Industrial Chemistry Flashcards

0
Q

From the data given, identify each of the following reactions as either exothermic or endothermic.
a​ CO2(g) + C(s) ® 2CO(g); DH = +161 kJ mol–1

b​ N2(g) + 3H2(g) ® 2NH3(g); DH = –91 kJ mol–1

c​ 6CO2(g) + 6H2O(l) ® C6H12O6(aq) + 6O2(g); DH = +2803 kJ mol–1

d​ H+ (aq) + OH–(aq) ® H2O(l); DH = –57 kJ mol–1

A

a​ endothermic

b​ exothermic

c​ endothermic

d​ exothermic

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
1
Q

Natural gas begins to burn when lit with a match. Why does it continue to burn when the match is taken away?

A

Once it has commenced, the reaction continues because its activation energy is continually being supplied by thermal energy released as the match burns.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

Consider the examples of reactions mentioned on page 251—wood burning on a camp fire, bathroom tiles being cleaned, a cake baking, and a tomato plant growing.

a​ How would you speed up the rates of these reactions?

b​ Explain why the methods you suggested would produce an increase in the reaction rate.

A

a​ To increase the rate at which wood burns you could use smaller pieces of wood. Bathroom tiles could be cleaned more rapidly using a more concentrated cleaning agent. Cakes bake more rapidly if the temperature in the oven is increased. Increased hours of sunlight make tomato plants grow faster.

b​ The frequency of collisions between the reacting particles is increased by increasing the surface area (in the case of the wood burning) and by increasing concentration (when bathroom tiles are being cleaned). Increasing the temperature at which a cake bakes causes more reactant particles to have sufficient energy to overcome the activation energy of reactions and also increases the rate of collisions between reacting particles. Increasing the hours of sunlight experienced by a tomato plant results in more reactant particles having sufficient energy to continue the reaction for a longer period.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

Explain the following observations in terms of the behaviour of particles.

a​ There have been many explosions in coal mines.

b​ Refrigeration slows down the browning of sliced apples.

c​ Bushfires often start during lightning storms.

d​ Iron anchors from shipwrecks can show little corrosion after years in the sea.

e​ A burning match is used to light a candle, but the candle continues to burn when the match is extinguished.

A

a​ If coal dust in a coal mine is accidentally ignited an explosion may occur. Coal dust has a high surface area and therefore burns rapidly, producing gases so quickly that pressure in the coal mine mounts and an explosion eventually occurs.

b ​At low temperatures, the rate of the reactions involved in food spoilage is slow because few particles have energy equal to or greater than the activation energy of the spoilage reactions. This effect is also compounded by the lower frequency of collisions between reactants at low temperatures.

c ​A lightning flash can provide the activation energy needed for particles to undergo a combustion reaction.

d​ The low concentration of oxygen at great depths in sea water results in a low frequency of collisions between oxygen molecules and iron metal and, as a consequence, very little corrosion.

e ​A match initially provides the reactant particles with the energy needed to overcome the activation energy of the reaction. Once the reaction has commenced, the heat released as a result is sufficient to maintain the process.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Some reactions are called explosions. What are the features of these reactions?

A

In an explosion, the reaction occurs rapidly and a large amount of energy is released. The products are gaseous and a large change in volume occurs.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

Decide, giving reasons for your answers, whether the following processes are endothermic or exothermic:

a ​burning of wood

b ​melting of ice

c ​recharging of a car battery

d​ decomposition of plants in a compost heap

A

a​ Exothermic, because heat and light energy are released to the surrounding environment by the combustion of wood.

b​ Endothermic, because thermal energy is absorbed from the surrounding environment to melt the ice.

c​ Endothermic, because electrical energy is consumed from a power supply as the battery is recharged.

d​ Exothermic, because heat energy is released to the surrounding environment as organisms in the compost heap decompose the plant material. The temperature of the heap rises as a consequence.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

The activation energy for the reaction A + B ® C is greater than the activation energy for the reverse (opposite) reaction C ® A + B. Is the reaction A + B ® C exothermic or endothermic? Explain.

A

Endothermic. As the diagrams demonstrate, the activation energy of an endothermic reaction is larger than the activation energy of the reaction in reverse.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

Hydrogen reacts explosively with oxygen to form water.
a​ What chemical bonds are broken in the reaction?
b​ What chemical bonds are formed?
c​ Explain how the energy changes during bond-breaking and bond-forming affect the energy change for the reaction.

d​ Why is there no reaction until the reaction mixture is ignited?

A

a​ The single H–H bond in each hydrogen molecule and the double O=O bond in each oxygen molecule are broken in the course of this reaction.

b ​Two H–O bonds are formed in each new water molecule during the reaction.

c​ The energy change for the reaction is the difference between the energy absorbed to break the bonds in the H2 and O2 reactants, and the energy released when the bonds in the H2O product are made.

d​ No reaction occurs until sufficient energy is supplied to overcome the activation energy.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

The combustion of butane gas in portable stoves can be represented by the equation:
2C4H10(g) + 13O2(g) ® 8CO2(g) + 10H2O(l); DH = –5772 kJ mol–1
a​ How does the overall energy of the bonds in the reactants compare with the energy of the bonds in the products?

b​ Draw an energy profile diagram for the reaction, labelling DH and the activation energy.

c​ Describe how your diagram would change for the equation:
C4H10(g) + 6.5O2(g) ® 4CO2(g) + 5H2O(l)

A

a​ higher energy in bonds of reactants
b

c​Activation energy and H values are halved.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Many major car makers have unveiled hydrogen-powered cars. In the engines of these cars, hydrogen reacts with oxygen from the air to produce water.
2H2(g) + O2(g) ® 2H2O(g)
Energy changes for the reaction are shown in Figure 15.22.

a​ What is the magnitude of the activation energy of this reaction?

b ​What is DH for this reaction?

c​ Several groups of scientists have claimed to have split water into hydrogen and oxygen using a molybdenum catalyst:
2H2O(g) 2H2(g) + O2(g)
​Sketch energy change graphs for this reaction with and without the presence of a catalyst.

d​ What is the value of DH for this water-splitting equation?

A

a ​1370 kJ mol–1
b​ –572 kJ mol–1
c​

d. ​+572 kJ mol–1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

When one mole of methane gas burns completely in oxygen, the process of bond breaking uses 3380 kJ of energy and 4270 kJ of energy is released as new bonds form.

a​ Write a balanced chemical equation for the reaction.

b ​Calculate the value of the heat of reaction, DH, for the reaction.

c​ Draw a diagram to show the changes in energy during the course of the reaction.

A

a​ CH4(g) + 2O2(g) ® CO2(g) + 2H2O(g)
b ​–890 kJ mol–1
c​

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

The formation of hydrogen iodide from its elements is represented by the equation:
H2(g) + I2(g) ® 2HI(g)
This endothermic reaction has an activation energy of 167 kJ mol–1 and the heat of reaction, DH, is +28 kJ mol–1. What is the activation energy for the reverse reaction, the decomposition of two mole of hydrogen iodide?

A

139 kJ mol–1

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

Account for the following observations with reference to the collision model of particle behaviour.
a​ Surfboard manufacturers find that fibreglass plastics set within hours in summer but may remain tacky for days in winter.

b​ A bottle of fine aluminium powder has a caution sticker warning that it is ‘highly flammable, dust explosion possible’.

c​ A potato cooks much more slowly in a billy of boiling water on a trekking holiday in Nepal than a potato boiled in a similar way in the Australian bush.

A

a​ At higher temperatures the molecules that react to form fibreglass plastics have greater energy. They collide more frequently and are more likely to have a total energy exceeding the activation energy of the reaction involved, increasing the rate of reaction.

b​ Fine particles have a large surface area, resulting in a high frequency of collisions of aluminium particles with gas molecules (such as oxygen) in the air and hence rapid reaction rate. The aluminium can burn vigorously and release a large quantity of heat.

c ​At high altitude, such as in Nepal, air pressure is considerably lower than at any location in the Australian bush and so the water boils at a lower temperature in Nepal (up to 30ºC lower!). Thus, the average kinetic energy of the molecules in the potato is lower and the reactions involved in cooking a potato occur more slowly.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

a​ Explain why surface properties are important to the operation of catalysts.

b​ Many industrial catalysts are made into porous pellets. What is the reason for this?

A

a​ Reactions involving a heterogeneous catalyst take place at the surface of the catalyst. Reactants form bonds with the catalyst, lowering the activation energy of reactions and allowing them to proceed more rapidly.

b ​A porous pellet has a much larger surface area than a solid lump. More reactants may be in contact with the surface of a porous pellet at any instant, producing a faster rate of reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Explain the meaning of the terms:
a​ catalyst
b​ activation energy

A

a​ Catalyst – a substance that changes the rate of a chemical reaction without itself undergoing permanent change.

b ​Activation energy – the energy required by the reactants in order to form products in a reaction.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

If a sugar cube is held in the flame of a candle, the sugar melts and browns but does not burn. However, the cube will burn if salt is first rubbed into it, even though the salt does not react. Explain the effect of the salt on the activation energy of this combustion reaction.

A

When salt is mixed with sugar, the salt acts as a catalyst and lowers the activation energy of the combustion reaction between sugar and oxygen.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Perform a literature search using an Internet search engine to find details about three new catalysts that are being developed.

A

Individual student response required.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Figure 15.23
Energy profiles at 40°C and 60°C.
a​ Figure 15.23 shows the kinetic energy profile of particles at two different temperatures, 40°C and 60°C. Indicate the temperatures represented by graph A and graph B.

b​ Draw a graph of number of particles versus kinetic energy that shows the effect of a catalyst on a reaction.

c​ Use the diagram you have drawn in part b to explain in terms of collision theory how a catalyst increases the rate of a reaction.

A

A18.
a

b​

c​When a catalyst is present a different reaction intermediate is formed with bonds that require less energy to break (E´) than in the uncatalysed reaction (E´´). At a given temperature (average kinetic energy) more molecules have a kinetic energy E´ than E´´. As more reactants have sufficient energy to react, the rate of reaction increases.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

Lumps of limestone, calcium carbonate, react readily with dilute hydrochloric acid. Four large lumps of limestone, mass 10.0 g, were reacted with 100 mL 0.100 M acid.
a​ Write a balanced equation to describe the reaction.

b​ Which reactant is in excess? Use a calculation to support your answer.

c​D escribe a technique that you could use in a school laboratory to measure the rate of the reaction.

d​ 10.0 g of small lumps of limestone will react at a different rate from four large lumps. Will the rate of reaction with the smaller lumps be faster or slower? Explain your answer in terms of collision theory.

e​ List two other ways in which the rate of this reaction can be altered. Explain your answer in terms of collision theory.

A

a​ CaCO3(s) + 2HCl(aq) ® CaCl2(aq) + CO2(g) + H2O(l)

b​ n(CaCO3) = 10.0 g/100.1 g mol–1 = 0.0999 mol
​n(HCl) = 0.1 M ´ 0.1 L = 0.01 mol
​\CaCO3 is in excess

c​ The rate of reaction can be measured by:
​·​a decrease in mass of reaction mixture as CO2(g) escapes to the atmosphere
​·​an increase in pH with a pH probe as acid is consumed.

d​ The rate of reaction with the smaller lumps will be faster. The smaller lumps have a larger surface area so more collisions can occur per second.

e ​Increase temperature; increase concentration of hydrochloric acid.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

The reaction of hydrogen and iodine to form hydrogen iodide:
H2(g) + I2(s) F 2HI(g)
is shown on the energy level diagram Figure 15.24.

Figure 15.24
Energy level diagram for the production of hydrogen iodide.
a​ Copy Figure 15.24 and label the following: H2(g) and I2(s); HI(g); DH; activation energy.

b ​Is the reaction endothermic or exothermic?

c ​Draw on the diagram the energy profile that would result if a catalyst was used in the reaction.

A

a, c

b​ The reactant is endothermic.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

In 1996, while the Turkish ship MV B. Onal was riding at anchor in Delaware Bay, near Philadelphia in the USA, a 2 tonne hatch cover suddenly blew off. As the ship was carrying a cargo of iron, the surprised crew asked themselves, ‘Can iron explode?’
As you may be aware, traditionally iron oxide (Fe2O3) is reduced to molten iron in a blast furnace […]
A new process which uses less energy has been developed. Iron oxide is converted directly to solid iron without having to heat the reactants to the melting point of iron. Iron oxide is heated to 550°C in the presence of carbon monoxide and hydrogen gas. The iron oxide is reduced to iron by both gases with the formation of carbon dioxide or water.
Fe2O3(s) + 3CO(g) ® 2Fe(s) + 3CO2(g) (1)
Fe2O3(s) + 3H2(g) ® 2Fe(s) + 3H2O (g) (2)
The pellets of pure iron that are formed are extremely porous and full of many tiny holes, in contrast to the solid formed when the molten iron from a blast furnace cools. Under the right conditions the iron pellets can be oxidised back to iron oxide.
In most cases, iron is oxidised slowly by oxygen back to iron oxide and the resulting heat can readily escape. If the pellets are more than one metre deep, as in the hold of a ship, the heat cannot escape quickly enough and the temperature rises. This speeds up the reaction rate. If the temperature increases sufficiently and water is present, another reaction occurs and the oxidation rate is speeded up 100-fold, with the release of more heat:
Fe(s) + H2O(g) ® FeO(s) + H2(g) (3)
Any spark or fire will set off an explosion of hydrogen gas, and that is what happened on the MV B. Onal.
a​ What is the main reason the new reduction process uses less energy than the old process?

b​ Write equations showing the oxidation of iron by oxygen to form iron(II) oxide and iron(III) oxide.

c ​If water is present the oxidation reaction is speeded up 100-fold. Is water acting as a catalyst? Explain your answer.

d​ Is the reaction shown in equation 3 endothermic or exothermic?

e ​List the factors that increased the rate of reaction in equation 3.

f​ Firefighters were not able to use water to put out the fire in the cargo hold. Why not? Suggest how they could put out the fire.

A

a​ The temperature of the new process (550ºC) is much lower than the temperature of 1800ºC in the blast furnace.
b​2Fe(s) + O2(g) ® 2FeO(s)
4Fe(s) + 3O2(g) ® 2Fe2O3(s)

c ​No. The reaction of iron with water is different to the reaction of iron with dry oxygen and proceeds at a different rate.

d ​exothermic

e ​High surface area of iron pellets, high temperature caused by trapped heat that was unable to escape rapidly.

f​ Water would have caused the production of more hydrogen and increased the fire. The method used by the firefighters to extinguish the fire was to flood the hold with liquid nitrogen, which extinguished the surface fire but did not stop the deeper burning. A crane and clamshell bucket was then used to unload the iron into piles less than 1 m deep so the heat could escape.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

Chemical reactions in the body normally take place at 37°C. Explain how the rate of chemical reactions in the body can account for the following facts.
a​ The body often responds to illness by an increase in temperature, accompanied by a higher pulse rate and faster breathing.

b ​People rescued from drowning after 20 or 30 minutes in freezing water can sometimes survive and recover with no brain damage.

A

a​ Higher body temperature increases the rate of reactions. Increased pulse and breathing rate increases the concentration of reactants.

b ​Lower body temperature decreases rate of metabolic reactions in the body.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

Write an equation to show the equilibrium that exists between NaI(s) and Na+(aq) and I–(aq).

A

NaI(s) F Na+(aq) + I–(aq)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

a​ Sketch a graph of the change in the radioactivity of the solution over time.

b​ On the same axes, sketch a graph of the change in the radioactivity of the solid over time.

A

a, b​

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

a​ By referring to the equilibrium:
H2O(l) F H2O(g)
​explain what is meant by the ‘dynamic nature’ of chemical equilibrium and why wet clothes in a closed laundry bag do not dry.

b​ When the bag in part a is opened the clothes begin to dry. Is this due to an equilibrium process? Explain your answer.

A

a​ Chemical equilibrium is ‘dynamic’ because both forward and reverse reactions occur at the same rate. An equilibrium develops between water vapour and water when wet clothes are in a sealed bag, with water evaporating as rapidly as water vapour condenses, so the clothes remain wet.

b​ When the bag is opened, water vapour escapes and the rate of evaporation exceeds the rate of condensation. The system is not in equilibrium and the clothes dry.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

Write an expression for the equilibrium constant for the following equations.
a​ 2NH3(g) F N2(g) + 3H2(g)

b ​4NH3(g) + 3O2(g) F 2N2(g) + 6H2O(g)

c​ Cu2+(aq) + 4NH3(aq) F Cu(NH3)42+(aq)

d ​Ni2+(aq) + 6NH3(aq) F Ni(NH3)62+(aq)

A

a ​K =
b ​K =
c​ K =
d​ K =

26
Q

At 255°C the equilibrium constant is 100 M2 for the reaction:
2NH3(g) F N2(g) + 3H2(g)
a​ Write an expression for the equilibrium constant for the equation:
N2(g) + 3H2(g) F 2NH3(g)

b​ Calculate the equilibrium constant for the equation in part a.

c​ Write an expression for the equilibrium constant for the equation:
NH3(g) F N2(g) + H2(g)

d ​Calculate the equilibrium constant for the equation in part c.

e​ Use your answers to parts a to d to state the effect on the value of equilbrium constant when:
i​) the equation is reversed
ii​) the coefficients of the equation are halved

A

a​ K =

b ​Since = 100 M2
​= = 0.01 M–2

c​ K =

d ​The expression in part c is equal to the square root of
​Hence K = = =10 M

e​i​ When an equation is reversed the value of the equilibrium constant is the reciprocal of the original constant.
ii​ When the coefficients are halved the new equilibrium constant has a value equal to the square root of the original constant.

27
Q

A chemist investigated three different reactions and determined the value of the equilibrium constant for each. In which of the reactions are the products favoured over the reactants? Explain.
a​ Reaction 1: K = 0.0057

b Reaction 2: K = 2.5 × 109

c​ Reaction 3: K = 3.1 × 10–3

A

Reaction 2. The largest value of K indicates that the reaction favours the products.

28
Q

Q5.
State whether the equilibrium constants for each of the following would be increased, decreased or unchanged by a rise in temperature:
a​ 2NH3(g) F N2(g) + 3H2(g); DH = +91 kJ mol–1

b​ 4HCl(g) + O2(g) F 2H2O(g) + 2Cl2(g); DH = –113 kJ mol–1

c​ H2(g) + CO2(g) F H2O(g) + CO(g); DH = +42 kJ mol–1

d​ 2CO(g) + O2(g) F 2CO2(g); DH = –564 kJ mol–1

A

a increase
b decrease
c increase
d​ decrease

29
Q

Q6.
For each of the equilibria in Question 5, would the amounts of products increase, decrease or remain unchanged as temperature increases?

A

a ​increase

b ​decrease

c ​increase

d ​decrease

30
Q

Q7.
Calculate the equilibrium constant for the reaction represented by the equation:
N2O4(g) F 2NO2(g)
if an equilibrium mixture in a 3.0 L container was found to consist of 0.120 mol of N2O4 and 0.500 mol of NO2.

A
A7.
Step 1​From the equation write the expression for the equilibrium constant.
​K = 
Step 2​Calculate equilibrium concentrations of reactants and products.
​[NO2]​= 0.500 mol/3.0 L
​= 0.1667 M
​[N2O4]​= 0.120 mol/3.0 L
​= 0.0400 M
Step 3​Substitute equilibrium concentrations in the equilibrium expression and calculate the value of the equilibrium constant.
​K​= 
​= 0.69470 M
​= 0.69 M (two significant figures)
31
Q

Q8.
The equilibrium constant for the reaction given by the equation:
2HI(g) F H2(g) + I2(g)
is 48.8 at 455°C. An equilibrium mixture in a 2.0 L vessel at this temperature contains 0.220 mol of H2 and 0.110 mol of I2.
a ​Calculate the concentration of HI in this mixture.

b​ Another mixture was prepared by placing 4.0 mol of HI in a 2.0 L vessel at 330°C. At equilibrium 0.44 mol of H2 and 0.44 mol of I2 were present. Calculate the value of the equilibrium constant at this temperature.

c ​A third mixture consisted of 1.0 mol of HI, 0.24 mol of H2 and 0.32 mol of I2 in a 2.0 L container at 330°C. Decide if the mixture is at equilbrium and, if not, predict the direction the reaction will shift to reach equilibrium.

A

Q8 chapter 16
a​ Step 1​From the equation write the expression for the equilibrium constant.
​K​=
​​​= 48.8 (no units here because they cancel)
​Step 2​Calculate equilibrium concentrations of reactants.
​[H2]​= 0.220 mol/2.0 L
​​= 0.110 M
​[I2]​= 0.110 mol/2.0 L
​​= 0.055 M
​Step 3​Substitute equilibrium concentrations in the equilibrium expression and calculate the concentration of HI.
​48.8​=
​[HI]2​=
​[HI]​=
​​= 0.01113 M
​​= 0.011 M (two significant figures)

b ​Step 1​This type of question is best done by using a table as shown below.
​​The entries in the first column indicate the initial amount that was in the closed vessel, the amount that was used that is calculated from stoichiometry and the amount that remains (or is in excess) and is present at equilibrium, calculated by subtraction (initial – used).
​​Equilibrium constants are calculated from concentrations in M, so it is important to remember to divide the mole value by the volume to convert to molarity, M.

Volume = 2.0 L
2HI (g)F
H2(g)+
I2(g)
Initial
4.0 mol
0.0 mol
0.0 mol
Used
0.88 mol
At equilibrium
As 0.44 mol of H2 was formed, the equation tells us that 0.88 mol of HI must have been used. Hence, (4.0 – 0.88) = 3.12 mol is present at equilibrium.
3.12 mol/2.0 L = 1.56 M
0.44 mol
0.44 mol/2.0 L = 0.22 M
0.44 mol
0.44 mol/2.0 L = 0.22 M

​Step 2​Substitute equilibrium concentrations in the equilibrium expression and find the value of the equilibrium constant.
​K​=
​​​=
​​​= 0.01989
​​​= 0.020 (no unit here, two significant figures)

c ​The reaction quotient for this reaction is

Calculate equilibrium concentrations of reactants.
​[H2]​= 0.240 mol/2.0 L = 0.120 M
​[I2]​= 0.320 mol/2.0 L = 0.160 M
[HI]​= 1.00 mol/2.0 L = 0.500M
The value of the reaction quotient is
= 0.077
The value of the reaction quotient is greater than the value of the equilibrium constant at the same temperature as calculated in part b. Thus the system is not in equilibrium. A net backward reaction will occur. As the reaction moves towards equilibrium the concentrations of H2 and I2 will decrease and the concentration of HI will increase until the value of the reaction quotient is equal to the equilibrium constant.

32
Q

Death by carbon monoxide poisoning has been likened to suffocation. Explain why.

A

Death from suffocation occurs if insufficient oxygen enters the lungs, causing too little oxygen to be supplied to cells in tissues which need oxygen for respiration.
Like suffocation, carbon monoxide poisoning results when an inadequate supply of oxygen reaches body tissues. With carbon monoxide poisoning, however, the transport of oxygen in the bloodstream is prevented by carbon monoxide binding strongly to haemoglobin.

33
Q

Suggest how a person suffering from carbon monoxide poisoning should be treated.

A

People suffering from carbon monoxide poisoning are removed from the toxic atmosphere and administered oxygen to increase the concentration of oxyhaemoglobin in the blood.

34
Q

Use Le Chatelier’s principle to predict the effect of adding more hydrogen gas to the following equilibria:
a​ H2(g) + I2(g) F 2HI(g)

b​ 2NH3(g) F N2(g) + 3H2(g)

c ​H2(g) + CO2(g) F H2O(g) + CO(g)

A

a​ net forward reaction

b​ net back reaction

c​ net forward reaction

35
Q

Predict the effect of the following changes on the position of each equilibrium.
a​ addition of SO3 to the equilibrium:
2SO2(g) + O2(g) F 2SO3(g)

b​ removal of CH3COO– from the equilibrium:
CH3COOH(aq) + H2O(l) F H3O+(aq) + CH3COO–(aq)

c​ halving the volume (doubling the pressure) of the equilibrium:
N2(g) + 3H2(g) F 2NH3(g)

d​ increasing the pressure on the equilibrium:
H2(g) + I2(g) F 2HI(g)

e ​increasing the temperature of the endothermic equilibrium:
N2(g) + O2(g) F 2NO(g)

A

a​ net back reaction

b ​net forward reaction

c ​net forward reaction

d ​no effect

e​ net forward reaction

36
Q

Consider the following equilibria.
i​H2(g) + CO2(g) F H2O(g) + CO(g); DH = +42 kJ mol–1
ii​N2O4(g) F 2NO2(g); DH = +58 kJ mol–1
iii​H2(g) + F2(g) F 2HF(g); DH = –536 kJ mol–1
How would you alter:
a​ the temperature of each equilibrium mixture in order to produce a net forward reaction?

b​ the volume of each mixture in order to produce a net forward reaction?

A

a​i​)increase
ii​)increase
iii​decrease

b​i​cannot cause forward reaction
ii​)increase
iii​cannot cause forward reaction

37
Q

An equilibrium mixture consists of the gases N2O4 and NO2:
N2O4(g) F 2NO2(g)
The volume of the container was increased at constant temperature and a new equilbrium was established. Predict how each of the following quantities would change at the new equilibrium compared with the initial equilibrium:
a​ concentration of NO2

b​ mass of NO2

A

An increase in volume will cause a decrease in pressure. The equilibrium system will respond by favouring the direction that increases pressure, i.e. more NO2 will be formed.

38
Q

The magnitude of an equilibrium constant is dependent on temperature. Explain why it is referred to as a constant.

A

At a given temperature, when a particular reaction mixture achieves an equilibrium, the value of the concentration fraction is constant. This value is referred to as the equilibrium constant.

39
Q

Q14. Chapter 16
The equilibrium constant is 10–10 at 25°C for the reaction
Sn4+(aq) + 2Fe2+(aq) F Sn2+(aq) + 2Fe3+(aq)
a​ Write an expression for the equilibrium constant for this reaction.

b​ Would significant reaction occur when solutions of tin(IV) chloride and iron(II) chloride are mixed? Explain.

c ​Determine the value of the equilibrium constant for the reaction:
Sn2+(aq) + 2Fe3+(aq) F Sn4+(aq) + 2Fe2+(aq)

d​ Would significant reaction occur when solutions of tin(II) chloride and iron(III) chloride are mixed? Explain.

A

a​K =
b​ No; K is very small.

c​ 1010

d​ Yes; K is very large (provided the rate is sufficiently fast).

40
Q

Q15. Chapter 16
Write balanced equations for the reactions with the following equilibrium laws:
a ​K =

b​ K =

c​ K =

A

a​ CH3OH(g) F 2H2(g) + CO(g)

b ​S2(g) + 2H2(g) F 2H2S(g)

c​ NO2(g) F N2O4(g)

41
Q

Q16. C16
Acetone (C3H6O) is used to remove nail polish. It can be prepared from propan-2-ol using a copper–zinc catalyst, according to the equation:
C3H8O(g) F C3H6O(g) + H2(g)
If an equilibrium mixture of these gases consists of 0.018 mol of propan-2-ol, 0.082 mol of acetone and 0.082 mol of hydrogen in a 20 L vessel, calculate the value of the equilibrium constant.

A
Step 1​Write the expression for the equilibrium constant.
​K = 
Step 2​Calculate equilibrium concentrations of reactants and products.
​[C3H6O]​= 0.082 mol/20 L
​= 0.0041 M
​[H2]​= 0.082 mol/20 L
​​= 0.0041 M
​​[C3H8O]​= 0.018 mol/20 L
​​​​= 0.0009 M
Step 3​Substitute equilibrium concentrations in the equilibrium expression and find the value of the equilibrium constant.
​K​= 
​= 0.0187 M
​= 0.019 M (two significant figures)
42
Q

Q17 c16
Consider the equilibrium:
PCl5(g) F PCl3(g) + Cl2(g)
A 3.00 L vessel contained 6.00 mol of PCl3, 4.50 mol of PCl5 and 0.900 mol of Cl2 at equilibrium at 250°C.
a​ Write an expression for the equilibrium constant for this reaction.

b​ Calculate the equilibrium constant for the reaction at 250°C.

c​ Another equilibrium mixture contains 0.0020 M PCl5 and 0.0010 M PCl3 at 250°C. What is the concentration of chlorine in this mixture?

d​ Determine the equilibrium constant at 250°C for the reaction:
PCl3(g) + Cl2(g) F PCl5(g)

A

a ​K =

b ​Step 1​Calculate equilibrium concentrations of reactants and products.
​​[PCl5]​= 4.50 mol/3.00 L
​​​= 1.50 M
​​[Cl2]​= 0.900 mol/3.00 L
​​= 0.0300 M
​​[PCl3]​= 6.00 mol/3.00 L
​​= 2.00 M
​Step 2​Substitute equilibrium concentrations in the equilibrium expression.
​​K​= 
​= 0.400 M (three significant figures)

c​ Step 1​Use the expression for the equilibrium constant from part a.
​K​=
​​= 0.400 M
​Step 2​Rearrange the expression to find [Cl2].
​[Cl2] = K ´
​Step 3​Substitute equilibrium concentrations in the equilibrium constant expression and find the value of [Cl2].
​[Cl2]​= 0.400 ´
​= 0.80 M (two significant figures)

d ​Note that the equation is reversed and the temperature is constant. For the reverse reaction the equilibrium constant is the reciprocal of the forward reaction.
​K​= M
​= 2.5 M–1

43
Q

Q18. C16
The reaction used to manufacture ammonia is represented by:
N2(g) + 3H2(g) F 2NH3(g)
The equilibrium constant for the reaction is 0.052 M–2 at 400°C. Each of the following gas mixtures is contained in a 1.0 L vessel at 400°C. Decide if each mixture is in equilibrium. If not, predict the direction the reaction will shift to reach equilibrium.
a ​0.20 mol of N2, 0.20 mol of H2, 0.20 mol of NH3

b ​1.00 mol of N2, 1.00 mol of H2, 0.25 mol of NH3

c​ 0.050 mol of N2 and 0.050 mol of H2 only.

A

a​Step 1​Write the expression for the equilibrium constant.
​K​=
​= 0.052 M–2
​Step 2​As a 1 L vessel is used, the concentrations are the same as the amount in moles. Substitute the concentration values into concentration fraction (CF).
​Concentration fraction​=
​​CF​= 25
​CF > K, therefore the system is not at equilibrium. The system will shift in the direction that makes the CF = K. Since CF > K, the reaction must shift to left, resulting in a net back reaction.
b​As a 1 L vessel is used, the concentrations are the same as the amount in moles. Substitute the concentration values into concentration fraction.
​CF​=
​= 0.0625 M–2
​CF > K, therefore the system is not at equilibrium. The system will shift in the direction which makes the CF = K. Since CF > K, the reaction must shift to left, resulting in a net back reaction.
c​As a 1 L vessel is used, the concentrations are the same as the amount in moles. Substitute the given concentration values into concentration fraction.
​CF​=
​= 0.0 M–2
​CF

44
Q

Q19. C16
The equilibrium constant for the reaction
CO(g) + 3H2(g) F CH4(g) + H2O(g); DH = +208 kJ mol–1
is 0.667 M–2 at a particular temperature. A mixture of 0.100 M CO, 0.200 M H2, 0.300 M CH4 and 0.400 M H2O is heated to this temperature. Copy the table below and mark with a tick what would happen to the concentration of each chemical as the reaction approaches equilibrium.

Concentration increases
Concentration does not change
Concentration decreases
CO

H2

CH4

H2O

A

K = = 0.667
​For the initial mixture, the ratio is equal to
​ = 150
​So the back reaction will occur to reduce this ratio and reach equilibrium.

Concentration increases
Concentration does not change
Concentration decreases
CO
P

H2
P

CH4

P
H2O

P

45
Q

Q20. C16
At one step during the synthesis of nitric acid, dinitrogen tetroxide is in equilibrium with nitrogen dioxide:
N2O4(g) F 2NO2(g)
0.540 mol of N2O4 was placed in a 2.00 L vessel. When equilibrium was achieved, 0.280 mol of NO2 was present. Calculate the value of the equilibrium constant at this temperature.

A

Step 1​This type of question is best done by using a table as shown below.
​The entries in the first column indicate the initial amount that was in the closed vessel, the amount that was used that is calculated from stoichiometry and the amount that remains (or is in excess) and is present at equilibrium, calculated by subtraction (initial – used).
​Equilibrium constants are calculated by using concentrations in M, so it is important to remember to divide the mole value by the volume to convert to concentration in molar, M.

Volume = 2.0 L
N2O4 (g)  F
2NO2 (g)
Initial
0.540 mol
0.0 mol
Used
0.140 mol
At equilibrium
As 0.280 mol of NO2 was formed, the equation tells us that 0.140 mol of N2O4 must have been used. Hence, (0.540 – 0.140) = 0.400 mol is present at equilibrium.
0.400 mol/2.0 L = 0.20 M
0.280 mol
0.280 mol/2.0 L = 0.140 M
Step 2​Substitute equilibrium concentrations in the equilibrium expression and find the value of the equilibrium constant
​K​= 
​= 
​= 0.0980 M (three significant figures)
46
Q

Q21. C16
5.89 mol of N2 and 8.23 mol of H2 are mixed in a 5.0 L vessel and react according to the equation:
N2(g) + 3H2(g) F 2NH3(g)
When equilibrium was reached, it was found that 0.48 mol of NH3 had been formed. Calculate the value of the equilibrium constant.

A

Step 1​This type of question is best done by using a table as shown below.
​The entries in the first column indicate the initial amount that was in the closed vessel, the amount that was used that is calculated from stoichiometry, and the amount that remains (or is in excess) and is present at equilibrium, calculated by subtraction (initial – used).
​Equilibrium constants are calculated by using concentrations in M, so it is important to remember to divide the mole value by the volume to convert to concentration in molar, M.

Volume = 5.0 L
N2+ (g)
3H2 (g)  F
2NH3 (g)
Initial
5.89 mol
8.23 mol
0.0 mol
Used
0.24 mol
0.72 mol

At equilibrium
As 0.48 mol of NH3 was formed, the equation tells us that 0.24 mol of N2 must have been used. Hence, (5.89 – 0.24) = 5.65 mol is present at equilibrium.
5.65 mol/5.0 L = 1.13 M
As 0.48 mol of NH3 was formed, the equation tells us that ´ 0.48 = 0.72 mol of H2 must have been used. Hence, (8.23 – 0.72) = 7.51 mol is present at equilibrium.
7.51 mol/5.0 L = 1.502 M
0.48 mol
0.48 mol/5.0 L = 0.096 M
Step 2​Substitute equilibrium concentrations in the equilibrium expression and find the value of the equilibrium constant.
​K​=
​=
​= 0.0024 M–2 (two significant figures)

47
Q

Q22. C16
Consider the reaction:
A + 3B F 2C + D
Analysis of an equilibrium mixture in a 2.0 L container shows that 2.0 mol of A, 0.50 mol of B and 3.0 mol of D are present. If the equilibrium constant of the reaction is 0.024 M–1, calculate:
a​ the concentration of A, B and D at equilibrium

b ​the concentration of C in the equilibrium mixture

c​ the amount of C, in mol, in the equilibrium mixture

A
a​ Step 1​Write the expression for the equilibrium constant.
​K = 
​Step 2​Calculate equilibrium concentrations of reactants and products.
​[A]​= 2.0 mol/2.0 L
​= 1.0 M
​[B]​= 0.50 mol/2.0 L
​= 0.25 M
​[D]​= 3.0 mol/2.0 L
​= 1.5 M
b ​Substitute equilibrium concentrations in the equilibrium expression and calculate the equilibrium concentration of C.
​K​= 0.024 = 
​[C]2​= 
​[C]​= 
​= 0.0158
​= 0.016 M (two significant figures)

c​ Convert the equilibrium concentration of C into a mole value.
​[C]​= 0.016 M
​n(C)​= 0.016 M ´ 2.0 L
​= 0.032 mol

48
Q

Q23. C16
2.0 mol of sulfur dioxide gas and 2.0 mol of nitrogen dioxide gas are placed in a 1.0 L vessel, which is then sealed and heated. After equilibrium has been reached, analysis of the mixture shows that 1.3 mol of nitrogen dioxide is present. The reaction that occurred is described by the equation:
SO2(g) + NO2(g) F SO3(g) + NO(g)
a​ What amount of sulfur trioxide is present in the equilibrium mixture?

b​ What amount of sulfur dioxide is present in the equilibrium mixture?

c​ Calculate the value of the equilibrium constant for the reaction under these conditions.

A

This type of question is best done by using a table as shown below.
The entries in the first column indicate the initial amount that was in the closed vessel, the amount that was used that is calculated from stoichiometry, and the amount that remains (or is in excess) and is present at equilibrium, calculated by subtraction (initial – used).
Equilibrium constants are calculated from concentrations in M, so it is important to remember to divide the mole value by the volume to convert to concentration in molar, M. In this case the volume is 1 L, so the mole values are the same as the concentrations in molar.

Volume = 1.0 L
SO2(g) +
NO2(g) F
SO3(g) +
NO(g)
Initial
2.0 mol
2.0 mol
0.0 mol
0.0 mol
Used
0.7 mol
0.7 mol

At equilibrium
As 0.7 mol of NO2 was formed, the equation tells us that 0.7 mol of SO2 must have been used. Hence, (2.0 – 0.7) = 1.3 mol is present at equilibrium which is 1.3 M.
1.3 mol
The amount of NO2 used = 2.0 – 1.3 mol
= 0.7 mol
The equation shows that for every mole of NO2 used, 1 mol of SO3 is formed, therefore n(SO3) at equilibrium must be 0.7 mol and the concentration is 0.7 M
The equation shows that for every mole of NO2 used, 1 mol of NO is formed. Hence 0.7 mol of NO must be present at equilibrium, and the concentration is 0.7 M.

a​ 0.7 mol
b​ 1.3 mol

c ​Substitute equilibrium concentrations in the equilibrium expression and find the value of the equilibrium constant.
​K =
​=
​= 0.29 (no unit here, two significant figures)

49
Q

How will the concentration of hydrogen gas in each of the following equilibrium mixtures change when the mixtures are heated and kept at constant volume?
a​ N2(g) + 3H2(g) F 2NH3(g); DH = –91 kJ mol–1

b​ CH4(g) + H2O(g) F CO(g) + 3H2(g); DH = +208 kJ mol–1

A

a​ increase

b ​increase

50
Q

The following equations represent reactions that are important in industrial processes. Predict the effect on the equilibrium position if each reaction mixture was compressed at constant temperature.
a​ C3H8O(g) F C3H6O(g) + H2(g)

b ​CO(g) + 2H2(g) F CH3OH(g)

c​ N2(g) + O2(g) F 2NO(g)

A

a​ net back reaction

b ​net forward reaction

c​ no effect

51
Q

Some elderly people, especially women, become very susceptible to bone breakages. It is thought that as people age they absorb Ca2+ ions from food inefficiently, reducing the concentration of these ions in body fluids. An equilibrium exists between calcium phosphate in bone and calcium ions in body fluids:
Ca3(PO4)2(s) F 3Ca2+(aq) + 2PO43–(aq)
Explain why inefficient absorption of Ca2+ ions could cause weakness in bones.

A

If Ca2+ ions were inefficiently absorbed from food, decreased concentrations of these ions could occur in body fluids. As a consequence, a net forward reaction would occur, raising the concentration of dissolved Ca2+ ions and resulting in decreased amounts of calcium phosphate in bones.

52
Q

Carbon monoxide is used as a fuel in many industries. It reacts according to the equation:
2CO(g) + O2(g) F 2CO2(g)
In a study of this exothermic reaction, an equilibrium system is established in a closed vessel of constant volume at 1000°C.
a ​Predict what will happen to the equilibrium position as a result of the following changes:
i​decrease in temperature
ii​addition of a catalyst
iii​addition of more oxygen

b ​What will happen to the equilibrium constant as a result of each of the changes above?

c​ If carbon monoxide can be used as a fuel, comment on the magnitude of the equilibrium constant for the reaction.

A

a​i​net forward reaction
ii​no change
iii​net forward reaction

b​i​)increase
ii​no change
iii​no change

c​likely to be large

53
Q

Sealed bottles of fizzy drinks, such as lemonade and sparkling wines, contain carbon dioxide gas in equilibrium with dissolved carbon dioxide:
CO2(g) F CO2(aq)
The forward reaction is exothermic.
a​ Use Le Chatelier’s principle to explain why bubbles appear when the bottles are opened.

b​ Why are the drinks usually cooled before they are carbonated?

A

a​ When the bottles are opened, carbon dioxide gas escapes. From Le Chatelier’s Principle, loss of a chemical on the left side of an equation will cause a net back reaction. As carbon dioxide is evolved from solution, bubbles appear.

b​ The value of K will increase as the temperature decreases, so carbon dioxide will be more soluble at low temperatures.

54
Q

Q29. C16
A step during nitric acid production is the oxidation of nitrogen oxide to nitrogen dioxide.
2NO(g) + O2(g) F 2NO2(g); DH = –114 kJ mol–1
Nitrogen dioxide is a brown coloured gas and nitrogen oxide and oxygen are colourless. An equilibrium mixture was prepared in a 1 L container at 350°C. Copy the table below, and for each of the following changes indicate if the reaction mixture would become darker or lighter, giving a reason for your choice.

Colour change
Explanation
lighter / darker

a​the temperature is increased to 450°C at constant volume
b​the volume of the container is increased at constant temperature
c​a catalyst is added at constant volume and temperature
d​more oxygen is added at constant volume and temperature

A

Colour change
Explanation

Lighter
Darker

a
P

Exothermic reaction, so K will decrease as temperature increases, so a net back reaction occurs.
b
P

As the volume increases the colour of the gas mixture will decrease. (A net back reaction will also occur, further lightening the colour.)
c

Addition of a catalyst would have not change the colour as the rate of the forward and back reactions are increased equally.
d

P
Addition of oxygen will cause a net forward reaction to occur.

55
Q

Q30. C16
Ethene gas is produced from ethane gas in an endothermic reaction represented by the equation:
C2H6(g) F C2H4(g) + H2(g); DH = +138 kJ mol–1
a​Copy the table below and place a tick in the appropriate box to indicate what will happen to the equilibrium percentage yield of ethene when each of the following changes is made.

Increased ethene yield
No change in ethene yield
Decreased ethene yield

i​the volume is reduced at constant temperature
ii​more hydrogen gas is added at constant temperature and volume
iii​the temperature is increased at constant volume
iv​a catalyst is added
v​argon gas is added at constant temperature and volume.
b​How will each of the changes in part a affect the rate at which the reaction achieves equilibrium?

A

a

Increased ethene yield
No change in ethene yield
Decreased ethene yield
i

P
ii

P
iii
P

iv

P

v

P

b​i​increase
​ii​increase
​iii​increase
​iv​increase
​v​no change
56
Q

Q31. C16
Methanol is manufactured for use as a fuel for racing cars. It can be made by reaction between carbon monoxide and hydrogen:
CO(g) + 2H2(g) F CH3OH(g); DH = –103 kJ mol–1
a​What conditions of temperature and pressure would be required for:
i​a fast reaction rate?
ii​a high equilibrium yield of methanol?
b​Will a compromise be needed in the choice of temperature or pressure?
c​Suggest another method that could be employed at a manufacturing plant to increase the rate of methanol production.
d​As part of an investigation of this process, the concentrations of a mixture of CO, H2 and CH3OH were monitored continuously. The mixture was intially at equilibrium at 400°C and constant volume. After 10 minutes additional CO was added to the mixture, as shown in the graph in Figure 16.22.

Figure 16.22
Graph of concentration versus time.
i​Sketch a graph to show how concentrations would change as a consequence of the addition of CO.
ii​Following the addition of the CO the mixture again reaches equilibrium. Sketch a second graph to show the effect on the concentrations if the temperature were then increased to 450°C.

A
a​i​high temperature; high pressure
ii​low temperature; high pressure
b​temperature
c​use a catalyst
d​i​
​
​ii​
​
57
Q

During the contact process for sulfuric acid manufacture, sulfur dioxide is converted to sulfur trioxide at temperatures of 400–500°C:
2SO2(g) + O2(g) F 2SO3(g); DH = –197 kJ mol–1
a​What would be the effect of increasing the pressure on:
i​reaction rate?
ii​equilibrium yield?
b​In practice, this step is usually performed at atmospheric pressure. Suggest why.
c​During the process, sulfur trioxide is removed from the reaction mixture by converting it to sulfuric acid. The remaining gases are then recycled to the reaction vessel. Explain the reason for recycling the gases.
d​What factors would have influenced the choice of the reaction temperature?

A

a​i​increase
ii​increase
b​In practice, satisfactorily high equilibrium yields of sulfur trioxide can be obtained without needing to use high pressures, which would involve more expensive equipment.
c​The purpose of recycling the gases from the absorption tower is to increase the yield of sulfur trioxide and to minimise emissions of sulfur oxides, which act as pollutants.
d​The rate of reaction and the magnitude of K.

58
Q

Write a paragraph about chemical equilbrium that uses the terms ‘forward reaction’, ‘reverse reaction’, ‘equilibrium law’, ‘equilibrium constant’, and ‘Le Chatelier’s principle’.

A

In a chemical equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. The concentrations of the chemicals involved in the equilibrium obey a mathematical ratio known as the equilibrium law. This ratio is equal to K, the equilibrium constant. Le Chatelier’s Principle can be used to determine predict the effect of a change on an equilibrium.

59
Q

Explain the difference between the terms ‘concentration fraction’ and ‘equilibrium constant’.

A

The concentration fraction is the ratio of the concentrations of the products to the concentration of the reactants, with the index of each concentration the same as the coefficient of the substance in the reaction equation.
The value of the concentration fraction becomes equal to the equilibrium constant at equilibrium.

60
Q

Q35. C16
a​The equilibrium constant is 0.67 M–2 at a particular temperature for the reaction:
​CO(g) + 3H2(g) F CH4(g) + H2O(g)
​A mixture of 0.100 M CO, 0.200 M H2, 0.300 M CH4 and 0.400 M H2O is heated to this temperature. Copy the table below and indicate with a tick what would happen to the concentration of each chemical as the reaction approaches equilibrium.

Concentration increases
Concentration does not change
Concentration decreases
CO

H2

CH4

H2O

b​When the temperature of the reaction mixture is increased by 10°C, the equilibrium constant for the reaction becomes 0.71 M–2. What conclusion can you make about the enthalpy change of this reaction?

A

a​For the initial mixture, the ratio = = 150 M–2
​A net back reaction will occur so this ratio becomes equal to K.

Concentration increases
Concentration does not change
Concentration decreases
CO
P

H2
P

CH4

P
H2O

P
b​endothermic

61
Q

Q36. C16
Carbon disulfide gas (CS2), which is used in the manufacture of rayon, can be made using an endothermic gas phase reaction between sulfur trioxide gas (SO3) and carbon dioxide. Oxygen gas is also produced in the reaction.
a​Write a balanced chemical equation for the reaction.
b​Write an expression for the equilibrium constant of the reaction.
c​An equilibrium mixture of these gases was made by mixing sulfur trioxide and carbon dioxide. It consisted of 0.028 mol of CS2, 0.022 mol of SO3 and 0.014 mol of CO2 in a 20 L vessel. Calculate the value of the equilibrium constant at that temperature.
d​Predict how each of the following changes to an equilibrium mixture would affect the yield of CS2.
i​removal of O2 (at constant total volume)
ii​increasing the temperature
iii​adding a catalyst
iv​increasing the pressure by decreasing the volume of the reaction vessel (at constant temperature)
v​increasing the pressure by introducing argon gas into the reaction vessel (at constant volume)

A
a​2SO3 (g) + CO2 (g) F CS2 (g) + 4O2 (g)
b​K = 
c​​=  = 0.0014 M
​​= 4 × 0.0014 = 0.0056 M
​​=  = 0.0011 M
​​=  = 0.000 70 M
​K =  =  = 1.6 × 10–3 M2
d​i​increase
ii​increase
iii​no effect
iv​decrease
v​no effect
62
Q

Q37.
Sulfur dioxide gas and oxygen gas were mixed at 600°C to produce a gaseous equilibrium mixture:
2SO2(g) + O2(g) F 2SO3(g)
A number of changes were then made, including the addition of a catalyst, resulting in the formation of new equilibrium mixtures. Figure 16.23 shows how the concentrations of the three gases changed.

Figure 16.23
Graph of concentration versus time.
a​Write an expression for the equilibrium constant, K, of the reaction.
b​During which time intervals was the reaction at equilibrium?
c​Calculate the value of K at 600°C.
d​At what time was the catalyst added? Explain your reasoning.
e​What change was made to the system at:
i​20 minutes?
ii​30 minutes?

A

a​K =
b​15–20 minutes; 25–30 minutes; 35–40 minutes
c​1.56 M–1
d​10 minutes; before 10 minutes the concentrations were changing slowly but the catalyst caused equilibrium to be reached more rapidly
e​i​increased pressure by reducing volume
ii​oxygen added