Equilibrium Reactions and Equilibrium Constants(1)

When does this system reach equilibrium/

Chemical EquilibriumThe state where the concentrations of all

reactants and products remain constant with time therefore macroscopic properties such as colour, temperature and pressure are constant. This can only occur in a closed system.On the molecular level, there is frantic activity.

Equilibrium is not static, but is a highly dynamic situation therefore the rate of the forward and reverse reactions are the same.

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2

Explain how it is possible to use this graph to determine the time at which equilibrium is reached. Suggest an equation for the reaction being investigated.

Can a physical change reach an equilibrium?

Are there reversible reactions occurring? In the closed system will the rate of evaporation eventually equal the rate of condensation?

Will the system reach equilibrium?

Which statement(s) is/are true for a mixture of ice and water at equilibrium? I. The rates of melting and freezing are equal. II. The amounts of ice and water are equal. III. The same position of equilibrium can be reached by cooling water and heating ice. A. I only B. I and III only C. II only D. III only

B

A sealed container at room temperature is half full of water. The temperature of the container is increased and left for equilibrium to re-establish. Which statement is correct when the equilibrium is reestablished at the higher temperature? A. The rate of vaporization is greater than the rate of condensation. B.The amount of water vapour is greater than the amount of liquid water. C. The amount of water vapour is greater than it is at the lower temperature. D. The rate of condensation is greater than the rate of vaporization.

C

Vapour pressure The magnitude of vapour pressure depends on the

concentration of the vapour present This indicates the position of equilibrium High vapour pressure indicates that equilibrium lies to the right Low vapour pressure indicates that equilibrium is to the left.

It is possible to determine the equilibrium pressure for any enclosed vapour at RT and 1 Atm. This pressure is defined as the Vapour Pressure of that substance.

Factors that affect vapour pressure 15 mins to make notes on the effect of Temperature Intermolecular forces

Brown & Ford p 254 259, and/or Derry HL p 168 172

17.1.1

Describe the equilibrium established between a liquid and its own vapour and how it is affected by temperature changes.

17.1.2

Sketch graphs showing the relationship between vapour pressure and temperature and explain them in terms of the kinetic theory.State and explain the relationship between enthalpy of vaporization, boiling point and intermolecular forces.

17.1.3

Vapour Pressure depends on Temperature:

If these graphs represent the same sample of a particular substance , which sample is at a higher temperature? In which case will the Vapour Pressure be greater? Is Vapour Pressure directly or inversely proportional to temperature?

Vapour Pressure depends on the strength of the Intermolecular Forces in the Liquid:

At any given temperature there will only be a certain number of molecules which will have sufficent energy to overcome the intermolecular forces and escape from the liquid into the vapour phase. Predict the difference between the vapour pressures of substances A and B, at a given temperature , if A has stronger Intermolecular forces than B.

The Relationship between Vapour Pressure and Boiling Point:This liquid is boiling. Consider the situation occurring inside one of the bubbles of vapour. Consider a bubble just below the surface.

A.P.The bubble contains vapour in equilibrium with its liquid it exhibits the Vapour Pressure for that liquid at that temperature. So why isnt the bubble squashed out of existence by the Atmospheric Pressure?

Boiling Point The temperature at which the Vapour Pressure of a liquid is equal to or greater then the Atmospheric Pressure ( ie. the bubble survives!!)

To correctly interpret this graph it is necessary to focus on the lines drawn to represent various changing atmospheric pressures. What happens to the boiling point of water on the top of Mount Everest?

Using Vapour Pressure Graphs:

Use this graph to predict the boiling point of each substance at 1 Atm.

State which substance has the highest Intermolecular Forces at RT. Explain your reasoning. Describe the relationship between temperature and Vapour Pressure. Use data from the graph to justify your description.

A Summary:

Write a paragraph to explain the trends shown on this table.

A liquid and its vapour are at equilibrium inside a sealed container. Which change will alter the equilibrium vapour pressure of the liquid in the container? A. Adding more liquid B. Adding more vapour C. Decreasing the volume of the container D. Decreasing the temperature

D

Equilibrium Expression: Write an equilibrium expression for the following

equation: 4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

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22

Equilibrium Expression:4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

NO2 H 2 O K 4 7 NH 3 O2

4

6

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23

17.2.1

Solve homogeneous equilibrium problems using the expression for Kc.

EQUILIBRIUM GENERAL QUESTIONSThere are several types of equilibrium questions. TYPE 1: Using K expressions for any equilibrium: QUESTION: This question refers to the equation: 2SO2 (g) + O2 (g) 2SO3 (g) If the equilibrium concentrations of SO2 and SO3 are 2.00M and 10.0M, find the equilibrium concentration for O2. (K = 800.0) GIVEN: Equilibrium concentrations and a K value. TO FIND: Other equilibrium concentrations. ANSWER: Always follow these initial steps 1. Write the balanced equation 2. Write the equilibrium expression 3. Fill in known values.

This question refers to the equation: 2SO2 (g) + O2 (g) 2SO3 (g) If the equilibrium concentrations of SO2 and SO3 are 2.00M and 10.0M, find the equilibrium concentration for O2. (K = 800.0)2SO2 (g) + O2 (g) 2SO3 (g)[ SO3 ]2 K 800 [ SO2 ]2 [ O2 ]

10.0 2 800.0 2.00 2 [O2 ]

10.0 2 [O2 ] 2 2.00 800.0

[O2] = 0.0313M

(Using 3 sig figs)

For the reaction belowH (g) + I (g) 2HI(g)2 2

at a certain temperature, the equilibrium concentrations are (in mol dm )3

[H ] = 0.30, [I ] = 0.30, [HI] = 3.02 2

What is the value of K? A. 5.0 B. 10 C. 15 D. 100

D

Notes on Equilibrium Expressions (EE) The

Equilibrium Expression for a reaction is the reciprocal of that for the reaction written in reverse.

When the equation for a reaction

is multiplied by n,

EEnew = (EEoriginal)n The units for K

depend on the reaction being

considered.

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29

The value of the equilibrium constant for the reaction 2HI(g) H2(g) + I2(g) is 0.25 at 440C. What would the value of the equilibrium constant be for the following reaction at the same temperature? H2(g) + I2(g) 2HI(g)

A. B. C. D.

0.25 0.50 2.0 4.0

D

EQUILIBRIUM GENERAL QUESTIONSTYPE 2: Using an ICE (initial, change, equilibrium) table to organize initial and equilibrium data:QUESTION: A2 (g) + B2 (g) 2AB(g) The initial concentrations of A2 and B2 were 1.00M and 0M respectively. The final concentrations of A2 and B2 are 1.50M and .50M. If the K value for this equilibrium is 4.0; find the initial and final concentrations of AB. GIVEN: Initial and Equilibrium concentrations and/or K value. TO FIND: Initial and Euilibrium and/or K concentrations. N.B. Use an ICE table to organise your data

Solving Equilibrium Problems1. 2. Balance the equation. Write the equilibrium expression. 3. List the initial concentrations. 4. Use an ICE table to determine the equilibrium concentrations 5. Calculate K

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33

A2 (g) + B2 (g) 2AB(g) The initial concentrations of A2 and B2 were 1.00M and 0M respectively. The final concentrations of A2 and B2 are 1.50M and .50M. If the K value for this equilibrium is 4.0; find the initial and final concentrations of AB.ANSWER: A2 (g) + B2 (g) 2AB (g)K [ AB] 2 4.0 [ A2 ][ B2 ]

Note that this equilibrium reacted in the reverse direction!!

Initial Change Equilibrium

EQUATION A2 B2 Before 1.00(given) 0 (given) During 0.50 .50 (This line must have the (therefore made) same ratios as the equation) After 1.50 (given) .50 (given) (This line can be used in the K expression)K X2 4.0 1.5 .50

2AB ?

1.0 (therefore used)

X

= X2 = 4.0 x .75 X2 = 3.00

Final Conc. of AB = X = 3.0 = 1.73M the initial concentration of AB must equal 1.73 - 1.00 =0 .73M

A 1.0 dm3 reaction vessel initially contains 6.0 mol of P and 6.0 mol of Q. At equilibrium 4.0 mol of R is present. What is the value of Kc for the following reaction? P(g) + Q(g) R(g) + S(g) A. 0.11 B. 0.25 C. 0.44 D. 4.00

D

Le Chteliers Principle . . . if a change is imposed on a

system at equilibrium, the position of the equilibrium will shift in a direction that tends to partially counteract that change.

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37

Effects of Changes on the System 1.

Concentration: The system will shift away from the added component K will remain the same.

2.

Temperature: K will change depending upon the temperature (treat the energy change associated with the reaction as a reactant or produce eg. if the reaction is endothermic show energy on the reactant side). If the movement predicted by Le Chateliers Principle favours the products then the value for K will increase.

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38

Effects of Changes on the System(continued) 3.

Pressure K will remain the same: a. Addition of inert gas does not affect the equilibrium position. b. Decreasing the volume shifts the equilibrium toward the side with fewer gaseous moles.

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39

In the reaction belowN (g) + 3H (g) 2NH (g) H = 92 kJ2 2 3

which of the following changes will increase the amount of ammonia at equilibrium? I. Increasing the pressure II. Increasing the temperature III. Adding a catalyst A. B. C. D. I only II only I and II only II and III only

A

Use Le Chateliers principle to predict the shift in the given equilibrium system for each of the given imposed changes.

In which of these cases will the value for K alter?

In each of these identified cases, will K become larger or smaller

The Haber Process:

Consider the Haber Reaction:

State two equilibria concepts that can be illustrated by this data.

State and explain two trends shown by this data. Use Le Chateliers Principle in your answer.

Suggest an explanation for the change in the observed K value.

A Summary of the conditions affecting the equilibrium position of the Haber Process:Describe and explain two trends shown by this graph.

The equation for the main reaction in the Haber process is: N2(g) + 3H2(g) 2NH3(g) H is negative (i) Determine the equilibrium constant expression for this reaction. (1) (ii) State and explain the effect on the equilibrium yield of ammonia with increasing the pressure and the temperature. (4) (iii) In practice, typical conditions used in the Haber process involve a temperature of 500C and a pressure of 200 atm. Explain why these conditions are used rather than those that give the highest yield. (2) (iv) At a certain temperature and pressure, 1.1 dm3 of N2(g) reacts with 3.3 dm3 of H2(g). Calculate the volume of NH3(g), that will be produced. (1) (v) Suggest why this reaction is important for humanity. (1) (vi) A chemist claims to have developed a new catalyst for the Haber process, which increases the yield of ammonia. State the catalyst normally used for the Haber process, and comment on the claim made by this chemist. (2) (Total 11 marks)

(i) (K =) (ignore units); 1 (ii) Increasing the pressure: Yield increases/equilibrium moves to the right/more ammonia; 4 gas molecules 2/decrease in volume/fewer gas molecules on right hand side; Increasing the temperature: Yield decreases/equilibrium moves to the left/less ammonia; Exothermic reaction/OWTTE; 4 (iii) Higher temperature increases rate; Lower pressure is less expensive/lower cost of operating at low pressure/reinforced pipes not needed; 2c

Do not award a mark just for the word compromise.

(iv)

2.2 (dm );3

1 1 max

Penalize incorrect units.

(v) Fertilizers/increasing crop yields; Production of explosives for mining; (vi) Fe/iron; Allow magnetite/iron oxide.

Claim is not valid since catalysts do not alter the yield/position of equilibrium/only increase the rate of reaction; 2 [11]

The Contact Process

Consider the following reaction in the Contact process for the production of sulfuric acid for parts (a) to (d) in this question. 2SO + O 2SO (a) Write the equilibrium constant expression for the reaction. (1)2 2 3

(b)(i) State the catalyst used in this reaction of the Contact process.

(1)(ii) State and explain the effect of the catalyst on the value of the equilibrium constant and on the rate of the reaction.

(4) (c)Use the collision theory to explain why increasing the temperature increases the rate of the reaction between sulfur dioxide and oxygen. (2) (d) Using Le Chateliers principle state and explain the effect on the position of equilibrium of(i) increasing the pressure at constant temperature.

(2)(ii) removing of sulfur trioxide. using a catalyst.

(2)(iii)

(2) (Total 14 marks)

(a) K/K = [SO ] [SO ] [O ]; Accept correct K expression.c 3 2 2 2 2 p

12 5

(b) (i)

vanadium(V) oxide/(di)vanadium pentaoxide/V O ;

1

Allow just vanadium oxide but not correct formula. (ii) catalyst does not affect the value of Kc;

forward and reverse rates increase equally/by the same factor; catalyst increases the rate of the reaction;

(by providing an alternative path for the reaction with) lower activation energy; 4

(c) more energetic collisions/more molecules have energy greater than activation energy; more frequent collisions; 2

Do not accept more collisions without reference to time. (d) (i)

shifts equilibrium position to the products/right;

to the side with fewer gas molecules or moles/lower volume of gas; (ii) shifts equilibrium position to the products/right;

2

to compensate for loss of SO /produce more SO ;3 3

2 2

(iii)

no effect;

forward and backward rates increased equally/by the same factor; [14]

The equation for another reaction used in industry is CO(g) + H O(g) H (g) + CO (g) H = -42 kJ (i) Under certain conditions of temperature and pressure, 2.0 mol of carbon monoxide and 3.2 mol of steam were left to reach equilibrium. At equilibrium, 1.6 mol of both hydrogen and carbon dioxide were present.2 2 2

Calculate the amounts of carbon monoxide and steam at equilibrium and the value of Kc.

(3) (ii) Under the same conditions of temperature and pressure, 2.0 mol of carbon monoxide and 2.0 mol of steam were left to reach equilibrium.Calculate the amounts of each reactant and product at equilibrium. (If you were unable to calculate a value for Kc in (i) use the value 9.0, although this is not the correct value.)

(2) (Total 5 marks)

(i) CO = 0.4 (mol); H O = 1.6 (mol); K (= 1.6 0.41.6) = 4.0/4;2 c 2

3

Apply ECF from K expression. Ignore units.c

(ii) H and CO /products = 1.33/1.3 (mol); CO and H O/reactants = 0.67/0.7 (mol); 22 2 2

Using K = 9.0, values for H and CO are 1.5 and values for CO and H O are 0.5.c 2 2 2

[5]

The equilibrium between nitrogen dioxide (dark brown) and dinitrogen tetroxide (colourless) is represented by the following equation.2NO (g) N O (g)2 2 4

H = negative K = 1 at 328Kc

(a) (b)

Write the equilibrium constant expression, K .c

(1) State and explain the effect of an increase in temperature on the value of K .c

(c)

(d)

(2) State and explain the visible change that takes place as a result of a decrease in pressure, after equilibrium is re-established. (2) Two moles of NO (g) and two moles of N O (g) were placed in an empty 1 dm container and allowed to come to equilibrium at 328 K. Predict, with reference to the value of K , whether the equilibrium mixture would contain more or less than two moles of NO (g).2 2 4 c 2

3

(2) (Total 7 marks)

No ECF throughout this question. (a) 1 (b) Kc decreases; forward reaction is exothermic/H is negative/equilibrium moves to left/OWTTE; 2 (c) (mixture will get) darker/darker than expected; equilibrium position moves to the left/towards reactants as there is an increase in the number of moles of gas from right to left; 2 (d) (equilibrium mixture contains) less (than 2 moles NO2); [N O ] given values make k= [NO ] 2 4 2 1 2 2

i.e. too much NO2/OWTTE; 2 [7]