ) for the new equation (K2 in this case) is the old equilibrium constant (K1) raised to the power of the multiplication factor.

Whenever the stoichiometric coefficients ( 化學計量系數 ) of a balanced equation are multiplied by some factor, the equilibrium constant ( 平衡常數 ) for the new equation (K2 in this case) is the old equilibrium constant (K1) raised to the power of the multiplication factor.

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Consider the gas-phase ( 氣相 ) reaction:

H2(g) + I2(g) 2 HI(g)

If a flask initially containing 0.025 mol/L of H2 and 0.025 mol/L of I2 is heated to 400oC, the concentrations of H2 and I2 decline and the concentration of HI increases. The concentration of HI at equilibrium is: [HI] = 0.039 mol/L. Calculate the equilibrium constants ( 平衡常數 ).

Exercise 1

Answer K = 50

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0.0500 mol of acetic acid ( 醋酸 ) was dissolved in 1.00 L of distilled water, and it was found that 3.05% of the acetic acid was ionized into CH3COO- ions and H+ ions. Calculate the equilibrium constant for ionization ( 電離作用 ) of acetic acid.

Exercise 2

Ans = 4.80 x 10-5

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The Meaning of the Equilibrium Constant

The value of the equilibrium constant ( 平衡常數 ) tells how far a reaction has proceeded by the time equilibrium has attained.

K >>1 Reaction is product-favored; equilibrium concentrations of products are greater than equilibrium concentrations of reactants.

NO(g) + O3(g) NO2(g) + O2(g) K =[NO2][O2]

[NO][O3]= 6 x 1034

K <<1 Reaction is reactant-favored. Equilibrium concentrations of reactants are greater than equilibrium concentrations of products.

3 O2(g) 2 O3(g)[O3]2

[O2]3K = = 6.25 x 10-58

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Calculating Equilibrium Concentrations

Exercise 3

Hydrogen ( 氫 ) reacts with carbon dioxide ( 二氧化碳 ) at high temperature and gives water and carbon monoxide ( 一氧化碳 )

A flask containing 0.200 mol/L of H2 and 0.200 mol/L of CO2 is heated at 420oC until equilibrium is attained. What are the concentrations of reactants and products at equilibrium?

H2(g) + CO2(g) H2O(g) + CO(g) K = 0.10 (at 420oC)

Answer: [H20] = [CO] = 0.048 mol/L; [H2] = [CO2] = 0.152 mol/L

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Calculating Equilibrium Concentrations

Exercise 4

The dimerization ( 二聚作用 ) of nitrogen dioxide ( 二氧化氮 ) to dinitrogen tetraoxide ( 四氧化二氮 ) has an equilibrium constant of 1.7 x 102 at 25oC.

2 NO2(g) N2O4(g) K = 1.7 x 102 (at 25oC)

If 1.00 mol N2O4 and 0.500 mol NO2 are initially placed in a container whose volume is 4.00L, calculate the concentrations of N2O4(g) and NO2(g) present when equilibrium is achieved at 25oC.

Ans: [NO2] = 0.414 mol L-1, [N2O4] = 0.292 mol L-1

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Shifting a Chemical Equilibrium:Le Chatelier’s Principle

The Le Chatelier’s Principle (勒沙得利爾原理 ):

If a system is in equilibrium and the conditions are changed so that the system is no longer at equilibrium, the system will adjust to a new equilibrium state such that the effect of the change in conditions is partially counteracted or compensated for.

Le Chatelier’s principle applies to changes in conditions such as:

• the concentrations of reactants or products that appear in the equilibrium constant expression

• the pressure/volume of the equilibrium system• the temperature

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Shifting a Chemical Equilibrium:Le Chatelier’s Principle

Change Effect on Equilibrium Change in Kc?

Increase concentration

Shifts to the opposite side No

Decrease concentration

Shifts to that side No

Increase pressure Shifts to side with lease moles of gas

No

Decrease pressure Shifts to side with most moles of gas

No

Increase temperature

Shifts in endothermic ( 吸熱 ) direction

Yes

Decrease temperature

Shifts in exothermic ( 放熱 ) direction

Yes

Add a catalyst ( 催化劑 )

No change No

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Effect of Concentration

If the concentration of a species is increased, then the equilibrium moves towards the other side causing the concentration to fall to a value between the original concentration and the increased value.

Fe(H2O)63+(aq) + SCN-(aq) [Fe(H2O)5SCN]2+(aq) + 6 H2O(l)

Yellow-brown Colourless Blood-red

Addition of Fe(H2O)63+(aq) - equilibrium shifts to product side

Addition of SCN-(aq) - equilibrium shifts to product side

Addition of [Fe(H2O)5SCN]2+(aq) - equilibrium shifts to reactant side

The values of K remain unchanged.The values of K remain unchanged.

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Effect of Pressure

If the total pressure of a system is increased then the equilibrium shifts to the side with least moles ( 摩爾 ) of gas, so causing the pressure to fall to a value between the original pressure and the increased value.

2 SO2(g) + O2(g) 2 SO3(g) 3 moles gas go to 2 moles gasIncreased P ; decrease P

C(s) + H2O(g) CO(g) + H2(g) 1 mole gas go to 2 moles gasIncreased P ; decrease P

H2(g) + I2(g) 2 HI(g) 2 moles gas go to 2 moles gasChanging P has no effect

The values of K remain unchanged.The values of K remain unchanged.

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Effect of Temperature

If the temperature of a system is increased then the equilibrium shifts in the direction of the endothermic ( 吸熱 ) change, so absorbing heat and causing the temperature to fall to a value between the original temperature and the increased value.

N2(g) + O2(g) 2 NO(g) H = +180 kJ mol-1

(forward reaction - endothermic)Increased T, K increases, equilibrium Decreased T, K decreases, equilibrium 2 SO2(g) + O2(g) 2 SO3(g) H = -197 kJ mol-1

(forward reaction - exothermic)Increased T, K decreases, equilibrium Decreased T, K increases, equilibrium

endothermic reaction – 吸熱反應 ; exothermic reaction – 放熱反應

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A Classical Example of EquilibriumThe Haber Process (哈柏法 )

The Haber process involves the direct combination of nitrogen ( 氮 ) and hydrogen ( 氫 ) to produce ammonia ( 氨 ).

N2(g) + 3 H2(g) 2 NH3(g)

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What is the Effect of Adding More N2 gas to the System?

http://www.cdli.ca/courses/chem3202/unit01/section02/lesson03/3-lesson-a.htm

According to the Le Châtelier’s Principle ( 勒沙得利爾原理 ), the system would counteract the adding of N2 by producing more NH3 – shifting the equilibrium position ( 平衡位置 ) to the product side.Adding starting materials favours formation of products.

N2 (g) + 3 H2 (g) 2 NH3 (g)

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What is the Effect of Reducing the Volume of the System?

N2 (g) + 3 H2 (g) 2 NH3 (g)

4 moles of gases 2 moles of gas

According to the Le Châtelier’s Principle ( 勒沙得利爾原理 ), the system would counteract the increased pressure by forming more ammonia ( 氨 ) (reducing the number of molecules) – shifting the equilibrium position to the product side.Increasing the pressure of the reaction also favours formation of ammonia.


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What is the Effect of Increasing the Temperature of the System?

N2 (g) + 3 H2 (g) 2 NH3 (g)

exothermic reaction(heat is given out)

H = - 92 kJ

The reverse reaction is endothermic (heat absorbing). According to the Le Châtelier’s Principle ( 勒沙得利爾原理 ), the system would counteract the increased temperature by shifting the equilibrium position to the reactant side.Increasing the reaction temperature reduces the reaction yield ( 反應產率 ).


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What is the Effect of Increasing the Temperature of the System?

N2 (g) + 3 H2 (g) 2 NH3 (g)

exothermic reaction(heat is given out)

H = - 92 kJ

A Balance Between Reaction Rate ( 反應速率 ) and Yield ( 產率 )A Balance Between Reaction Rate ( 反應速率 ) and Yield ( 產率 )

Although carrying out the reaction at a lower temperature can increase the product yield, the reaction rate is decreased as well. It would take a very long time to attain the equilibrium.

To attain a compromise between reaction rate and yield, the Haber process ( 哈柏法 ) is usually carried out at 450 – 500 °C.

Rule of Thumb: Reaction rate becomes double as the temperature is increased by 10°C.

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The Role of Catalyst (催化劑 )

The Haber Process ( 哈柏法 ) can be speeded up by adding a catalyst.

A catalyst provides an alternative route of reaction where the activation energy ( 活化能 ) is lower than the original chemical reaction.

As the activation energy is lowered, more molecules possess sufficient energy to overcome the barrier, hence the reaction is accelerated.

Usually, the catalyst is not consumed by the overall reaction.

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What are the Best Conditions for the Haber Process

Keq=[NH3]2

[N2][H2]3

Temp (°C) Keq

25 6.4 x 102

200 4.4 x 10-1

300 4.3 x 10-3

400 1.6 x 10-4

500 1.5 x 10-5

http://www.ausetute.com.au/haberpro.html

N2(g) + 3 H2(g) 2 NH3(g)

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Chemistry of the Haber ProcessChemical Equilibrium

Typical Industrial Conditions for Manufacturing Ammonia ( 氨 )Typical Industrial Conditions for Manufacturing Ammonia ( 氨 )

Temperature: 450 – 500 °CPressure: about 250 atmospheres ( 大氣壓 )Catalyst: iron ( 鐵 )Yield: about 10 – 20%

These conditions achieve a balance between the yield and production rate, as well as the costs involved in the building and operation of the manufacturing facility, and safety concerns.

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Chemical Equilibrium