Unit 4 Industrial Chemistry

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Learningby Lisachem

Materials

VCE CHEMISTRY 2010 YEAR 12 UNIT 4

Industrial Chemistry

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Time allowed: 50 minutes Total marks: 47

12 Multiple Choice Questions

4 Short Answer Questions

An Answer Sheet is provided for Section A. Answer all questions in Section B in the space provided.

To download the Chemistry Data Book please visit the VCAA website: http://www.vcaa.vic.edu.au/vce/studies/chemistry/chem1_sample_2008.pdf Page 20

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Learning Materials by Lisachem VCE Chemistry 2010 Year 12 Unit 4 Industrial Chemistry

Student Name..................................................................... VCE Chemistry 2010 Year 12 Unit 4 Industrial Chemistry Student Answer Sheet Instructions for completing test. Use only a 2B pencil. If you make a mistake erase and enter the correct answer. Marks will not be deducted for incorrect answers. Write your answers to the Short Answer Section in the space provided directly below the question. There are 12 Multiple Choice questions to be answered by circling the correct letter in the table below.

Question 1 A B C D

Question 2 A B C D

Question 3 A B C D

Question 4 A B C D

Question 5 A B C D

Question 6 A B C D

Question 7 A B C D

Question 8 A B C D

Question 9 A B C D

Question 10 A B C D

Question 11 A B C D

Question 12 A B C D

Learning Materials by Lisachem VCE Chemistry 2010 Year 12 Unit 4 Industrial Chemistry 1

VCE Chemistry 2010 Year 12 Unit 4 Industrial Chemistry

Multiple Choice Questions - Section A

Section A contains 12 multiple choice questions. This section is worth approximately 26 per cent of the total marks available.

For each question choose the response that is correct or best answers the question. Indicate your answer on the answer sheet provided.

(Choose only one answer for each question.)

Question 1 A 0.0100 M aqueous solution of a weak monoprotic acid has a pH of 3.6. The numerical value of acidity constant for this acid would be A. 2.510-6 B. 6.310-6 C. 2.510-2 D. 7.710-4 Question 2 Which one of the following waste management strategies would be the more sound practice for a chemical company to implement? A. Treat all waste produced on site in accordance with best practice procedures. B. Recycle waste materials produced either in the process or in other processes. C. Design and implement processes that reduce the amount of waste produced. D. Dispose of waste materials produced in accordance with the environment protection

guidelines. Question 3 The formation of methanol from carbon monoxide and hydrogen gas can be described by the chemical equation

12 3CO(g) 2H (g) CH OH(g) H 90 kJ mol

+ = The most significant increase in the rate of methanol formation will occur when A. the temperature is increased and the pressure decreased. B. both the temperature and pressure are increased. C. both the temperature and pressure are decreased. D. the temperature is decreased and the pressure increased.


Question 4 The substitution reaction between trichloromethane and chlorine can be represented by the chemical equation

13 2 4CHCl (g) Cl (g) CCl (g) HCl(g) H 100 kJ mol

+ + = Which one of the following diagrams would best represent how the concentration of hydrogen chloride would vary in an equilibrium mixture as the temperature is changed?

Question 5 2.641 g of solid sodium hydroxide was dissolved in 400.0 L of deionised water. The pH of this solution at 25 C would be A. 13.2 B. 3.8 C. 10.2 D. 0.8


Question 6 The diagram shows the energy profile for a chemical reaction.

For this reaction, A. the energy required to break the bonds in the reactants is greater than the energy

released in the formation of the bonds in the products. B. the energy required to break the bonds in the reactants is less than the energy released

by the reaction. C. the energy released in the formation of the bonds in the products is greater than the

energy required to break the bonds in the reactants. D. the energy released by the reaction is greater than the energy released in the formation

of the bonds in the products. Question 7 The dissociation of dinitrogen tetroxide can be described by the chemical equation

1 32 4 2N O (g) 2NO (g) H 57 kJ mol K(25 C) 6 10 M

= + = A sample of dinitrogen tetroxide was allowed to reach equilibrium in a 1.0 L piston. Which one of the following best describes the changes that would occur, after the system was allowed to re-equilibrate, when the volume of the piston was halved while keeping the temperature constant? [NO2] at new equilibrium n(NO2) at new equilibrium Change in position of equilibrium

A. increase increase shift towards products

B. decrease increase shift towards products

C. decrease decrease shift towards reactants

D. increase decrease shift towards reactants


Question 8 The energy profile for a two stage reaction process is shown in the diagram.

The enthalpy change, H, for the reverse process will be A. +780 kJ mol-1 B. +360 kJ mol-1 C. -780 kJ mol-1 D. -360 kJ mol-1 Questions 9 and 10 refer to the following information. The Fisher-Tropsch process is a method that can be used to produce hydrocarbon compounds using a feedstock containing carbon monoxide and hydrogen gases. The synthesis of propane using this process could be represented by the chemical equation

12 3 8 26CO(g) 11H (g) 2C H (g) 6H O(g) H 1023 kJ mol

+ + =

Question 9 Which one of the following would not increase the yield of propane using the process above? A. Increasing the pressure at which the reaction was carried out. B. Removing water vapour from the gas mixture in the reactor. C. Increasing the amount of hydrogen gas in the feed gas mixture. D. Increasing the temperature at which the reaction was carried out. Question 10 Which one of the following would increase the rate of propane formation in the above process? A. Using higher pressures. B. Using a lower temperature. C. Using a lower pressure. D. Not using a catalyst.


Question 11 Petrol is classified as being a dangerous goods product because A. exposure to it can lead to short-term health problems. B. it will burn. C. exposure to it can lead to long-term health problems. D. it has the potential to pose immediate danger to people or the environment. Question 12 The label shown in the diagram was attached to a chemical bottle used in the laboratory. This label should be attached to any bottle containing A. vinegar. B. ethanol. C. hydrogen peroxide. D. hydrochloric acid.

End of Section A


VCE Chemistry 2010 Year 12 Unit 4 Industrial Chemistry

Short Answer Questions - Section B Section B contains four questions, numbered 1 to 4.

This section is worth approximately 74 per cent of the total marks available. The marks allotted are shown at the end of each part of each question.

All questions should be answered in the spaces provided.

Question 1 Sodium cyanide, a highly poisonous material, is extensively used industrially in mineral extraction and electroplating. It can be manufactured from ammonia, methane, air and sodium hydroxide in a two stage process known as the Andrussow process. A schematic for this process is shown below.

a. The reactor usually operates at 1050 C and the reaction that occurs can be described

by the chemical equation 1

3 4 2 22NH (g) 2CH (g) 3O (g) 2HCN(g) 6H O(g) H 939 kJ mol+ + + =

i. What effect would increasing the temperature have on the equilibrium yield of HCN in this reaction?

(1 mark)


ii. What could be one reason why such a high temperature is used in this process?

(1 mark) iii. What would be the benefit of using a platinum catalyst in this process?

(1 mark) b. The methane used in the process is obtained from natural gas. Before it can be used in

the reactor any trace amounts of sulfur containing compounds, normally present in natural gas, must be removed. What could be the one likely reason for this?

(1 mark) c. In the plant, a unit (indicated by the coloured rectangle in the schematic diagram) is

placed between the reactor and the absorber. What would be the main function of this unit and how could its use form part of the overall waste management plan for the plant?

(2 marks) d. Write an appropriate chemical equation for the reaction that would occur in the

absorber.

(1 mark)


e. The waste gas stream exiting the absorber contains small amounts of HCN which is destroyed by burning the gases and then treating any products produced to ensure that no noxious gases are discharged into the atmosphere. What would be one better possible alternative to this disposal of the HCN?

(1 mark) f. The high toxicity and lethality of the materials involved would require a number of

special work site safety practices. What would be one possible safety practice that would be mandatory for a sodium cyanide plant?

(1 mark)

g. What would be the main message conveyed by the HAZCHEM sign on a drum

containing solid sodium cyanide?

(1 mark)


Question 2 a. The reaction between hydrogen and iodine gases can be described by the chemical

equation

2 2H (g) I (g) 2HI(g)+ The graph below shows the concentrations of hydrogen, iodine and hydrogen iodide at

various times after a mixture of hydrogen and iodine were combined and allowed to reach equilibrium. Single changes to this equilibrium mixture were made at t1, t2 and t3 and the system was allowed to re-equilibrate following each change.

i. Write an expression for the equilibrium constant, Keq, for this reaction.

(1 mark) ii. Determine the value for the equilibrium constant, Keq, at te.

(1 mark) iii. What change was made to the system at t1?

(1 mark)


iv. What change was made to the system at t2?

(1 mark) v. The temperature of the system was increased at t3. What do the changes in the

concentrations of the reactants and products indicate about the thermochemical nature of the reverse reaction?

(1 mark) b. The reaction between chlorine and phosphorous(III) chloride can be described by the

chemical equation 1

2 3 5 eCl (g) PCl (g) PCl (g) K (398 K) 100 M+ =

i. A 3.0 L sample of a gas mixture at 398 K contained all three gases, such that the amounts of Cl2, PCl3, and PCl5 were 0.030, 0.030 and 0.045 respectively. In which direction would the rate of reaction be higher?

(2 marks) ii. What would be the effect on the position of equilibrium if the pressure on a

sample of these three gases at equilibrium was increased while the temperature remains constant?

(1 mark)


Question 3 a. Hypochlorous acid is a weak acid that plays a key role in the chlorination of

swimming pools to keep the levels of bacteria and algae under control. i. Write an appropriate chemical equation for the ionisation of hypochlorous acid

in water.

(1 mark) ii. Write an expression for the acidity constant for hypochlorous acid.

(1 mark) iii. Calculate the pH of a 5.010-4 M aqueous solution of hypochlorous acid.

(1 mark) iv. Humans experience skin and eye irritation when swimming pool water is either

less than 7.2 or greater than 7.8. How could adjusting the pH of the above solution to these levels affect the [HOCl] in the above solution?

(1 mark) v. A swimming pool had a pH of 7.5 and [OCl-] of 9.610-5 M. What would be

the [HOCl] in the water?

(1 mark)


b. The colour of acid-base indicators depends on whether they are protonated, HIn, or deprotonated, In-, essentially making them weak acids. The lower value of pH range for an indicator is where the indicator clearly displays its acid colouration, while the upper value is for the base colouration. One commonly used indicator in VCE chemistry is methyl red.

2 3HIn(aq) H O(l) In (aq) H O (aq) ++ +

i. Write an expression for the acidity constant for this indicator based on the above chemical equation.

(1 mark) ii. Calculate the value for the [In ]

[HIn]

for both the lower and upper values of the

pH range and thereby determine which is the predominant form of the indicator in solution.

(4 marks)


Question 4 a. The change in enthalpy, H, for a chemical reaction is -800 kJ mol-1. The chemical

energy for the reactants is -75 kJ mol-1 and the activation energy for the reverse reaction is 1200 kJ mol-1.

i. On the grid provided, draw an energy profile diagram for this reaction.

(2 marks)

ii. Determine the chemical energy for the products.

(1 mark) iii. Determine the activation energy for the forward reaction.

(1 mark) b. The combustion of methanol can be described by the chemical equation

3 2 2 22CH OH(l) 3O (g) 2CO (g) 4H O(l)+ + i. Calculate the H for this reaction.

(1 mark)


ii. 12 2H O(g) H O(l) H 44 kJ mol =

Calculate the H for the reaction described by the chemical equation 3 2 2 22CH OH(l) 3O (g) 2CO (g) 4H O(g)+ +

(2 marks)

End of Section B

End of Unit Test

Learning Materials by Lisachem Suggested Answers VCE Chemistry 2010 Year 12 Unit 4 IC 1

Suggested Answers VCE Chemistry 2010 Year 12 Unit 4 Industrial Chemistry

Multiple Choice Questions Section A (1 mark per question) Q1 B Consider a monoprotic acid has HA, its dissociation in aqueous solutions can

be represented by the chemical equation 2 3HA(aq) H O(l) H O (aq) A (aq)

+ + + [H3O+(aq)] can be simplified as [H+] pH = -log10[H+] [H+] = 10-pH = 10-3.6 = 2.510-4 M [A-] = [H+] = 2.510-4 M For a weak acid the [HA] can be assumed to be the same as the acid

concentration [HA] = 0.0100 M

Ka = [H ][A ]

[HA]

+

= 4 4(2.5 10 )(2.5 10 )

0.0100

= 6.310-6 M.

Q2 C The order of waste management strategies are: Prevention Elimination Reduction Recycling Treatment Disposal Designing processes to reduce the amount of waste produced would be the

more sound practice of the alternatives given in the responses. Q3 B The rate of reaction for all reactions is increased when the temperature is

increased, because a higher proportion of the reacting species will have sufficient energy (equal to or greater than the activation energy) for a fruitful reaction. Increasing the pressure of a gas phase reaction increases the likelihood that the particles will collide, thereby increasing the chance of particles with sufficient energy colliding and leading to a fruitful reaction and increasing the rate of reaction.

Q4 A The forward reaction is an exothermic reaction, therefore based on LeChateliers Principle, an equilibrium system will shift its position to oppose any change. Increasing the temperature would favour the reverse reaction, therefore the equilibrium [HCl] would decrease as the temperature is increased. The change would be continuous and not have any abrupt changes as seen in response D.

Q5 C M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol-1 n(NaOH) = m / M = 2.641 / 40.0 = 6.6010-2 mol c(NaOH) = n / V = 6.6010-2 / 400.0 = 1.6510-4 M [OH-] = c(NaOH) = 1.6510-4 M At 25 C Kw = [H+][OH-] = 10-14 M2 [H+] = 10-14 / [OH-] = 10-14 / 1.6510-4 = 6.0610-11 M pH = -log10[H+] = -log10(6.0610-11) = 10.2


Q6 C The energy required to break the bonds of the reactants is the difference between the energy of the reactants and the activation energy. The energy released in forming the bonds of the products is the difference between the energy of the products and the activation energy. In this energy profile the energy released in forming the bonds is greater than the energy required to break the bonds, hence this energy profile is for an exothermic reaction.

Q7 D Using LeChateliers Principle that the system will shift in an attempt to oppose the change, a decrease in volume will initially increase the number of particles occupying the volume so, to oppose this, the position of equilibrium will shift to the side of the reaction with the lesser number of particles. In this case, the shift will be towards the reactant, dinitrogen tetroxide, therefore the amount of NO2 will decrease. However because this now occupies a smaller volume, its concentration will be higher.

Q8 B The total enthalpy change for a reaction sequence is the sum of the individual enthalpy changes. Therefore for the forward process

HTotal = -570 + 210 = -360 kJ mol-1 Therefore the enthalpy change for the reverse process will have the opposite

sign. H = +360 kJ mol-1 Q9 D LeChateliers Principle is the basis for this answer because to increase the

yield of propane, the position of equilibrium needs to be shifted in favour of the products.

The reaction has fewer particles on the product side of the equation, therefore increasing the pressure will shift the position of equilibrium to favour the forward reaction, thereby increasing the equilibrium yield of propane.

Removing the water vapour from the system in the reactor will result in the position of equilibrium having to shift to produce more, thereby increasing the equilibrium yield of propane.

Increasing the amount of hydrogen in the feed gas mixture will also shift the position of equilibrium to favour the products.

The forward reaction, leading to the formation of propane, is exothermic, therefore increasing the temperature of the system will favour the reverse reaction, shifting the position of equilibrium to the reactants and lowering the propane yield.

Q10 A The rate of a reaction is increased by increasing the temperature, because this will give more particles sufficient energy to overcome the activation energy barrier and result in a fruitful reaction. Increasing the pressure that a reaction is carried out at will increase the likelihood of the particles colliding, and provided that they have sufficient energy, result in a fruitful collision. Catalysts provide an alternative pathway that has a lower activation energy, so more particles at a given temperature will have this lower amount of energy, therefore the likelihood of a fruitful collision is increased and a faster rate results.


Q11 D Dangerous goods are defined as substances that because of their properties can cause serious injury to people and can severely damage property and the environment. Therefore the best response is D. While the others may apply, they are not specific enough. Paper will burn but it is not a dangerous good. Many substances that we ingest can cause both short- and long-term health risks but again these are not dangerous goods. Hazardous materials are classified on the effects that exposure to them have on the health and safety of people, particularly in the workplace.

Q12 D The label is for a hazardous substance that is corrosive. Of those listed only hydrochloric acid fulfils this criterion. Vinegar is not classed as a hazardous substance, even though it contains an aqueous solution of ethanoic acid. Ethanol is a flammable liquid so bottles containing this should have the Class 3 label. Hydrogen peroxide is an oxidising agent so bottles containing it should have the Class 5 label.

Short Answer (Answers) Section B Question 1 a. i. Since the forward reaction is exothermic any increase in temperature will

lower the equilibrium yield of HCN. (1 mark) ii. The high temperature is required to achieve a satisfactory reaction rate in the

process. (1 mark) The higher the temperature, the higher the reaction rate, because more particles have sufficient energy to overcome the activation energy barrier and result in a fruitful reaction.

iii. The catalyst would provide an alternative reaction pathway with a lower activation energy, thereby increasing the rate of reaction since more particles will have this energy at a given temperature and more fruitful reactions can occur. (1 mark)

b. Possible answers include: (Total marks 1 mark) The sulfur compounds could poison the catalyst thereby reducing its efficiency. (This

is the actual reason.) The sulfur compounds could lead to other side reactions occurring thereby reducing

the yield of HCN in the products. The sulfur compounds could react with the HCN and form unstable and dangerous

compounds. c. The unit would be a heat exchanger to cool the gases before they enter the

absorber, since the gases are going into a unit that uses an aqueous solution of sodium hydroxide, the gas mixture will need to be about 100 C. (1 mark) The heat exchanger would be used to heat water and produce steam that could be used in other parts of the plant, thereby reducing the waste in energy within the plant. (1 mark)


d. 2NaOH(aq) HCN(g) NaCN(aq) H O(l)+ + (1 mark) e. Since the material requiring disposal is the product required, then using the sound

waste management protocols prevention and elimination are not feasible alternatives. Possible alternatives include: (Total marks 1 mark) The next best protocol is reduction. This could be achieved by passing the waste gas

through a subsequent absorber which would further reduce the levels of HCN in the gas stream.

Recycling is the next protocol and this would involve passing the waste gas back into the reactor. (This is not possible in practice due to potential explosive gas mixtures forming.)

f. Possible alternatives include: (Total marks 1 mark) Training all employees on the emergency procedures required in the event of a

malfunction. All employees being equipped with, and trained in, the use of specialised personal

protection equipment. Training of employees in emergency medical procedures to deal with an incident. Having the plant within a bunded area, contained within small walls, to contain any

liquid spills. Gas detectors installed around the plant and linked to an alarm system. g. The main message on the HAZCHEM sign for solid sodium cyanide would be that it

is toxic or poisonous. (1 mark)

Question 2 a. i. The expression for the equilibrium constant is the concentration fraction for the

reaction.

Keq = 2

2 2

[HI][H ][I ]

(1 mark)

ii. The concentrations of the species at te are: [HI] = 0.074 M [H2] = 0.023 M [I2] = 0.0025 M

Keq = 2

2 2

[HI][H ][I ]

= 2(0.074)

(0.023)(0.0025) = 95 (1 mark)

Since the concentrations of the species have to be read from the graph, a value between about 80 and 110 would be acceptable.

iii. At t1 the concentrations of all three species decreased, which suggests that either the volume of the containing vessel was increased or the pressure was reduced. (1 mark) The concentrations at t1 are about one-half the concentrations prior to the change, which would indicate a doubling of the volume or halving of the pressure. Because this is an equimolar reaction (the number of reactant and product particles are the same) the position off equilibrium is not altered by this type of change.


iv. At t2 the concentration of iodine increases and this is followed by a change in the concentrations of the species so that when the system has re-equilibrated, the concentrations of the hydrogen and iodine decrease and the hydrogen iodide increases. Therefore the change resulted in a net forward reaction which is indicative, using LeChateliers Principle, of the addition of one of the reactants, in this case the addition of some iodine. (1 mark)

v. Following the change at t3 the concentrations of the hydrogen and iodine have increased while the hydrogen iodide concentration has decreased, which indicates a shift in the position of equilibrium to the reactants side of the reaction. Therefore a temperature increase favours the reverse reaction, indicating that the enthalpy change for this reaction would be an endothermic one, H>0. (1 mark)

b. i. The volume is 3.0 L n(Cl2) = 0.030 [Cl2] = 0.030 / 3.0 = 0.010 M n(PCl3) = 0.030 [PCl3] = 0.030 / 3.0 = 0.010 M n(PCl5) = 0.045 [PCl5] = 0.045 / 3.0 = 0.015 M

Concentration fraction = 52 3

[PCl ][Cl ][PCl ]

= (0.015)(0.010)(0.010)

= 150 M-1 (1 mark)

Since the value of the concentration fraction is greater than the value of the equilibrium constant, the rate of the reverse reaction must be the greater of the two, as the system is moving towards the left in order to establish equilibrium. (1 mark)

ii. Since the number of particles on the left hand side of the reaction is greater than those on the right, an increase in pressure using LeChateliers Principle will favour the forward reaction and the position of equilibrium will shift in that direction, to the right. (1 mark) Therefore prior to re-establishment of equilibrium, the rate of the forward reaction will be greater than the rate of the reverse reaction. At equilibrium the rates of both the forward and reverse reactions are the same.

Question 3 a. i. Hypochlorous acid, HOCl, is listed in Table 12 of the VCE Chem Data

Booklet. 2 3HOCl(aq) H O(l) OCl (aq) H O (aq)

++ + (1 mark) ii. The acidity constant, Ka, does not include the [H2O] as this is essentially

constant.

Ka = 3[OCl ][H O ]

[HOCl]

+

or [OCl ][H ][HOCl]

+

(1 mark)

iii. It is assumed that because it is a weak acid very little of the HOCl has ionised, so the [HOCl] = 5.010-4 M

[OCl-] = [H3O+] = x

Ka = 4(x)(x)

5.0 10 = 2.910-8

x2 = 5.010-4 2.910-8 = 1.510-11

[H3O+] = x = 111.5 10 = 3.910-6 [H3O+] = 3.910-6 M pH = -log10[H3O+] = -log10(3.910-6) = 5.4 (1 mark)


iv. Increasing the pH from 5.5 to between 7.2 and 7.8 would result in lowering the [H3O+] or removing H3O+ form solution. Using LeChateliers Principle, the removal of a product from an equilibrium system will cause the system to shift its position to oppose the change, and in this case move to the right thereby decreasing the [HOCl]. (1 mark)

v. [H3O+] = 10-pH = 10-7.5 = 3.110-8 M

Ka = 3[OCl ][H O ]

[HOCl]

+

= 5 8(9.6 10 )(3.1 10 )

[HOCl]

= 2.910-8

[HOCl] = 5 8

8

(9.6 10 )(3.1 10 )2.9 10

= 1.010-4 M (1 mark)

b. i. Ka = 3[In ][H O ]

[HIn]

+

(1 mark)

ii. From Table 11 of the VCE Chem Data Booklet Methyl Red: pH Range 4.2 6.3 Ka = 810-6 Lower pH value (red): [H3O+] = 10-4.2 = 6.310-5 M

[In ][HIn]

= a3

K[H O ]+

= 6

5

8 106.3 10

= 0.1 (1 mark)

At this pH most of the indicator is in the HIn form. (1 mark) Higher pH value (red): [H3O+] = 10-6.3 = 5.010-7 M

[In ][HIn]

= a3

K[H O ]+

= 6

7

8 105.0 10

= 20 (1 mark)

At this pH most of the indicator is in the In- form. (1 mark) Question 4 a. i. The graph should be clearly labelled. (2 marks)

0

500

-500

-1000

Reactants

Products

H = -800 kJ mol-1

E (Forward)A

E (Reverse)A1200 kJ mol-1

Ener

gy [k

J mol

]-1

-75

325

-875


ii. H = E(products) E(reactants) E(Products) = H + E(reactants) = -800 + (-75) = -875 kJ mol-1. (1 mark) iii. From the energy profile diagram it can also be determined that H = EA(forward) EA(reverse) EA(forward) = H + EA(reverse) = -800 + 1200 = 400 kJ mol-1. (1 mark) Alternatively: From the diagram the activation barrier maximum is at 1200 + (-875) = 325 kJ mol-1 The difference between this and the energy of the reactants 325 (-75) = 400 kJ mol-1. b. i. Table 13 of the VCE Chem Data Booklet gives the molar enthalpy of

combustion for various fuels, this is for 1 mol of the fuel. The equation involves 2 mol of methanol, therefore

H = 2 -725 = -1450 kJ mol-1. (1 mark) ii. The thermochemical equation for the reaction in i. above is 13 2 2 22CH OH(l) 3O (g) 2CO (g) 4H O(l) H 1450 kJ mol

+ + = For the reaction described by the chemical equation 3 2 2 22CH OH(l) 3O (g) 2CO (g) 4H O(g)+ + 4 mol of H2O(l) must be converted into 4 mol H2O(g). 12 2H O(g) H O(l) H 44 kJ mol

= Therefore for the reverse reaction 12 2H O(l) H O(g) H 44 kJ mol

= + (1 mark) The reaction involves 4 mol of water H = -1450 + (4 44) = -1274 kJ mol-1. (1 mark)

End of Suggested Answers

VCE Chemistry 2010 Year 12 Unit 4 Industrial ChemistryStudent Answer SheetVCE Chemistry 2010 Year 12 Unit 4 Industrial ChemistryMultiple Choice Questions - Section AEnd of Section A

VCE Chemistry 2010 Year 12 Unit 4 Industrial ChemistryShort Answer Questions - Section BEnd of Section BEnd of Unit Test

Suggested AnswersVCE Chemistry 2010 Year 12 Unit 4 Industrial ChemistryMultiple Choice Questions Section AShort Answer (Answers) Section BEnd of Suggested Answers

Unit 4 Industrial Chemistry

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