代做CHEM 191 MODULE 6 REACTIONS 4: CHEMICAL EQUILIBRIUM AND ACIDS AND BASES代做留学生Matlab程序

日期：2025-02-12 04:54

CHEM 191

MODULE 6

REACTIONS 4: CHEMICAL EQUILIBRIUM AND ACIDS AND BASES

By the end of this module you should be able to:

• Describe the nature of equilibrium reactions

• Discuss the effect on equilibrium systems from changes to concentration, pressure or temperature

• Use equilibrium constants to determine the position of an equilibrium or concentrations of species in an equilibrium mixture

• Discuss the nature of acidic and basic solutions including conjugate acids and bases, and strong and weak acids.

• Calculate pH, pOH, [H3O+ ] and [OH- ] for solutions of strong acids and bases

Reference: ESA Chapters 18, 19, 20

HOW FAR DOES A REACTION GO?

A consideration for chemists, along with the issue of how fast a reaction will occur, is whether the reaction actually goes to completion. Some reactions, especially those in open containers,   will keep going until one or all of the reactants are used up. Other reactions reach a state of chemical equilibrium where the reverse reaction starts to occur before the forward reaction is complete. A fast reaction might reach a state of chemical equilibrium quickly   but not produce many products because of the reverse reaction begins before the forward reaction in complete. The formation of stalactites and stalagmites in limestone caves is dependent on an  equilibrium reaction between the water in the cave and carbon dioxide in the atmosphere, Equilibrium reactions can be a problem in industry when a synthesis reaction appears to stop before much of the reactants have been converted to products. Learning to work with equilibrium reactions has been an important factor in the chemical industry. Placing a stress on an equilibrium system, whether it is chemical, biological, societal, environmental, or personal, causes the position of the equilibrium to change.  Le Chatelier’s Principle allows us to predict the results that follow from changing the conditions of a system at    chemical equilibrium.  This means that scientists are able to develop techniques to control chemical reactions in natural and industrial settings in order to obtain desired products.

CHEMICAL EQUILIBRIUM

A chemical equilibrium occurs because both the forward and reverse reactions proceed simultaneously. When reversible reactions reach a point where the rate of the forward reaction is equal to the rate of the   backward reaction a state of dynamic equilibrium is said to exist. The amounts of both reactants and products will remain the constant, eventhough both forward and reverse reactions continue to occur. Hence no change will be observed in a reaction at equilibrium. An equilibrium mixture can be recognised by the presence of both reactants and products in the reaction mixture.

For example

Activity 6.1 - A Model for Equilibrium

QUESTIONS:

1.    How many employees move in and out of the factory building during each hour?

2.    Are the employees who move in and out of the building each hour the same people? Explain your answer.

3.    Does the number of employees in the building change from hour to hour? Explain your answer.

4.    Over the course of a day, the employees in the Acme Manufacturing Plant are said to be in a "dynamic equilibrium". Based on your understanding of how the staff move in and out of the plant, explain what is meant by the term "dynamic equilibrium".

Like the Acme Manufacturing Plant, chemical reactions can also reach equilibrium. Answer the following questions about the chemical equation in table above by applying the insight you gained from the Acme Manufacturing Plant questions.

5.    When the reaction between hydrogen and oxygen reaches equilibrium:

(a)   Does the number of molecules in the reaction vessel change? Explain.

(b)   Is the reaction still proceeding in the forward direction?

(c)   Is the reaction still proceeding in the reverse direction?

(d)   Are the concentrations of the products and reactants changing?

(e)   Are the rates of the forward and reverse reactions the same?

(f)   Does the heat content of the system become constant? Explain your answer

EXAMPLE: The Ammonia Equilibrium

To synthesise ammonia, x moles of nitrogen gas, 3x moles of hydrogen gas and a catalyst such as platinum, in a closed vessel and the temperature increased to 500oC. The hydrogen and nitrogen will react together to form. ammonia.

N2(g)   +   3H2(g)    →    2NH3(g)      (1)

The rate of the reaction is rapid at first but slows down as the amounts of nitrogen and hydrogen decrease. Eventually the reaction appears to cease and the vessel contains a mixture of ammonia, nitrogen and hydrogen.

Suppose we now place pure ammonia in a similar vessel with the same catalyst and again increase the temperature to 500oC. The ammonia is decomposed according to the reaction 2NH3(g)    →   N2(g)   +   3H2(g)     (2)

Although initially fast, the rate of this reaction will slow down as the concentration of the ammonia is decreased. However, as N2  and H2  are formed they will begin to react to form. ammonia as we have already seen in reaction (1) above. They react slowly at first but as their concentrations increase the rate of formation of the ammonia increases. So in this vessel we have a decreasing rate of decomposition of the ammonia and an increasing rate of formation of ammonia.

When the two rates become equal a state of equilibrium will exist. The rate of combination equals the rate of decomposition - no net reaction occurs, there is no change in any concentration and the reaction appears to stop. Since neither reaction actually stops, the system is said to be in a state of dynamic equilibrium. If we measured the concentrations in the two vessels at equilibrium we would find that the concentrations of NH3, H2 and N2 were exactly the same.

To show that both forward and reverse reactions are occurring, double arrows are used for the reaction equation.

N2(g)   +   3H2(g)    ⇌    2NH3(g)

In principle, all chemical reactions are equilibrium reactions but for many the mixture at equilibrium has so little of the reactant(s) that the single-ended arrow that we have been using can be regarded as an accurate description of the experimental facts. We would say that the position of equilibrium is so far to the right that the reaction can be regarded as going to completion.  If we chose to write the reaction the other way around, we would have a reaction that did not happen!

Focussing Questions 1:

1.    What does it mean when a double headed arrow, ⇌, is used in a chemical equation?

2.    What is happening to the products of a chemical reaction when a system is at equilibrium?

3.    What is the relationship of the rate of the forward and reverse reactions when equilibrium is established?

4.    What would be observed in a system that is at equilibrium (macroscopic level)?

5.    What would be observed at particle level when a system is at equilibrium?

6.    Why would a system at equilibrium be a problem for an industry relying on a chemical reaction as part of its production line?

7.     What has the picture of the escalator or conveyor belt got to do with equilibrium?

CHANGING THE POSITION OF AN EQUILIBRIUM - LE CHÂTELIER’S PRINCIPLE

When a reaction mixture is at equilibrium it appears to have stopped. It is  possible to “disturb” the equilibrium and make more reactants or products by changing the reaction conditions. Le Châtelier’s Principle provides a    useful way of making qualitative predictions about the effects that changes in reaction conditions will have on an equilibrium. It states that if a stress is applied to a system at equilibrium, the equilibrium will tend to  shift in a direction to relieve that stress.

A stress is a change in conditions, such as concentration, pressure, or temperature.  A shift is an increase in   the rate of the forward reaction to form. products or the reverse reaction to form. reactants until equilibrium is re-established.

We will consider the effect of some different stresses on the reaction:   N2(g)   +   3H2(g)     ÷  2NH3(g)

1. Changing concentration

Concentration can be changed by adding or removing any of the reactants or products. (Each of these actions is a stress). Adding more nitrogen will cause the equilibrium to shift to the right to relieve the stress of an added reactant. After the shift there will be more NH3 present, less H2  (as some of it will have reacted with the added N2) and more N2  than in the original mixture.

On the other hand if we were to add more NH3  then the equilibrium will shift to the left to use up the added product. After the shift, there would be more N2  and H2  and less NH3  than there would have been if no shift had occurred.  There will be more NH3  in the container than was present in the initial mixture.

2. Changing pressure

Increasing pressure will move the equilibrium to the side with the smaller number of gas molecules. So increasing the pressure on the ammonia equilibrium will result in a shift to the right giving more ammonia. This is because on the lefthand side of the equation there are 4 moles of gas particles (1 ‘molecule’ of N2 and 3 ‘molecules’ of H2). But on the right hand side there are just 2 moles of ammonia gas particles. Pressure can be increased by reducing the volume of the system. This will result in more molecules per unit volume and the equilibrium will shift in a direction that reduces the number of molecules of gas in the new volume. In the case of the reaction of nitrogen with hydrogen to form ammonia, more ammonia is formed because    that reduces the number of gas particles.

Note that changes in pressure of gaseous reactions do not always lead to a shift in the equilibrium.  Consider the equilibrium

N2(g)   +   O2(g)    ÷  2 NO(g)

Since there are two molecules of gas on each side of the equation, neither a shift to the left nor the right will change the total pressure. A reduction in volume of the reaction vessel would increase the total pressure but would not shift the equilibrium.

3. Changing temperature

The effect of a change in temperature is dependent on whether the reaction is endothermic or exothermic. The formation of ammonia is an exothermic reaction i.e. the thermodynamic equation is:

N2(g)   +   3 2(g)   ÷ 2 NH3(g)   +   heat

If weremove heat from the system by lowering the temperature the system will react to oppose this stress by creating more heat, which means it moves to the right causing more ammonia to be formed. For an endothermic system, a decrease in temperature will cause the reaction to move to the left as this is the direction that will replace heat.

4. Using a catalyst

Catalysts speedup the rate at which the reaction reaches equilibrium, but does not alter the position of the equilibrium.

Focussing Questions 2:

1.    What does an equilibrium system do when subjected to some kind of change or stress?

2.    Will an equilibrium move to favour reactants or products when more reactant is added? Does this mean that it will move to the right or the left?

3.    How does reducing the volume of a system affect the pressure? Explain this using collision theory.

4.    How does increasing the pressure of a system affect an equilibrium if both reactants and products are gasses? Will increasing the pressure always change the position of an equilibrium?

5.    How does the equilibrium change when an exothermic system is heated?

Activity 6.2- Le Chatelier’s Principle

INFORMATION:

Reactant: Increase (↑) causes the equilibrium to shift to the right (→)

Decrease (↓) causes the equilibrium to shift to the left (←)

Product: Increase (↑) causes the equilibrium to shift to the left (←)

Decrease (↓) causes the equilibrium to shift to the right (→)

Temperature: A change in temperature corresponds to a change in energy therefore by using the ‘energy’ term in the equation, energy can be treated like a reactant or product (see above).

Pressure: An increase (↑) in pressure causes the equilibrium to favour the "smaller number of moles of gas".

A decrease (↓) in pressure causes the equilibrium to favour the "larger number of moles of gas"

Note: If the number of moles of gas is the same on both sides, then a change in pressure has no effect in the equilibrium.

The following equation describes a system that is at equilibrium:

2H2 (g) + O2 (g) ⇌ 2H2O (g) + energy (heat)

QUESTIONS:

1.   For each of the stresses given below, apply Le Chatelier's Principle and indicate the direction of the shift in the equilibrium.  (The first one is completed for you.)

Stress	Shift   Direction
(a) Concentration H2 increases	→   shifts to the right
(b)   Concentration H2 decreases
(c) Concentration of O2 increases
(d) Concentration of O2 decreases
(e) Concentration of H2O increases
(f) Temperature increases
(g)   Temperature decreases
(h)   Pressure increases

The following questions are based on your answers to question 1.

(a)  In general terms, describe the direction of the equilibrium shift when the concentration of a reactant is increased.

(b)  If an equilibrium shifts to the right, which reaction is favoured, the forward or the reverse?

(c)  What happens to the concentrations of the reactants H2 and O2 when the reaction above shifts to the right?

(d) What happens to the concentration of the product H2O when the reaction above shifts to the right? (e)  If an equilibrium shifts to the left, which reaction is favoured, the forward or the reverse?

(f)  What happens to the concentrations of the reactants H2 and O2 when the reaction above shifts to the left?

(g) What happens to the concentration of the product H2O when the reaction above shifts to the left?

(h) What is true of the reaction rates for the forward and reverse reactions when a new equilibrium is established?

Application: The Industrial Preparation of Ammonia

About 100 million tonnes of ammonia are produced annually using the Haber process. In NZ this takes place at the ammonia-urea plant in Kapuni in Taranaki. Most of the ammonia is used in the production of nitrogen fertilisers which are needed to improve the yields of wheat and other grains. It has been suggested that the discovery of this process for making ammonia is largely responsible for the survival of 40% of the world’s population.

The Ammonia-urea plant in Taranaki

The chemical engineers operating a manufacturing plant need to understand the factors that influence the position of the equilibrium. We have already learnt the following:

(a)   Increasing the temperature of the reaction mixture speeds up the rate of approach to equilibrium but

shifts the equilibrium in favour of the formation of N2  and H2.

(b)   Increasing the pressure favours the formation of NH3.

(c)   Removing NH3(g), a product, favour the formation of NH3

(d)   Adding a catalyst will mean that equilibrium is reached quickly This is taken into account in the following ways:

•     Lower temperatures and higher pressures favour the formation of NH3.

•     At lower temperatures the rate of reaction is slow so equilibrium is reached slowly so ammonia plants   are normally operated between 380-450 oC so that the reaction proceeds at a reasonable rate. A catalyst must be present in order to speedup the rate even at these temperatures.

•     Normal operating pressures for the formation of NH3 are about 200 atmospheres, although higher

pressures favour the production of more NH3because it is expensive to build containers that can cope with these high pressures.

•     The gaseous ammonia product is liquefied as it forms to help shift the equilibrium further to the right.

If we were able to find a better catalyst, which would increase the rate of reaction, then it should be possible to operate at both lower temperatures and maybe lower pressures. Such a catalyst we could make considerable savings in the cost of production of ammonia. In nature the enzyme nitrogenase, which occurs in bacteria that live in the root nodules of legumes (peas and clover) converts N2(g) to NH3(g) at 1 atm pressure and room temperature. We do not know how this enzyme works but we do know that the bacteria expend large amounts of energy in order to drive this reaction.

Exercise 6.1

(a)      Consider the equilibrium:   N2O(g)  +  NO2(g)   ÷ 3 NO(g)     ΔrH = 156 kJ mol-1

How will the amount of NO be affected by the following changes? Explain your answer.

(i)  Addition of NO2(g)

(ii)  Removal of N2O(g)

(iii) Removal of NO(g)

(iv) Increasing the temperature of the reaction mixture

(v) Increasing the volume of the container

(b)     Consider the equilibrium:  C(s)  +  2H2(g)   ÷ CH4(g)    ΔrH = -75 kJ mol-1

In which direction will the equilibrium be shifted by the following changes? Explain your answer using Le Chatelier’s Principle (i)   Addition of CH4(g)

(ii)  Removal of H2(g)

(iii) Decreasing the temperature of the reaction mixture

(iv)  Decreasing the volume of the container

(v)   Adding a catalyst to the system

EQUILIBRIUM CONSTANTS

Reversible reactions can proceed in either direction, depending on the conditions. In a reaction mixture, once equilibrium is reached, reactions in opposite directions proceed at the same rate sono change in concentrations of reactants and products is observed. However, on the initial mixing of the reactants, some reactions will proceed much further than others before the reverse reaction kicks in and equilibrium is established.

The equilibrium constant, Kc, provides information about the extent to which a reaction has occurred when equilibrium is reached by comparing the concentrations of reactants and products.

The entire equation is called the equilibrium constant expression. The square brackets mean “the

concentration of ” the substance they enclose. No matter what the initial concentration of reactants or products at a given temperature the ratio of the concentrations at equilibrium will be equal to the

constant K. NOTES:

•   Solids, pure liquids and solvents (usually water) are not included in the equilibrium constant expression

•   The numerical value of the equilibrium constant does not have any associated units

•   In equilibrium constant expressions the pressures of gaseous components are expressed in atmospheres (abbreviation atm) not in Pascal.

•   The equilibrium constant is for the forward reaction as written in the equation.

For the reverse reaction K' = K/1