Industrial Chemistry
Short Description
Own notes, from various sources....
Description
HSC Chemistry Option Module: Industrial Chemistry Summary
Industrial Chemistry: 1. Replacements for natural products
x
Identify data, gather and process information to identify and discuss the issues associated with the increased need for a named natural resource that is not a fossil fuel and evaluate the progress currently being made to solve the problems identified.
x
Discuss the issues associated with shrinking world resources with regard to one identified natural product that is not a fossil fuel, identifying the replacement materials used and/or current research in place to find a replacement for the named material.
Natural product: Rubber
Natural rubber is an elastomer (elastic hydrocarbon polymers) originally derived from latex, the milky colloid found in some plants. The commercial source of rubber is the Para rubber tree- rubber therefore a renewable resource. The uses of rubber are widespread, from household items to many uses in medicine and industry: tires (the majority is used for this), hoses, belts, matting, flooring, gloves, toy balloons, rubber bands, pencil erasers, textiles, even spacecrafts. In addition, the rubber wood can be used as timber for quality furniture and construction once the tree becomes unviable for rubber production. It is strong, flexible, resistant to microbes and easy to work with. Flow Chart for conventional rubber production: Rubber tree cultivation: tropical countries i.e. South-East Asia, Africa, South America provide optimum growing conditions, with up to 25 year productive phase Tapping: Outer layer of bark is cut off and latex is collected in containers supported by wires attached to the trunk of the tree Collection: Latex is transferred to air-tight containers for ammoniation (ammonia prevents latex from coagulating) Processing: processed into latex concentrate for dipped products, or coagulated under clean conditions using formic acid Grading: Rubber is process into various grades (higher grades have size reduction and cleaning process to remove contaminants) Final processing and packaging: Rubber is dried, then baled, pelletized and packaged ready for shipping
Robert Lee Chin
Page | 1
HSC Chemistry Option Module: Industrial Chemistry Summary
Problems with natural rubber The production of rubber trees has environmental and political issues relating to the smallholders who grow the crop. Over 90% of natural rubber production occurs in developing nations with tropical climates including Thailand, Malaysia, Indonesia, South America and most recently, Africa. A “slashand-burn” approach is used, leading to land degradation, air pollution and deforestation. End of season burn-offs are used to increase plantation area which has seen the emergence of new diseases previously isolated within the forest. Rubber growers have little economic incentive to improve the quality of their product as they have little bargaining power compared to the intermediate traders. As a result, they often leave contaminants in unrefined rubber in an attempt to be paid more for the weight of the rubber. Scarcity of land for rubber plantations has led to escalating land prices, leaving ownership of viable land to a few elite farmers. There are food shortages as the result of land being used for Rubber plantations rather than crop staples. Reasons for solutions: *Some people are severely allergic to proteins in latex products, which can cause occupational problems esp. in the healthcare industry. *Rubber plantations in Asia are affected by a fungal disease that could potentially decimate current production. *There are also predicted shortages of supply due to economic developments in China and India *Today, 75% of rubber production is a synthetic product derived from petroleum. Although this is a cheap way to produce rubber, petroleum is a non-renewable resource which is rapidly depleting. Therefore, there is great need for an alternative source of rubber. * Synthetic rubber is poorer quality and cannot be used for high impact purposes such as aeroplane tyres and tyres carrying heavy loads. Solutions to resource shortage Several substitute plants are being trialled to enhance global production of rubber, the most promising is a Mexican shrub called guayule. Guayule has the benefits of: -the potential to reduce environmental effects as it can tolerate more arid conditions -less likely to cause allergic reactions than latex from rubber trees
2. Equilibrium processes in industrial processes
x
Identify data, plan and perform a first-hand investigation to model an equilibrium reaction
Robert Lee Chin
Page | 2
HSC Chemistry Option Module: Industrial Chemistry Summary x
Choose equipment and perform a first-hand investigation to gather information and qualitatively analyse an equilibrium reaction
First-hand Investigation: Systems at equilibrium Aim: To examine the effects of changing reaction conditions on a system at equilibrium Materials: Dilute solutions (0.1 M) of: - Iron (III) chloride -potassium thiocynate (KSCN) -potassium chloride (KCl) -Iron (III) nitrate (Fe[NO3]2) -sodium hydroxide (NaOH) Method: 1. Place equal volumes of iron(III) chloride and potassium thiocyanate into a test tube and allow equilibrium to be established; a blood-red solution will form. 2. Dilute solutions with distilled water until colour lightens 3. Place equal volumes of the diluted solution into five separate test tubes 4. Add a small volume of KCl to the first test tube 5. Repeat step 4 with Iron(III) nitrate for the second test tube 6. Repeat with KSCN for the third test tube 7. Repeat with NaOH for the forth test tube. Results: The reaction between Iron(III) chloride and Potassium thiocyanate (KSCN): FeCl 3( aq ) KSCN ( aq ) Fe(SCN ) 3 ( aq ) 3KCl 3( aq ) Chemical added to FeCl3 and KSCN KCl
Chemical disturbing equilibrium
Observation regarding colour
Shift in equilibrium and explanation
KCl
Lighter red
Fe[NO3]2
darker red
KSCN
KSCN
Darker red
NaOH
NaOH
Orange (precipitate)
Increasing the concentration of one of the products shifts equilibrium to the right Adding Fe3+ ions increases the concentration of reactants; shift to right Increasing the concentration of one of the reactants shifts equilibrium to the right Fe3+ ions react with OH- ions to form a yellow orange ion complex; shifts left???
Fe[NO3]2
Robert Lee Chin
Page | 3
HSC Chemistry Option Module: Industrial Chemistry Summary Conclusion: Changing the concentrations of reactants and products in an equilibrium, or adding a substance that will react with either products or reactants, will have an effect on the equilibrium position. Increasing the concentration of a reactant or a substance that reacts with a product will shift the equilibrium to favour the products. Conversely, Increasing the concentration of a product or a substance that reacts with a reactant will shift the equilibrium to favour the reactants. x
Explain the effect of changing the following factors: -Concentration -Pressure -Volume -temperature
Le Chatelier’s Principle When a system that is at equilibrium is changed by altering temperature, concentration of products and/or reactants or the volume/pressure of gas; the equilibrium shifts to counteract the effect and return the system to equilibrium Concentration An increase in temperature will cause: -An exothermic reaction to reverse. Products will decompose into reactants i.e. concentration of products will decrease while concentration of reactants will increase. The reverse occurs if the temperature of the system is lowered. -An endothermic reaction to proceed as written. The reverse occurs if the temperature of the system is increased. Pressure and volume An increase in pressure and volume will cause -changes to occur in gases only -same effect as decreasing volume of the reaction vessel, as the particles are forced to occupy less space -equilibrium favours the side with least number of moles of gas Concentration An increase in concentration will cause: -equilibrium to shift to maintain the molarities of products and reactants i.e. if more reactants are added, it will shift to produce more product and vice versa. x
Process and present information from secondary sources to calculate K from equilibrium conditions The extent to which an equilibrium reaction can be quantitatively defined using the equilibrium constant, K. When the value of K is large, the equilibrium is towards the product side. When K is small, it is towards the reactant side.
Robert Lee Chin
Page | 4
HSC Chemistry Option Module: Industrial Chemistry Summary If we define the general homogenous (same chemical state i.e. solid, liquid or gas) equilibrium reaction as: aA bB cC dD , then K is defined as: [products] [C] c [D] d K , where [ ] denotes concentration in molL-1 [reac tan ts] [A] a [B] b
System at equilibrium Examples: 1) Consider the equilibrium: 3I 2(g) 6F2(g) 2IF5(g) I 4 F2(g) At a certain temperature 3.0 moles of F2 and 2.0 moles of I2 are introduced into a 10.0 L container. At equilibrium, the [I4F2] is 0.02 M. Calculate the K for the reaction.
K
[IF5 ] 2 [I 4 F2 ] [F2 ]6 [I 2 ]3
Concentrations (molL-1) Initial concentration Δ concentration Equilibrium concentration x
[0.04] 2 [0.02] [0.18]6 [0.14]3
I2
342.87106... # 343 (3 s.f.)
Reactants
Products
F2
IF5
I4F2
0.2
0.3
0
0
-3(0.02) = -0.06 0. 1 4
-6(0.02) = -0.12 0.18
+2(0.02) = 0.04 0.04
+0.02 0.02
Interpret the equilibrium constant expression (no units required) from the chemical equation of equilibrium reaction If K is large, the equilibrium lies to the right, so the concentration of products is high and the reaction almost goes to completion. If K is small, the equilibrium lies to the left, so there is very little reaction. The size of the K value is an indication of the relative strength of acids and bases, or the solubility of a substance. Points to remember: *When calculating K, disregard any reactants or products in solid or liquid (solvents) state *the concentrations used refer to the concentrations of reactants and products at equilibrium
Robert Lee Chin
Page | 5
HSC Chemistry Option Module: Industrial Chemistry Summary Identify that temperature is the only factor that changes the value of the equilibrium constant (K) for a given equation
x
The value of K for a given equilibrium is unaffected by changes in concentration or the addition of a catalyst, which simply increases the rate of both forward and reverse reactions. For exothermic reactions, K decreases with increasing temperatures as the equilibrium favours the reactants. Conversely, K increases with decreasing temperatures, as it favours the products For endothermic reactions, K increases with increasing temperatures because the equilibrium favours the products. Conversely, K decreases with decreasing temperatures, as it favours the reactants For example, in the oxidation of sulfur dioxide is an exothermic equilibrium:
2SO 2( g ) O 2( g ) 2SO 3( g ) heat ,
K=
[SO3 ]2 [O2 ][SO2 ]2
Heating the reaction vessel will shift the equilibrium to the left. This will increase the values of [O2] and [SO2] and decrease [SO3], therefore K will be smaller. However, if the vessel is cooled, equilibrium will shift to the right i.e. [SO3] will increase while [O2] and [SO2] will decrease, so K will be larger.
3. Sulfuric Acid x
Outline three uses of sulfuric acid in industry
Sulfuric acid is one of the most important industrial chemicals, with many applications: *Used by fertiliser industry (~70% total production) to make ammonium sulfate and superphosphate *Removes oxides and grease from steel and iron before galvanising or electroplating *Dehydration agent i.e. removes water liberated during manufacture of explosives, dyes, detergents, polymers, esters, electrolysis of sodium chloride solution *Electrolyte in vehicle batteries *Used in the production of nitroglycerine for explosives and as a vasodilator (treatment of heart conditions)
x
Describe the processes used to extract sulfur from mineral deposits, identifying the properties of sulfur which allow its extraction and analysing potential environmental issues associated with its use
Robert Lee Chin
Page | 6
HSC Chemistry Option Module: Industrial Chemistry Summary Elemental sulfur occurs in deposits near volcanoes, hot springs and underground. It is also found in ores e.g. PbS, hydrogen sulfide in fossil fuel sources and sulfates in the ocean Extraction of Sulfur Frasch Process: Sulfur is extracted from underground sources by the Frasch process. This process relies on the following properties of sulfur: -relatively low melting point (113°C) and boiling point (445°C) -low density -insolubility in water There are 4 main steps: 1. 3 concentric pipes are drilled to the deposit 2. Superheated steam (160°C) under is pumped through the outer pipe, melting the sulfur 3. Air under high-pressures is pumped through the inner pipe, pushing the low-density liquid sulfur foam around the middle pipe to the surface 4. The insoluble sulfur is easily separated without the need to evaporate the water. Slight cooling causes the sulfur to solidify and settle out. Hot, compressed air Molten sulfur
Superheated water
Sulfur deposit
Natural gas and petroleum: Extracted from hydrogen sulfide (H2S) in natural gas and petroleum deposits. The sulfur is produced in incomplete combustion of H2S: 3H 2 S ( g ) O 2 ( g ) o 2H 2 S ( g ) 3S ( g ) SO 2 ( g ) 1. The mixture is cooled to condense the sulfur and passed over a heated catalyst: 2H 2S( g ) SO 2 ( g ) o 2H 2 O ( g ) 3S( g )
Robert Lee Chin
Page | 7
HSC Chemistry Option Module: Industrial Chemistry Summary 2. The mixture is cooled again to condense the remaining sulfur. Smelting or metal ores Sulfur may also be released as sulfur dioxide when metal oxide ores are smelted. The general equation, using M to represent the metal is: MS(s ) O 2( g ) o M (s ) SO 2( g ) Environmental Issues Sulfur dioxide readily dissociates with water to form sulphurous and sulfuric acid, thus producing acid rain. Acid rain lowers the pH of soil and water sources, potentially damaging the environment and also contributes to chemical weathering of statues and metal. Most of the sulfur dioxide produced by metal smelting is used to make sulfuric acid and sulfur dioxide emissions into the atmosphere are strictly controlled by government regulations. Also produced during mining and smelting is hydrogen sulfide (H2S), a gas more toxic than cyanide. Soluble substances such as arsenic may be brought to the surface
x x x
Outline the steps and conditions necessary for the industrial production of H2SO4 from its raw materials Describe the reaction conditions for the production of SO2 and SO3 Gather, process and present information from secondary sources to describe the steps and chemistry involved in the industrial production of H2SO4 and use available evidence to analyse the process to predict ways in which the output of sulfuric acid can be maximised
Production of SO2 and SO3 Combustion of sulfur or metal sulfides occurs in a combustion furnace as an exothermic reaction. Conditions that favour high rate of reaction and yield are high temperature and surface area (by crushing rock). Molten sulfur from the Frasch process is filtered to remove impurities and lime (calcium hydroxide) is added to reduce the acidity, thus preventing corrosion. In the combustion chamber excess dry oxygen is used to maximise the yield of SO2. Dry air under pressure
Molten sulfur
Robert Lee Chin
Combustion furnace
SO2
Page | 8
HSC Chemistry Option Module: Industrial Chemistry Summary SO2 derived from smelting of metal ores must be purified and dried before going to the converter: O2 smelter Metal sulfide
SO2 Metal particles
Water scrubber
SO2 Steam particles
Electrostatic precipitator
SO2
Drying tower
Steam
Dry SO SO2
The production of SO3 from SO2 and O2 is an equilibrium reaction involving a compromise between reaction rate, product yield and economic factors. The following conditions produce a yield of over 98%: *High temperatures increase rate of reaction, but the equilibrium is exothermic and high temperatures may damage the catalyst. 450 - 600°C provides a compromise. *The vanadium pentoxide or platinum catalyst increases the rate of reaction by lowering the activation energy for the reaction. *A multi-stage process is used to convert the SO2 into SO3. With each new stage the temperature is reduced, thus shifting the equilibrium to the right. This produces a yield of over 98%. *Increased pressure shifts the equilibrium to the right, but this is expensive, so pressures of 12 atmospheres are used *Increasing the concentration of the reactants pushes the reaction to the right. From the reaction, the stoichiometic ratio of O2:SO2 is 1:2. Industrially, twice as much O2 is used i.e. 1:1 ratio. Production of H2SO4 Sulfuric acid is manufactured using the contact process. There are 4 main steps involved: 1. Combustion Chamber Pure sulfur or sulfide ore is combusted in air at 1000°C to produce sulfur dioxide: 000q C S (s ) O 2 ( g ) 1 o SO 2 ( g )
'H
-300 kJmol -1
2. Converter Sulfur dioxide is passed into a tower is passed into a tower made of stacked vertical beds of vanadium pentoxide (V2O5) or platinum catalyst. It is oxidised to sulfur trioxide in an exothermic equilibrium reaction, with temperatures of 450 - 600°C used as a compromise. High pressures also shift the reaction to the right. anadium pentoxide/platinum 2SO 2 ( g ) O 2 ( g ) v o 2SO 3( g )
'H
-100 kJmol -1
3. Absorption Tower Sulfur trioxide cannot immediately be mixed with water, as it forms a fine mist which is difficult to collect. Instead the gas is bubbled through very concentrated sulfuric acid in absorption towers to form Oleum (disulfuric acid), H2S2O7.
Robert Lee Chin
Page | 9
HSC Chemistry Option Module: Industrial Chemistry Summary
SO 3( g ) H 2SO 4 ( l ) o H 2S2 O 7 ( l ) The sulfuric acid used to dissolve the SO2 must be maintained at 70°C and 98% concentration to ensure maximum absorption. The concentration is important because at 98% concentration, the vapour pressure of SO3 above H2SO4 is a minimum. Concentrations greater than 98% and the water vapour pressure (high vapour pressure at normal temperatures = volatile) increases sharply, causing the resulting acid mist to evaporate. The acid is also concentrated at a rate to minimise an increase in concentration 4. Hydration of Oleum: Oleum is converted into concentrated sulfuric acid by the addition of dilute sulfuric acid and water. The reaction is highly exothermic, so the acid needs to be cooled before storage.
H 2S2 O 7 ( l ) H 2 O ( l ) o 2HSO 4 ( l ) Traditionally, mild steel was used for the manufacturing equipment. The trend is now to use Teflon coated piping to reduce contamination of the acid due to iron corrosion.
x
Apply the relationship between rates of reaction and equilibrium conditions to the production of SO2 and SO3
Condition Temperature
Increase reaction rate High temperatures
Increase Equilibrium yield Low temperatures
Pressure
n/ A
High pressures
Concentration of reactants
n/ A
Increasing reactant concentration
Concentration of products Other
n/ A
Removal of products n/A
Catalyst
Unreacted gases
Robert Lee Chin
Recycling back in reaction vessel
Economic factors High temp. Expensive and damages catalyst High pressures are expensive Increasing SO2 is more expensive than increasing O2 n/A Reduces activation energy and therefore, reaction temperature Efficient; maximises yield
Conditions used 450-600°C
1-2 atm. Excess O2 O2:SO2 is 1:1 Removing SO3 as it is produced Vanadium pentoxide/platinum
Recycling of Unreacted O2 and SO2 Page | 10
HSC Chemistry Option Module: Industrial Chemistry Summary x
Perform a first-hand investigation to observe the reactions of sulfuric acid acting as: -an oxidising agent -a dehydrating agent
Sulfuric acid as an oxidising agent The addition of sulfuric acid to zinc metal liberates hydrogen gas. Sulfuric acid as a dehydrating agent Addition of sulfuric acid to sugar (sucrose, C12H22O11) Aim: To observe the reactions of sulfuric acid acting as an oxidising agent and as a dehydrating agent Equipment: -Concentrated sulfuric acid -Granulated Zinc metal -matches -Table sugar (sucrose) -Glass beaker -Glass stirring rod Safety: concentrated sulfuric acid is highly corrosive and toxic and may cause severe burns to the body. ALWAYS add acid to water, stirring constantly. Use sodium bicarbonate to neutralise any spills. Wear acid-resistant safety glasses, gloves and aprons at all times. Avoid inhaling fumes, and use a well-ventilated area, or if possible, a fume cupboard. If contact with acid occurs, run under cold tap water for at least 15 minutes Method: As an oxidising agent: 1) Place a small amount of granulated zinc metal into a glass beaker 2) Slowly add a small volume of concentrated sulfuric acid and place a lid over the beaker. Zinc bubbles should begin forming on the zinc surface. 3) After several minutes, place a lit match into the beaker. You should hear a loud ‘pop’, indicating the presence of hydrogen gas As a dehydrating agent: 1) In a fume cupboard, add a small amount (50 g) of sucrose to a 250mL beaker 2) Slowly pour in concentrated sulfuric acid to just cover the sucrose. After a few minutes you should observe the beaker becoming hotter,the cusorse reacting to form a tower of black carbon and sulfur oxide fumes emerging from it. Results: As an oxidising agent: The addition of hot concentrated sulfuric acid to zinc metal liberates sulfur dioxide gas and water as shown: H 2SO 4 ( aq ) Zn (s ) o ZnSO 4 ( aq ) SO 2 ( g ) H 2 O ( l ) In the process zinc becomes zinc sulfate i.e. zinc is oxidised (oxidation state changes from 0 to +2). As a dehydrating agent:
Robert Lee Chin
Page | 11
HSC Chemistry Option Module: Industrial Chemistry Summary The addition of concentrated sulfuric acid to sucrose forms carbon, water and sulfuric acid in a highly exothermic reaction which causes most of the water to form steam: H 2 SO 4 ( aq ) C12 H 22 O11 (s ) o 12C (s ) 11H 2 O(l) H 2 SO 4 ( aq )
x
Describe and explain the exothermic nature of sulfuric acid ionisation
The ionisation of sulfuric acid is a rapid, highly exothermic reaction. It occurs in two steps: H 2 SO 4 ( aq ) o H ( aq ) HSO 4 ( aq ) Kc f HSO 4
-
(aq)
H (aq) SO 4
2-
(aq)
K c 1.2 u 10 2
In the first dissociation, H2SO4 acts as a strong acid as Kc approaches infinity. In the second dissociation, HSO4- is a weak acid, so Kc is small. x
Identify and describe safety precautions that must be taken when using and diluting concentrated sulfuric acid
Diluting sulfuric acid is potentially hazardous due to the vigorous exothermic nature during ionisation. Safety Precautions Wear protective equipment and clothing, especially safety glasses. ALWAYS ADD ACID TO WATER. Add a small amount at a time, stirring constantly (sulfuric acid is much denser than water and this may cause a layer of concentrated acid to form at the bottom, forming a steep temperature gradient). This produces a dilute solution, releasing only a small amount of heat which can be absorbed by the water. Any splatters that do occur are more likely to be dilute than concentrated. If spills do occur, clean up with large amounts of water and sodium hydrogen carbonate (bicarbonate soda). If water is added to concentrated acid, the heat released will make the water boil violently, splattering droplets of concentrated acid out of the container. The heat generated may be enough to crack the container. x
Use available evidence to relate the properties of sulfuric acid to safety precautions necessary for its transport and storage
Due to its corrosive nature (low pH), it must not be transported with foodstuffs, dangerouswhen-wet substances, oxidising agents, organic peroxides, toxic substances and radioactive substances. In high school science labs, dilute sulfuric acid is stored in glass/plastic containers with a tightly-fitting lid, so that it does not react with moisture in the air. Commercially, it is transported in its pure molecular form in steel containers as there is no water for it to ionise and react with the metal. It MUST NOT be stored with foodstuffs, oxidising agents and organic or combustible materials (it oxidises carbohydrates, causing them to char), or metals
Robert Lee Chin
Page | 12
HSC Chemistry Option Module: Industrial Chemistry Summary (liberates explosive hydrogen gas). Its high vapour pressure means it is volatile, so it must be stored in well-ventilated areas.
4. Sodium Hydroxide Explain the difference between galvanic cells and electrolytic cells in terms of energy requirements There are two types of electrochemical cells: galvanic cells and electrolytic cells. x
In galvanic cells a redox reaction takes place spontaneously to produce electricity. It consists of two half cells with separate electrolytes and a salt bridge to complete the circuit. chemical energy o electrical energy In galvanic cells, the anode is negative and the cathode positive In electrolytic cell a compound is decomposed by passing electricity. It is not spontaneous; it is forced by applying a voltage; electrical energy o chemical energy In electrolytic cells, the anode is positive potential and the cathode has the negative potential
x
Identify, plan and perform a first-hand investigation to identify the products of the electrolysis of an aqueous solution of sodium chloride
Investigation: Electrolysis of Aqueous Sodium Chloride Aim: To determine the products of the electrolysis of aqueous sodium chloride or brine Background: Electrolysis is a process that uses electricity to decompose ionic compounds. The ion are caused to migrate to the charged electrodes and pure elements will form. Electrolysis is also used in industry in the purification of active metals such as aluminium and sodium. Materials: -carbon electrodes -transformer -brine or salt solution -phenolphthalein in dropper bottle -voltameter
Robert Lee Chin
Page | 13
HSC Chemistry Option Module: Industrial Chemistry Summary Carbon Electrode Method: 1/ Place brine solution into a beaker 2/ Place two carbon electrodes into the beaker and connect to the transformer. 3/ Set to DC 8V and allow the current to run for a few minutes. 4/ Add drops at the cathode of the circuit, observe. 5/ Add a few drops at the cathode, observe.
Voltameter Method: 1/ Place a few drops of phenolphthalein indicator into a beaker containing brine solution. 2/ Fill the voltameter with the brine solution. 3/ Connect to the DC terminal of a transformer and allow 8V to flow through the voltameter. 4/ Observe the voltameter and test the gases collected above both the cathode and the anode.
cathode
anode
Results:
Equations: Anode (positive electrode) reaction:
C l ( aq ) o 1 2 C l 2 ( g ) e
Robert Lee Chin
EI
- 1 .3 6 V
Page | 14
HSC Chemistry Option Module: Industrial Chemistry Summary Cathode (negative electrode):
H 2 O (l) e - o 1 2 H 2 ( g ) OH ( aq )
EI
-0.83 V
Net ionic reaction:
Cl ( aq ) H 2 O ( l ) o 1 2 Cl 2 ( g ) 1 2 H 2 ( g ) OH ( aq ) Minimum voltage requirement: 1.36 0.83
x
2.19 V
Analyse information from secondary sources to predict and explain the different products of the electrolysis of aqueous and molten sodium chloride
Electrolysis of pure, molten sodium chloride If pure, molten NaCl undergoes electrolysis, the result is the decomposition of the compound into its element. Chloride ions are oxidised to form chlorine gas at the anode: Cl ( l ) o 1 2 Cl 2 ( g ) e E I -1.36 V At the cathode, sodium ions are reduced to form pure sodium metal: Na ( l ) e o Na (s ) E I -2.71 V
E I Total
4.07 V
As there are only 2 species involved, the only products that will form are chlorine gas and sodium metal. Electrolysis of aqueous sodium chloride In an aqueous sodium chloride solution, electrolysis may produce different products because water can also participate in reactions. There are 2 possible reduction reactions at the cathode: Na ( aq ) e o Na (s ) E I -2.71 V H 2 O (l) e - o 1 2 H 2 ( g ) OH ( aq )
EI
-0.83 V
The water reduction has a significantly lower voltage requirement, so it will proceed. If any sodium metal did form, it would instantly react with the water anyway. There are also 2 possible reactions at the anode: Cl ( aq ) o 1 2 Cl 2 ( g ) e H 2 O ( l ) o 1 2 O 2 ( g ) 2H
EI
( aq )
2e
E
I
-1.36 V -1.23 V
Because the voltage requirements are very similar, the reaction that occurs will depend on the concentration of the reactants. In a dilute solution, oxygen gas would be produced as there would be mostly water molecules. In a concentrated solution, however, a mixture of the two would be produced. In the industrial process to manufacture NaOH, the salt solution used is so saturated, that pure chlorine gas is produced.
Robert Lee Chin
Page | 15
HSC Chemistry Option Module: Industrial Chemistry Summary x x
Outline the steps in industrial production of sodium hydroxide from sodium chloride solution and describe the reaction in terms of net ionic and full formula equations Distinguish between the three electrolysis methods used to extract sodium hydroxide: -mercury process -diaphragm process -membrane process by describing each process and analysing the technical and environmental difficulties involved in each process
The electrolysis of sodium chloride to produce sodium hydroxide and chlorine is known as a chloralkali process. Electrolysis of sodium chloride to produce sodium hydroxide and chlorine is performed using 3 types of electrolytic cells: mercury, diaphragm and membrane cells. The main technical difficulty in all methods is preventing the chlorine produced contacting (and therefore, reacting) with the sodium hydroxide or hydrogen. A saturated brine (NaCl) solution is used in all 3 cells. Impurities are first removed as a sludge by precipitation reactions: -Calcium:
Ca 2 ( aq ) Na 2 CO 3( aq ) o 2 Na ( aq ) CaCO 3(s )
-Magnesium: Mg 2 ( aq ) NaOH ( aq ) o MgOH (s ) Na ( aq ) -Iron -sulfate:
Fe 2 ( aq ) NaOH ( aq ) o FeOH (s ) Na ( aq ) Fe 2 ( aq ) Na 2 CO 3( aq ) o 2 Na ( aq ) FeCO 3(s ) SO 4
2
( aq )
CaCl 2 ( aq ) o CaSO 4 ( s ) 2Cl ( aq )
Despite using different methods, all 3 cells achieve the same overall reaction: 2 NaCl ( aq ) 2H 2 O ( l ) o 2 NaOH Cl 2 ( g ) H 2 ( g ) The Mercury Process In the mercury cell, titanium coated with a rare earth metal is the anode. A saturated brine (NaCl) solution floats above cathode, which is a thin layer of mercury at the bottom. Chlorine gas is oxidised at the anode and pumped away: 2Cl ( aq ) o Cl 2 ( g ) 2e Sodium is reduced at the cathode: Na ( aq ) e o Na (s ) , where is forms an amalgam (Na/Hg) with the mercury and is constantly removed. Electrolysis prevents the amalgam reacting with the brine. The amalgam pumped to a decomposing vessel where it is reacted with water to form sodium hydroxide and water: 2 Na / Hg 2H 2 O ( l ) o 2 NaOH(aq) H 2 ( g ) 2Hg ( l )
Robert Lee Chin
Page | 16
HSC Chemistry Option Module: Industrial Chemistry Summary Although the mercury is recycled, the mercury cell is being phased out due to concerns of mercury pollution, particularly of aquatic environments. The NaOH produced is 50% (0.5 M) concentrated- suitable for commercial purposes. To Ti anode
+
Cl2(g)
Cl2(g) Brine
To Hg cathode
_
H2(g)
NaOH(aq) Cold water
H2O(l)
Na/Hg amalga, Decomposer
Mercury cooler and pump Hot water
Membrane cell
The Diaphragm Process In the diaphragm Cell the products of electrolysis are kept apart using a permeable asbestos diaphragm. This provides a lining for the steel-mesh cathode. The anode is a rod of solid graphite. Electrolysis occurs when the brine soaks through the asbestos to the cathode. Hydrogen gas is discharged at the cathode: 2H 2 O ( l ) 2e o H 2 ( g ) 2OH ( aq ) In an aqueous environment the reduction of sodium does not occur. Chlorine gas is discharged at the anode: 2Cl ( aq ) o Cl 2 ( g ) 2e
Sodium ions (spectator ions) from the brine are attracted to the cathode and with the hydroxide ions produced at the cathode, are washed to the bottom where a caustic brine (NaOH/NaCl) leaves the cell. The NaOH must be concentrated to 50% and the salt removed via evaporation. The salt may then be reused to form brine. The overall cell reaction: 2 NaCl ( aq ) 2H 2 O ( aq ) o 2 NaOH ( aq ) H 2 ( g ) Cl 2 ( g )
Robert Lee Chin
Page | 17
HSC Chemistry Option Module: Industrial Chemistry Summary Brine H2(g)
Cl2(g) Spend solution
Asbestos diaphragm
Steel-mesh cathode
Graphite anode
Steam NaOH solution
This cell is slowly being phased out due to the asbestos used, which poses a serios health hazard as it is carcinogenic and may cause life-threatening respiratory diseases such as asbestosis and mesothelioma. Another disadvantage is that is produces low purity NaOHonly about 12% pure. This is because some of the chloride ions can seep through, thus contaminating the NaOH.
Membrane Cell The membrane cell is the most recent cell due to the recent development of high-tech Teflon membrane. It has a titanium anode (will not react with chlorine gas) and a nickel cathode. Anode and cathode compartments are separated by an ion exchange that allows water and positive ions i.e. Na+ to diffuse through, but not negative ions. Brine is pumped into the anode compartment where chlorine ions are oxidised to form chlorine gas, which is pumped away: 2Cl ( aq ) o Cl 2 ( g ) 2e At the cathode, water is reduced to hydrogen gas and hydroxide ions: 2H 2 O ( l ) 2e o H 2 ( g ) 2OH ( aq ) Sodium ions migrate to the cathode compartment where they combine with the hydroxide ions to form NaOH. Unfortunately, the brine is able to migrate to the cathode compartment, thus reducing the purity of the NaOH obtained, so further purification is needed. Overall cell reaction is the same as for the diaphragm cell. The membrane cell is likely to replace the other two, as it has the least environmental impact and is very energy efficient as the electrodes can be placed very close to each other. It produces NaOH of about 30-40% purity (mixture contains NaCl from brine), so further purification is needed to increase purity to 50%.
Robert Lee Chin
Page | 18
HSC Chemistry Option Module: Industrial Chemistry Summary Cl2(g)
Ti anode
Ni cathode
H2(g)
Brine in
H2O(l) Na+(aq)
Brine out
Property Cathode Cathode product decomposer Anode Anode product NaOH purity Operating voltage Cell temperature (°C)
Environmental Issues
Technical Issues
Robert Lee Chin
NaOH(aq)
Mercury Cell Hg Na/Hg amalgam NaOH H2(g) Titanium coated with rare-earth metal Cl2(g) 50% 4.0 – 4.5 90 – 95
Mercury Cell Use of mercury and its disposal: -poses hazard to aquatic environments
Diaphragm Cell Steel-mesh NaOH(aq) NaCl(aq) H2(g)
Membrane cell Nickel NaOH H2(g) n/A
Titanium/titanium steel-alloy/graphite 12 – 15% 4.0 – 5.0 75 – 85
~40% 3.0 – 4.0 88 – 90
Diaphragm Cell Use of asbestos and its disposal: -expensive to remove -not readily decomposed -small fibres remain suspended in air for long periods
Membrane Cell n/A
Chlorine gas is toxic: -must wear protective breathing apparatus -check for leaks Chemicals used are corrosive to metal surfaces and bricks: -maintenance is expensive Hydrogen is highly reactive with chlorine and oxygen: - check for leaks esp. diaphragm and membrane cells Mercury is toxic: n/A -check for leaks -safe disposal Page | 19
HSC Chemistry Option Module: Industrial Chemistry Summary Use of electricity creates heat: -size of gap between electrodes must be controlled by computer. -size of gap determines operating voltage and therefore, production costs Quality control: -NaOH concentration (titration) -impurity metal ions in brine (AAS) -impurities in Cl2, H2, O2 (gas chromatography)
1. Saponification Background Triglycerides A triglyceride is the general name for a compound containing 3 “fatty acid” molecules joined to one “glycerol” molecule. Triglycerides are esters.
Fatty acids are long-chain hydrocarbons with a carboxylic (-COOH) group at one end, which is either saturated or unsaturated. Schematically, fatty acids can be represented by:
Non-polar and hydrophobic/lipophilic hydrocarbon chain
Polar and hydrophilic/lipophobic –COOH group
Glycerol is a ‘triple alcohol’ i.e. it contains 3 OH- functional groups. Systemically, it is named ‘propantriol’
Robert Lee Chin
Page | 20
HSC Chemistry Option Module: Industrial Chemistry Summary H | H C OH | H C OH | H C OH | H
Non-polar Hydrocarbon chain
polar –OH functional groups
Fats and oils are both types of triglycerides. The difference is that fats are solid at room temperature due to all single C–C bonds, therefore they are said to be saturated. Oils are liquid at room temperature due to the presence of many double C=C bonds i.e. it is ‘unsaturated’. x
Describe saponification as the conversion in basic solution of fats and oils to glycerol and salts of fatty acids
Saponification is the reaction in which an ester is ‘hydrolysed’ with a strong base (e.g. NaOH, KOH) to produce glycerol (alcohol) and salts of fatty acids (carboxylic acid). If the fatty acid was oleic acid CH 3 (CH 2 )7 CH CH (CH 2 )7 COOH and the base KOH, then the soap molecule produced would be potassium oleate, CH 3 (CH 2 ) 7 CH C18H33KO2 x
CH (CH 2 ) 7 COO K
describe the conditions under which saponification can be performed in the school laboratory and compare these with industrial preparation of soap
The saponification (SAP) value is a measure of the amount of base required to saponify 1 gram of an ester. This value is a measure of the average molecular weight of the all the fatty acids. Long-chain fatty acids, found in fats, have low SAP values due to fewer –COOH groups compared to short-chain fatty acids. Laboratory Saponification In the lab, soap can be made by heating a mixture of soap or oil with NaOH or KOH. The soap produced can be precipitated by using a saturated sodium chloride solution and washed to remove excess glycerol and base. Dilute acid can be used to neutralise and excess base. Industrial Saponification Industrially, soap can be prepared in one stage (Kettle Boiled Batch process) or two stages (fatty acid neutralisation process). Obvious differences in both of these processes as compared to laboratory scale are: -a mixture of fats and oils are used e.g. palm, olive and coconut Robert Lee Chin
Page | 21
HSC Chemistry Option Module: Industrial Chemistry Summary -huge reaction vessels are used, up to 120 tonnes -elevated temperatures and pressures (makes water more soluble in fats) -metal-based catalyst -the shape size and internal roughness of the reaction Bessel must be considered -glycerol is removed and used for other purposes Fatty-acid neutralisation process This process uses two steps: The first step hydrolyses the triglycerides to form fatty acids and glycerol, the reverse of esterification. A zinc-oxide catalyst improves the reaction rate. The fatty acids flow upwards to a vacuum dryer, while the glycerol/water mixture is pumped away and separated via distillation. The second step neutralises the fatty acids with precise amounts of a strong base (determined from SAP value) to form the sodium/potassium fatty-acid salt (soap). Kettle Boiled Batch Process As the name suggests, this process takes place in large, open steel tanks called ‘kettles’. The reaction mixture is kept constantly boiling via injection of high pressures, which also helps keep the reaction mixing. Soap from previous reactions is kept to help the water and oil emulsify. Batch process refers to a reaction where all the reactants (oil, base, salt, catalyst, water) are added at the beginning and the products removed at the end. At the end of this process, extra salt is added to help solidify the soap. Extra steam is added and the mixture settled to remove the glycerol. The main difference compared to the lab process is the use of high pressure steam to hydrolyse the fats/oils.
x x
account for the cleaning action of soap by describing its structure explain that soap, water and oil together form an emulsion with the soap acting as an emulsifier
What is an emulsion? An emulsion is a mixture formed when two or more normally immiscible substances form a stable mixture which does not separate. It is not to be confused with a solution, as non of the molecules are associated with each other. Instead, one of the liquids is evenly dispersed throughout the other, through the action of another chemical called an emulsifier. Each emulsion has two different types. For example, if the two liquids were oil and water, the two emulsions would be: -oil suspended in water e.g. sorbolene cream -water suspended in oil e.g. mayonnaise
Robert Lee Chin
Page | 22
HSC Chemistry Option Module: Industrial Chemistry Summary What are surfactants? Surfactants are chemicals which reduce the surface tension of liquids by adsorbing at the interface of two different liquids. Surfactants may act as: detergents, wetting agents, emulsifiers, foaming agents and dispersants. In soaps and detergents, surfactants make it easier for water to emulsify with oil/ grease. Soap as an emulsifier The use of soap as a cleaning agent is dependent on its ability to emulsify dirt/grease in water. Most dirt is non-polar and oils/grease is non-polar long-chain hydrocarbons. Water is polar, thus it is immiscible with dirt and grease. When soap is placed in water and agitated, its ions dissociate (the positive ion takes no part in the cleaning action). The negative fatty acid ions, called the surfactant (surface acting agent) do not disperse evenly. Instead, they form clumps with the polar –COOH heads sticking outwards. The –COOH groups form hydrogen bonds with water. This spherical structure is called a micelle. Hydrophilic -COOH group forms hydrogen bonds with water
Non-polar lipophilic hydrocarbon chains dissolve triglycerides
Non-polar grease/dirt molecules are encapsulated and dissolved by the non-polar lipophilic core of the micelle, thus forming an emulsion between water and oil. This emulsion keeps the grease molecules suspended, which can be carried away by water.
Suspended grease molecule
Robert Lee Chin
Page | 23
HSC Chemistry Option Module: Industrial Chemistry Summary x
Distinguish between soaps and synthetic detergents in terms of: -the structure of the molecule -chemical composition -effect in hard water
Detergents are artificial soaps, invented during the 1940’s and have largely replaced soap in cleaning because they are more powerful emulsifiers and because they can be modified to specialised applications. While soaps are made from animal and vegetable oils and manufactured via saponification (heating triglycerides with a strong base), detergents are derived from petroleum. An alkanol from petroleum is reacted with Sulfuric acid to a sulfur-containing acid, which is then reacted with a strong base. Composition and structure of detergents The composition of detergents is similar to soaps in that both contain long hydrocarbon chains. The difference lies in that while soaps are sodium/potassium salts, detergents are hydrocarbons with a sulfate or sulfonate functional end group. Structurally, they both contain an ionic or polar head and a non-polar hydrocarbon tail. However while soaps act only as anionic surfactants, detergents can act as anionic, cationic or non-ionic surfactants. Effect in hard water Hard water contains significant concentrations of magnesium and calcium ions. Soap will not ‘lather’ in hard water as it forms a precipitate ‘scum’ which accumulates on surfaces. Detergents will lather in hard water and they do not form precipitates. distinguish between anionic, cationic and non-ionic synthetic detergents in terms of: -chemical composition -uses Anionic Anionic detergents contain a negative ion (most commonly sulfate, SO42- or sulfonate SO3-) as the hydrophilic end of the molecule. The hydrocarbon chain contains a benzene ring at the ends. x
They are primarily used as cleaners for glass and ceramic dishes and clothing and are highly sudsing. This is due to the fact that these substances have negative surface charges, which repel the anionic detergents and can be easily washed away, carrying the emulsified grease/dirt with it. They are usually combined with ionic detergents to provide enhanced stability.
_
Robert Lee Chin
Page | 24
HSC Chemistry Option Module: Industrial Chemistry Summary Cationic Cationic detergents contain a positive ion (usually an ammonia compound). They emulsify fats well, but are unsuitable for glass and ceramics as they contain negative surface charges. They are commonly used as fabric softeners and hair conditioners because they reduce static, thus causing fabric and hair to appear ‘fluffy’. Other uses include germicides, mouthwash and antiseptic soaps owing to their germicidal properties.
+ Non-ionic Non-ionic detergents do not form ions in a solution at all. Instead, they contain hydrophilic groups (such as oxygen atoms, alcohol groups or polysaccharides) along the hydrocarbon chain which form hydrogen bonds with water. They are used for car-washing, cosmetics, dishwashing and froth flotation. One disadvantage is their loss of solubility in warm water.
O
δ+
δ+ O δ-
x
O δ-
solve problems and use available evidence to discuss, using examples, the environmental impacts of the use of soaps and detergents
Biodegradability Soaps consists of fatty acids and can therefore be decomposed by bacteria when it enters sewage or waterways. Therefore, it has no negative impact on the environment. Early detergents were non-biodegradable as they were made from branched-chain hydrocarbons which are difficult for bacteria to decompose. This was alleviated in the 1970’s when straight-chain hydrocarbon detergents were introduced. The sulfonic acid group found in anionic detergents and non-ionic detergents are highly biodegradable. Presence of phosphates The main environmental impact of detergents was due to the addition of phosphates, as it was discovered that they aided the effectiveness of detergents. The main problem with the use of phosphates is their contribution to eutrophication in waterways.
Robert Lee Chin
Page | 25
HSC Chemistry Option Module: Industrial Chemistry Summary Eco-friendly cleaning agents Eco-friendly cleaning agents use less non-essential chemical additives such as dyes and fragrances to reduce the toxicity of the detergent. x
gather, process and present information from secondary sources to identify a range of fats and oils used for soap-making Oil/fat Olive oil (oleic acid) Palm oil (palmitic acid) Coconut oil (lauric, myristic acid) Lard (palmitic, stearic, myristic acid) Tallow (palmitic, oleic, stearic) Soybean oil (mainly polyunsaturated) Safflower oil (polyunsaturated) Shea Oil Sunflower oil Rice bran oil Wheatgerm oil
x
Perform a first-hand investigation to carry out saponification and test the product
Experiment: Soap Production Aim: to make soap by reacting a fat or oil and an alkali Background: Soaps and other cleaning agents are often produced by the saponification process. Soap making has been an important industry for many centuries. This reaction involves the adding of a caustic alkali solution to fats or oils. The ratio of alkali used to fat/oil depends upon the chemical make-up of the oil.
Mass of alkali (KOH) Materials: -beaker -potassium hydroxide -boiling chips -stirring rod -heating apparatus -universal indicator -hydrochloric acid
Robert Lee Chin
mass of oil (g) u SAP value of KOH (0.1876) -olive oil -boiling chips -electronic scales -mould -universal indicator
Page | 26
HSC Chemistry Option Module: Industrial Chemistry Summary
Precautions: Clean up any spillage and make sure the laboratory is well ventilated. There is a risk of spitting when the mixture is being heated. Do not use the soap produced on your skin as there may be excess KOH.
Method: 1/ Accurately weight 10g of olive oil in a clean beaker. 2/ Calculate the mass of potassium hydroxide needed according to the equation given above. 3/ Add the potassium hydroxide to the olive oil. 4/ Mix the reactants with a stirring rod. 5/ Add boiling chips to the beaker and gently heat. 6/ Continuously stir the reaction mixture while it is being heated. 7/ When the soap has formed, wash it thoroughly to remove any excess hydroxide. 8/ Add 5 drops of universal indicator. 9/ Add hydrochloric acid drop wise till the mixture is neutral. 10/ Place into mould and place in incubator to dry the soap Conclusion: What evidence suggests a reaction is occurring? Macroscopic changes such as the reaction mixture forming a stable, foamy emulsion. Why was neutralisation necessary? To remove the excess hydroxide ions produced from the KOH used for the saponification process
Robert Lee Chin
Page | 27
HSC Chemistry Option Module: Industrial Chemistry Summary x Perform a first-hand investigation to gather information and describe the properties of a named emulsion and relate these properties to its use. x Perform a first-hand investigation to demonstrate the effect of soap as an emulsifier Experiment: The Emulsifying Power of Soaps and Detergents Aim: -To examine the production of a soap/stain emulsion -To compare the emulsifying ability of various commercially available soaps and detergent. Background: An emulsion is a mixture of two or more chemicals which under normal conditions would be immiscible. Another chemical called an emulsifying agent is added to the immiscible chemicals and allows them to mix. Detergents and soaps act as emulsifying agents when we use them to clean materials. Materials: test tubes test tube rack olive oil or other suitable fat or oil rubber stoppers or corks a variety of commercially available soaps and detergents Precautions: Clean up any spillage and make sure the laboratory is well ventilated. Be careful when shaking not to spill any of the chemicals. Method: Part A: Making an Emulsion 1. Place 10 mL of tap water into two test tubes. 2. Add 2 ml of oil to each test tube. 3. Add a small amount of soap to one of the tubes. 4. Place a rubber stopper or cork on each test tube and shake for 30 seconds. 5. Place both test tubes in a rack and allow to stand.
Robert Lee Chin
Page | 28
HSC Chemistry Option Module: Industrial Chemistry Summary Part B – Testing the Ability of Various Soaps and Detergents: 1. Place 10ml of tap water into as many test tubes as you have soaps and detergents. 2. Add 2mL of oil to each test tube. 3. Add equal mass of different soaps to each one of the tubes. If liquid detergents are used, equal volumes should be used. 4. Place a rubber stopper or cork on each test tube and shake for 30 seconds. 5. Place both test tubes in a rack and allow to stand. 6. Compare how long each mixture remains an emulsion by timing how long it takes to separate into layers.
Conclusion and Discussion: What evidence suggests an emulsion has been produced? The oil and water formed a homogenous mixture with no distinct oil-water layers What observations can be used to determine the emulsifying ability of the soaps and detergents? The longer it takes for the emulsion to separate back into water and oil layers, the more powerful its emulsifying ability How could we gain a better idea as to which of the detergents and soaps were the best emulsifiers? Repeat trials using a variety of different types of oils/fats, as well as varying the ratio of water to oil used. Why must equal masses or volumes of soaps and detergents be used in the experiment? To be able to compare the emulsifying abilities of different surfactants, it it necessary to use equal amounts of the surfactants
Robert Lee Chin
Page | 29
HSC Chemistry Option Module: Industrial Chemistry Summary
6. Solvay Process x
Identify the raw materials used in the Solvay process and name the products
Raw materials are sodium chloride (from seawater), ammonia and Calcium carbonate (limestone). Products are sodium carbonate and calcium chloride (waste product). The overall reaction: x
2 NaCl ( aq ) CaCO 3(s ) o Na 2 CO 3(s ) CaCl 2 (s ) salt brine
limestone
product
waste
Describe the uses of sodium carbonate
Uses of sodium carbonate: -manufacture of soap, glass, ceramics, paper, sodium hydroxide and sodium bicarbonate -petroleum refining -water softener (removes Mg2+ and Ca2+ ions) -pollution control (used to scrub exhaust gases to remove sulfur dioxide at power stations) -cheaper alternative to sodium hydroxide as a general base -cleaning agent in washing compounds x
Identify, given a flow chart, the sequence of steps used in the Solvay process and describe the chemistry involved in: -brine purification -hydrogen carbonate formation -formation of sodium carbonate -ammonia recovery
Brine Purification Seawater is pumped into shallow ponds, where water is evaporated to leave behind a 30% concentrated brine. This is a mixture of magnesium and calcium salts, which must be removed by precipitation. Ca2+ ions are removed by adding Na2CO3: Na 2 CO 3( aq ) Ca 2 ( aq ) o CaCO 3(s ) 2 Na ( aq ) Mg2+ ions are removed by adding NaOH: 2 NaOH ( aq ) Mg 2 ( aq ) o Mg (OH) 2 (s ) 2 Na ( aq ) A flocculent is added to allow the precipitate to be easily removed. Hydrogen Carbonate Formation Crushed limestone is heated in a kiln to produce carbon dioxide and calcium oxide: eat CaCO 3(s ) ho CO 2 ( g ) CaO (s )
Robert Lee Chin
Page | 30
HSC Chemistry Option Module: Industrial Chemistry Summary Coke (carbon) is also present in the kiln, producing more carbon dioxide when heated and providing heat to decompose the lime: C ( s ) O 2 ( g ) o CO 2 ( g )
The CaO is used in a later process. The purified brine from the previous step is added to ammonia (NH4OH). The carbon dioxide from the limestone is bubbled through:
NaCl ( aq ) NH 3( aq ) H 2 O ( l ) CO 2 ( g ) o NH 4 Cl ( aq ) NaHCO 3( aq ) As chloride and sodium are spectator ions, the overall ionic equation: Cl ( aq ) NH 3( aq ) H 2 O ( l ) CO 2 ( g ) o NH 4 ( aq ) HCO 3 ( aq ) Formation of Sodium carbonate The ammonium chloride-sodium hydrogen carbonate solution is cooled to 0°C to precipitate the sodium carbonate, as it is only slightly soluble at lower temperatures: NaHCO 3( aq ) o NaHCO 3( s )
The mixture is filtered, dried and heated to ~300C to decompose it into sodium carbonate, water and carbon dioxide: eat NaHCO 3(s ) ho CO 2 ( g ) H 2 O ( l ) Na 2 CO 3(s ) Ammonia Recovery Calcium oxide from Hydrogen carbonate formation is dissolved in water to form calcium hydroxide: CaO ( s ) H 2 O ( l ) o Ca (OH) 2 ( aq ) When ammonium chloride from hydrogen carbonate formation is added, ammonia gas is liberated to be reused: 2 NH 4 Cl ( aq ) Ca (OH) 2 ( aq ) o 2 NH 3( g ) H 2 O ( l ) CaCl 2 ( aq )
x
Discuss environmental issues associated with the Solvay process and explain how these issues are addressed
Compared to previous methods, the Solvay process is less polluting as it reuses the products ammonia and carbon dioxide. However, some environmental issues remain. Disposal of calcium chloride waste Calcium chloride has few practical uses in vast quantities. When discharged into the ocean, there are no noticeable effects. However, when disposed of into rivers/waterways, it can increase the chloride and calcium ion concentrations to such an extent that local ecosystems
Robert Lee Chin
Page | 31
HSC Chemistry Option Module: Industrial Chemistry Summary are affected. In some areas, the calcium chloride is evaporated and placed in land-fill. This does not seem to have any effects, but future effects on water supplies are unknown. Heat (thermal) pollution As the Solvay process is exothermic and the precipitation of NaHCO3 requires cooling. This has traditionally been done by using river water as the coolant and simply discharging the hot water back into the waterway. This can have major effects on aquatic environments as oxygen is less soluble at higher temperatures. Modern plants use air-cooling towers and recycle the water instead. Solid waste The roasting of limestone in the kiln produces large amounts of silicates, clays as well as unburnt limestone. Traditionally, this waste was discharged into waterways where it caused problems by forming a sludge. This waste may be used for bricks or simply as landfill. Dust Control Dust is problematic and must be controlled by keeping vehicles in asphalt roadways, using wetting solution to suppress dust in exposed areas, bag filters in the hydrogen carbonate plant and installation of dust scrubbing systems. Noise suppression Noise is reduced by enclosing noisy areas, using silencers to dampen noise and community monitoring to identify noise sources. x
Perform a first-hand investigation to assess risk factors and then carry out a chemical step involved in the Solvay process identifying any difficulties associated with the laboratory modelling of the step
Investigation: Sodium carbonate production Aim: To model the production of sodium carbonate by the Solvay process Materials: -Large beaker -sodium carbonate (Na2CO3) -Sintered glass filter -dilute ammonia (NH4OH) -dry ice/bottled CO2/CO2 from conc. HCl and CaCO3 -Retort stand and ring -ice -Measuring cylinder -Bunsen burner -cork/stoppers -glass/rubber/plastic tubing Safety: clean up any spillage and ensure good ventilation (ammonia solution is corrosive). As gas is used, be careful of pressure accumulation. Dry ice ‘burns in contact with skin. Wear safety glasses at all times. Method: 1/ Add sodium chloride to an ammonia solution until solution is saturated 2/ Filter the solution to remove excess solid salt Robert Lee Chin
Page | 32
HSC Chemistry Option Module: Industrial Chemistry Summary 3/ 4/
Place ammonia/salt solution into measuring cylinder Place marble chips into a conical flask, add concentrated HCl and use the tubing to bubble CO2 into the ammonia-salt solution.
Marble chips and concentrated HCl
5/
Measuring cylinder with ammonia-salt solution
Allow the gas to bubble through the solution, but monitor gas inlet as Sodium hydrogen carbonate may block it
Beaker with ice to aid precipitation of NaHCO3
6/ 7/ 8/ 9/ 10/
Place the measuring cylinder into a large beaker and cool by filling with ice Use sintered glass filter to filter sodium hydrogen carbonate that precipitates in flask Place sodium hydrogen carbonate into a clean, dry test tube/crucible Use Bunsen burner to gently heat the solid until water evaporates Allow the product to cool
Results:
Marble chips and concentrated HCl
Questions: 1. Write a balanced equation for the reaction
Robert Lee Chin
Page | 33
HSC Chemistry Option Module: Industrial Chemistry Summary NaCl ( aq ) salt brine
NH 3( aq ) H 2 O ( l ) CO 2 ( g ) ammonia
NH 4 Cl ( aq )
o
NaHCO 3(s )
water carbon dioxide ammonium chloride sodium hydrogen carbonate
2. Explain why the gas inlet tube must be monitored NaHCO3 may precipitate at the gas inlet and block it, thus causing an accumulation of CO2 pressure in the flask containing acid and marble chips. As concentrated acid is used, this is a potential safety hazard. 3. Write a balanced equation for the conversion of the sodium hydrogen carbonate into sodium carbonate 2NaHCO 3(s) O 2(g) o H 2 O(l) CO 2(g) Na 2 CO 3(s) sodium hydrogen bicarbonate
oxygen
water
carbon dioxide sodium carbonate
4. Why must the sodium hydrogen carbonate be heated gently? Carbon dioxide gas is produced, which may cause the mixture to bubble violently 5. Calculate the yield of sodium hydrogen carbonate if 5g of salt was initially used in the reaction. 5 moles NaCl moles 58.44 5 ? moles NaHCO 3 moles 58.44 5 mass NaHCO 3 u 84.008 7.187543... # 7.19 g(2 s.f.) 58.44 6. If 4.5 g of sodium chloride produced 0.9 g of solid sodium carbonate, calculate the efficiency of the reaction
2 NaCl ( aq ) CaCO 3(s ) o CaCl 2 (s ) Na 2 CO 3(s ) 4.5 moles 58.44 1 § 4.5 · 4.5 ? moles NaHCO 3 moles ¨ ¸ 2 © 58.44 ¹ 116.88 4.5 mass Na 2 CO 3 u 84.008 3.23439...g 116.88 moles NaCl
effeciency (%)
x
actual mass Na 2 CO 3 u 100 calclulated mass Na 2 CO 3
27.82592... # 28% (2 s.f.)
Process information to solve problems and quantitatively analyse the relative quantities of reactants and products in each step of the process
Robert Lee Chin
Page | 34
HSC Chemistry Option Module: Industrial Chemistry Summary
x
Use available evidence to determine the criteria used to locate a chemical industry using the Solvay process as an example
Factors which influence the location of a Solvay plant x Proximity to raw materials
Example – The Solvay plant in Osborne, SA x
x x x
Proximity to market for product/transport links to market
x
x
Proximity to a population centre to provide a labour force and the community infrastructure Proximity to site for waste disposal
x
x
Robert Lee Chin
x x
Osborne is located close to the coast. This allows easy access to sea water, which can be pumped into ponds for concentration and purification. The hot climate of Adelaide also helps concentrate the brine A limestone deposit is ~50 km away, with a rail line allowing a deposit each day Osborne has excellent rail, road and harbour connections to supply sodium carbonate to glass makers throughout Australia Proximity to Adelaide means there is a reliable workforce, schools and shops for workers Previously discharged into Port river Now being used for landfill, brick manufacture
Page | 35
HSC Chemistry Option Module: Industrial Chemistry Summary
Solvay Process – Flow Chart Purified Brine 30% NaCl, H2O
Ammoniated brine
Ammonia NH3(g)
Limestone CaCO3 Lime Kiln
CaCO 3( s ) o
Carbonating tower NaCl ( aq ) NH 3( aq ) CO 2 ( g ) H 2 O ( l )
CaO ( s ) CO 2 ( g )
o NaHCO 3 ( s ) NH 4 Cl ( aq )
heat
H2O
CaO
Lime Slaker CaO 3( s ) H 2 O ( l ) eat ho Ca (OH) 2 ( aq )
Filter Heat
Product Na2CO3(s) Sodium carbonate
Robert Lee Chin
Ammonia recovery Ca (OH) 2 ( aq ) 2 NH 4 Cl ( aq ) o
CaCl 2 ( s ) 2H 2 O ( l ) 2 NH 3( g )
Waste product CaCl2
Page | 36
View more...
Comments
