Getting the Most From Reactants

Getting the Most From Reactants
18/11/2018

Calculation of the mass
18/11/2018 Calculation of the mass I am able to use balanced equations to work out mole ratios of reactants and products. I can use balanced equations and formula mass to work out the mass of product from reactant and vice versa.

Calculations based on Equations
Balance Equation Underline Mole Ratio

N5 22.2g CaCO3(s) + 2HCl(aq)  CaCl 2 (aq) + CO 2(g) + H 2 O(l)
Worked example 1. The equation below shows the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq)  CaCl 2 (aq) CO 2(g) H 2 O(l) 20g of calcium carbonate reacts with excess hydrochloric acid. Calculate the mass of calcium chloride formed. N5 CaCO3(s) + 2HCl(aq)  CaCl 2 (aq) CO 2(g) H 2 O(l) 100g g 20g 22.2g

What mass of water is produced when 160 g of methane burns completely in air.
CH O2  CO H2O Calculate the mass of carbon dioxide produced when 5 g of calcium carbonate reacts with an excess of dilute hydrochloric acid. Calculate the mass of magnesium oxide produced when 100 g of magnesium carbonate is decomposed completely by heating. Ammonia and phosphoric acid can be reacted together to make the fertiliser ammonium phosphate: 3NH3 + H3PO4  (NH4)3PO4 What mass of ammonium phosphate fertiliser could be made from kg of ammonia?

Calculation of the mass
18/11/2018 I am able to use balanced equations to work out mole ratios of reactants and products. I can use balanced equations and formula mass to work out the mass of product from reactant and vice versa.

Calculations involving solutions.
18/11/2018 Calculations involving solutions. I can give the units for concentration I can work out quantities of reactants and/or products using one or more of the following: Balanced equations. Concentrations and volumes of solutions. Masses of solutes.

The graph below was obtained when 1
The graph below was obtained when 1.0g of powdered zinc was added to excess hydrochloric acid 1.0 mol l-1, copy the graph and sketch a line to show what you would expect if the reaction was repeated using a) 2.0 mol l-1 HCl and 1.0g Zn b) 1.0 mol l-1 HCl and 0.75g Zn Vol H2 cm3 time /s 8

Calculations involving concentration
n = C x V n=number of moles C=concentration mol l-1 V=volume in Litres

Standard solutions Standard solution is a solution of accurately known concentration. They are made in volumetric glassware and can be made from a solid or from another solution.

n = C x V 1 mole  123.5 g n = 0.1 x 100 0.01 moles  =(0.01/1)x123.5
What mass of solid should be dissolved in 100cm3 of water to prepare a 0.1 mol l-1 solution of copper(II) sulphate? n = C x V 1000 n = 0.1 x 100 n= 0.01 moles 1 mole  g 0.01 moles  =(0.01/1)x123.5 =1.235g

Making Standard Solutions
1) weigh out the solid 2) dissolve the solid in a small amount of solvent. 3) Swirl the flask and add washings. 4) Finally, fill the solvent to the line 18/11/2018

Calculations involving concentration
How many moles are in 100cm3 of sodium hydroxide concentration 0.1 mol l-1? 0.01 moles

Neutralisation Reactions/titrations
What volume hydrochloric acid 1.0 mol l-1 is needed to neutralise 50 cm3 of potassium hydroxide solution concentration 0.25 mol l-1? 12.5cm3

Calculations for you to try.
Calculate the concentration of potassium hydroxide (KOH ) if 14.8 cm3 is required to neutralise 20cm3 of 0.1 mol/l nitric acid (HNO3). Calculate the volume of 0.15 mol/l sulphuric acid(H2SO4) if it is neutralised by 25 cm3 of 0.25mol/l sodium hydroxide (NaOH). 0.135 mol/l 20.83 cm3

Calculations involving solutions.
I can give the units for concentration I can work out quantities of reactants and/or products using one or more of the following: Balanced equations. Concentrations and volumes of solutions. Masses of solutes.

Excess 18/11/2018 I can explain and identify the excess reactant(s).
I can use a balanced equation to work out the reactant in excess and therefore the limiting reactant, for a chemical reaction.

Calculations based on equations
2CO(g) + O2(g)  2CO2(g) What mass of carbon dioxide would be obtained by burning 56g of carbon monoxide? CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) What mass of water would be produced if 32g of methane was burnt? MnO2(s) + 4HCl(aq)  Mnl2(aq) + 2H2O(l) + Cl2(g) What mass of chlorine would be produced if 1000kg of manganese oxide reacted completely with hydrochloric acid? 88g 72g 817kg 18/11/2018

12 slices would be used and 4 would be left over.
Excess 2 Bread + 1 Ham  1 Sandwich 16 Bread + 6 Ham  6 Sandwiches ? The Bread is in EXCESS. The Ham is the Limiting Factor 12 slices would be used and 4 would be left over.

Excess As soon as one of the reactants in a chemical reaction is used up the reaction stops. Any other reactant which is left over is said to be ‘in excess’. The reactant which is used up determines the mass of product formed. 18/11/2018

Excess: Worked example
Which reactant is in excess when 10g of calcium carbonate reacts with 100cm3 of 1 mol l-1 hydrochloric acid? Write the balanced equation for the reaction and show mole ratio:- CaCO3 + 2HCl  CaCl CO H2O 1 mol mol n C V(l) Calculate the number of moles of each reactant:- Number of moles in 10g of CaCO3 = 10/100 = 0.1 Number of moles of HCl = 1 x 0.1 = 0.1 m n gfm From equation 0.1 mol of CaCO3 needs 0.2 mol of HCl and as we only have 0.1 mol of HCl the CaCO3 is in excess.

HCl is in excess by 0.1 moles
Worked example 1.2g of magnesium was added to 100cm3 2 mol l-1 hydrochloric acid. Calculate the reagent in excess? Mg HCl  MgCl H2 n = m/gfm = 1.2 / 24.3 = 0.05moles n = c x v = 2 x 0.1 = 0.2moles 1 2 0.05 0.1moles HCl is in excess by 0.1 moles

CaCO3 is in excess by 0.05 moles
Worked example. Which reactant is in excess when 10g of calcium carbonate reacts with 100cm3 of 1 mol l-1 hydrochloric acid? CaCO HCl  CaCl CO H2O n = m/gfm = 10 / 100 = 0.1moles n = c x v = 1 x 0.1 = 0.1moles 1 2 0.1 0.2moles CaCO3 is in excess by 0.05 moles

EXCESS CALCULATIONS For each of the following reactions calculate which reagent is in excess? 4.86g magnesium added to 250cm3 2 mol l-1 hydrochloric acid 2.7g aluminium added to 200cm3 1 mol l-1 hydrochloric acid 2.43g magnesium added to 200cm3 1 mol l-1 sulphuric acid 3.27g zinc added to 100cm3 0.2 mol l-1 hydrochloric acid. HCl Al H2SO4 Zn 24

H2SO4 (aq) + 2 NaOH(s) Na2SO4 (aq) + 2 H2O(l)
Calculating Mass of a Product 10g of sodium hydroxide pellets were reacted with 50ml of 5 mol l-1 sulfuric acid. Calculate the MASS of solid sodium sulfate would be obtained, if the product mixture was evaporated to dryness. H2SO4 (aq) NaOH(s) Na2SO4 (aq) H2O(l) n = c x v = 5 / 0.05 = 0.25moles n = m / gfm = 10 / 40 = 0.25moles 1 2 0.25 0.5moles H2SO4 is in excess m = n x gfm = x 142 = 17.75g 1 2 0.25 0.125moles

Calculations for you to try
What mass of calcium oxide is formed when 0.4 g of calcium reacts with 0.05 mole of oxygen? 2Ca O2  2CaO 2 mol mol mol n = m/gfm = 0.4 / 40 = 0.01 moles n = 0.05moles 2 1 0.01 0.005moles O2 is in excess. 2 2 m = n x gfm = 0.01 x 56 = 0.56g 0.01 0.01 moles

n = c x v = 2 x 0.025 n = m/gfm = 3.27 / 65.4 = 0.05moles
What mass of hydrogen is formed when 3.27g of zinc is reacted with 25cm3 of 2 mol l-1 hydrochloric acid? Zn HCl  ZnCl H2 1 mol mol 1mol n = c x v = 2 x 0.025 = 0.05moles n = m/gfm = 3.27 / 65.4 = 0.05moles 1 2 0.1moles 0.05 zinc is in excess. 2 1 m = n x gfm = x 2 = 0.05g 0.05 0.025 moles

AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq)
A student added 0.20g of silver nitrate, AgNO3, to 25 cm3 of water. This solution was then added to 20cm3 of mol l-1 hydrochloric acid. The equation for the reaction is: AgNO3(aq) HCl(aq)  AgCl(s) HNO3(aq) Calculate the mass of silver chloride formed. A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing g of silver nitrate. The reaction that occurs is Zn(s) + 2AgNO3(aq)  2Ag(s) + Zn(NO3)2(aq) Calculate the mass of silver formed.

Excess I can explain and identify the excess reactant(s).
I can use a balanced equation to work out the reactant in excess and therefore the limiting reactant, for a chemical reaction. Industrial Processes

Design of an industrial process
18/11/2018 I can state what industrial processes are designed to maximise and minimise. I can give the 7 factors that influence the design of an industrial process. I know what environmental issues need to be considered when designing a chemical process.

Excess Calculations 1 2 3 0.80g 3.2g 0.57L 18/11/2018

The Chemical Industry Task: In your group discuss the following questions: What? What kinds of chemicals do you think are manufactured in the UK? Where? Where do you think chemical manufacture and industry is located in the UK? What other factors? What needs to take into account when designing and running an industrial chemistry site?

Key Words Bingo Draw a bingo grid on a bit of paper. Choose any six of the words below and put one in each box. Listen carefully through the lesson and score the words off as they are said / appear on the slides. Good luck! raw materials Scaling Up feedstocks Fixed costs consumer products Variable costs Batch process labour intensive Continuous process capital intensive Capital costs Economy Energy costs Pollution Pilot Study

The Chemical Industry The chemical industry is one of the largest industries in the UK. It’s products are vital to many aspects of modern life and many are used for the benefit of society. The chemical industry involves the investment of large sums of money but employs relatively few people making it a capital intensive and not a labour intensive industry. The basis of the chemical industry is to make a profit by changing raw materials into feedstocks and consumer products.

Top 5 categories of ‘chemicals’ made
Basic inorganics and fertilisers Dyestuffs, paint and pigments Petrochemicals and polymers Pharmaceuticals Specialities (e.g. explosives)

Original substances as obtained from the earth in their “raw” form.
Raw Materials Original substances as obtained from the earth in their “raw” form. Fossil fuels Metal Ores Minerals Air Sulphur Sea salt Water

Feedstocks A substance or a mixture of substances from which other chemicals can be extracted or synthesised. Ethene Synthesis Gas (hydrogen + carbon monoxide) Copper, Zinc, Silver (from their oxides) Sulphuric Acid (from burning sulphur)

Products for the “consumer” synthesised from Feedstocks.
Consumer Products Products for the “consumer” synthesised from Feedstocks. Paints Detergents Fertillisers Food Additives Textiles Pesticides Cosmetics Solvents Plastics Acids Alkalis Medicines

Manufacturing Processes and Routes
A chemical manufacturing process usually involves a sequence of steps which start with the raw materials and end with consumer products. Raw material Feedstock Chemical processes and reactions Consumer product

Stages in Development of New Products
Production Running of full scale production plants Review Evaluation and improvements Pilot Study Find out new possible routes and processes of manufacture Research Find out new structure properties and method of preparation Scaling Up Best conditions for full scale-up yields etc

Batch and Continuous Processing
There are two main types of chemical processing. Batch and continuous. In batch processing the chemicals are loaded into the reaction vessel and the reaction is monitored. At the end of the reaction the product is collected and the reaction vessel is cleaned out ready for the next batch. In continuous processing the reactants are continuously added at one end of the reaction vessel and the products are removed at the other end. Each process has advantages and disadvantages.

Batch Processes Advantages
suited to smaller scale production (up to 100 tons per year) more versatile than continuous as they can be used for more than one reaction more suited for multi step reactions or when reaction time is long Disadvantages possibility of contamination from one batch to the next filling and emptying takes time during which no product, and hence no money, is being made safety – relatively large amounts of reactants may not be controllable in the event of an exothermic reaction going wrong. An example of a batch process is used in the making of pharmaceuticals

Continuous Process Advantages
suited to large scale production (>1000 tons per year) suitable for fast single step processes more easily automated using computer control smaller workforce operates round the clock 365 days per year tend to operate with relatively low volumes of reactants allowing easy removal of excess heat energy Disadvantages very much higher capital cost before any production can occur not versatile, can make only one product not cost effective when run below full capacity Some examples of a continuous process is the making of sulphuric acid, iron and poly(ethene) and ammonia.

Types of Process The choice is usually decided by the scale of production required. Batch Process Small amounts of product, e.g. Pharmaceuticals, cosmetics Raw materials are mixed together in a reactor vessel and the product separated from the mixture when complete. The plant is flexible and can be used to make several different products. But usually more labour intensive Continuous Process Large amounts of product, e.g. ammonia, sulphuric acid Raw materials are fed in at one end and the product is formed continually 24hrs a day Less labour intensive and more efficient. But the plant can only make 1 product and is difficult to shut down for maintenance.

Notes: Make your own note. It must include the following information:
What is meant by a batch process ? What is meant by a continuous process ? Give two advantages associated with each type of operation. Give one disadvantage associated with each type of operation. Give two chemicals manufactured by each type of operation.

Manufacturing Costs The main Costs in the chemical industry can be categorised under 3 headings: Capital costs – initial expenses to set up the process, e.g. purchase of land, buildings, machinery etc. Fixed costs – constant and do not change with the level of production, e.g. depreciation of the plant, maintenance, R&D and staff wages. Variable costs – increase as production increases, e.g raw materials, distribution and energy. The chemical industry is generally capital intensive vs labour intensive (expensive to set up but require few operators to maintain production).

Energy Costs Energy costs can be high therefore efficient use of energy is vital to the economic success of the Chemical Industry. Saving Energy Used heat from exothermic reactions elsewhere in the plant Use “waste” heat to generate electricity Sell energy to local area Pollution – low energy wastage is always desirable.

Manufacturing Routes and Conditions
Routes and conditions are chosen to maximise economic efficiency and minimise environmental impact. Factors influencing this are: Cost, availability and suitability of feedstock(s) Yield of products By-products – any economic use? Can reactants or by-products be recycled? Energy efficiency – catalyst availability/cost? Safety and environmental issues

Questions to Consider What are the costs of the different suitable feedstocks? What yield of product can we expect to obtain? Is a catalyst available and how much does it cost? Do the by products have any economic use Can by-products be recycled? Will Batch or Continuous processes be most suitable? Health and Safety?

Location of the Chemical Industry
The site of an industrial plant is chosen carefully with the following criteria in mind: Availability of raw materials Transport to and from the plant Skilled labour resource Historical reasons Pollution effects

7 principles of design process
Availability of feedstocks Cost of the feedstock Sustainability of the feedstock Opportunities for recycling Energy requirements Marketability of by-products Product yield (or atom economy)

Past Paper Questions

Design of an industrial process
I can state what industrial processes are designed to maximise and minimise. I can give the 7 factors that influence the design of an industrial process. I know what environmental issues need to be considered when designing a chemical process.

18/11/2018 Percentage Yield I understand what is meant by “percentage yield” and can give the formula. I can calculate percentage yield from balanced equations and masses of reactants and products. I can use given costs, and percentage yields to work out the cost of the feedstock(s) to produce a given mass of product.

Percentage Yield In chemical reactions we very rarely (if ever) get the total quantity of product that we can calculate from the chemical equation. The reasons for this can be: at the end of the reaction there may be reactant left unconverted to product some reactant may be converted into a by-product the isolation of the product may be difficult It is therefore important to know just how much product you can expect in comparison to the theoretical one. This is known as the percentage yield and is calculated in the following way: 18/11/2018

100 tonnes of limestone should produce 56 tonnes of quicklime.
Limestone (calcium carbonate) is used to make quicklime (calcium oxide) for cement making. CaCO  CaO CO2 Mole ratio: 1 mole mole mole GFM: g g g Theoretically: 100 tonnes of limestone should produce 56 tonnes of quicklime. But the Actual Yield is only 48 tonnes So the percentage yield is only 48 / 56 x 100 = 85.7%

1 C(s) + 2H2(g)  CH4 (g) AY 1mole 2 mole 1 mole 12g  16g TY
In the reaction below, 16g of methane was produced when 24g of carbon burned. What is the percentage yield? C(s) H2(g)  CH4 (g) 1mole mole mole 12g  g 24g  g TY % Yield = AY / TY x 100 = (16/ 32) x 100 = 50%

2 SO2 (g) + ½ O2 (g) → SO3(g) AY SO2 (g) + ½ O2 (g) → SO3(g)
Sulphur dioxide reacts with oxygen to form sulphur trioxide. SO2 (g) + ½ O2 (g) → SO3(g) Over a period of time, 3205g of SO2 is introduced to the reaction chamber with an excess of O g of SO3 is produced in the process. Calculate the percentage yield of the reaction. AY SO2 (g) + ½ O2 (g) → SO3(g) 1mole mole mole TY 64g  g 3205g  g % Yield = (2403/ ) x 100 = %

C3H7OH(l) + CH3COOH(l)  C3H7OOCCH3(l) + H2O(l)
%Y If the percentage yield of an esterification reaction is 65%, what mass of propyl ethanoate will be produced from 100g of propan-1-ol C3H7OH(l) CH3COOH(l)  C3H7OOCCH3(l) + H2O(l) 1 mole mole mole mole 60g  g 100g  g TY Actual Yield = (%Y x TY )/100 = (65 x 170) x 100 = 110.5g

4 %Y C3H7OH(l) + CH3COOH(l)  C3H7OOCCH3(l) + H2O(l) 588.23 g
If the percentage yield of an esterification reaction is 65%, what mass of reactants are needed to produce 1kg of ester? C3H7OH(l) CH3COOH(l)  C3H7OOCCH3(l) + H2O(l) 1 mole mole mole mole 60g  g  g g If %Y was 100% Mass of reactants required at 65% Yield = ( / 65) x 100 = 904.9g

2. Ethene reacts with steam to ethanol C2H4 + H2O  C2H5OH
1. In a reaction 320 kg of sulphur dioxide reacted with oxygen to form 360kg of sulphur trioxide. Calculate the percentage yield. 2. Ethene reacts with steam to ethanol C2H H2O  C2H5OH Calculate the percentage yield of ethanol when 140g of ethene reacts to form 170 g of ethanol. tonnes of hydrogen reacts with excess nitrogen to form tonnes of ammonia. N H2  2NH3 Calculate the percentage yield of ammonia. 18/11/2018

Calculations for you to try: ANSWERS
1. In a reaction 320 kg of sulphur dioxide reacted with oxygen to form 360kg of sulphur trioxide. Calculate the percentage yield. 2SO O2  2SO3 2 mol  mol 128.2 g  g 128.2 kg  kg 320 kg  kg percentage yield = x = % 360 399.9

2. Ethene reacts with steam to ethanol C2H H2O  C2H5OH Calculate the percentage yield of ethanol when 140g of ethene reacts to form 170 g of ethanol. C2H H2O  C2H5OH 1 mol  1 mol 28 g  46 g 140 g  /28 x 46 g  g percentage yield = (170 / 230) x = %

tonnes of hydrogen reacts with excess nitrogen to form tonnes of ammonia. N H2  2NH3 Calculate the percentage yield of ammonia. N H2  2NH3 3 mol  2 mol 6 g  34 g 6 tonnes  34 tonnes 0.5 tonnes  2.83 tonnes percentage yield = (0.66 / 2.83) x = %

Percentage Yield I understand what is meant by “percentage yield” and can give the formula. I can calculate percentage yield from balanced equations and masses of reactants and products. I can use given costs, and percentage yields to work out the cost of the feedstock(s) to produce a given mass of product.

Atom Economy 18/11/2018 I understand what is meant by “atom economy”.
I can calculate the atom economy of a reaction using the correct formula. I understand why some reactions that have a high percentage yield may have a low atom economy. I can relate percentage yield and atom economy to different routes taken in manufacturing products.

Atom Economy The higher the % yield the more efficient a reaction is.
However a more modern assessment of how environmentally friendly your reaction is, looks at the atom economy. This is a measure of what percentage of your reactants get turned into useful products. The better the atom economy the less unwanted by-products are formed and hence less of your reactants are wasted. 18/11/2018

Atom Economy In an ideal reaction, all reactant atoms end up within the useful product molecule. No waste is produced! Inefficient, wasteful reactions have low atom economy. Efficient processes have high atom economy and are important for sustainable development. They conserve natural resources and create less waste. Modern chemists design reactions with the highest possible atom economy in order to minimise environmental impact. A reaction may have a high percentage yield but a low percentage atom economy, or vice versa.

High atom economy All reactant atoms included in the desired product.

Low atom economy Some reactant atoms not included in the desired product.

Catalysts Catalysts have a crucial role in improving atom economy.
They allow the development of new reactions requiring fewer starting materials and producing fewer waste products. Catalysts can be recovered and re-used and allow reactions to run at lower temperatures, cutting energy requirements.

1 What is the percentage atom economy for the following reaction for making hydrogen by reacting coal with steam? C(s) H2O(g) → CO2(g) H2(g) 12 g (2 + 16) g [12 + (2 × 16)] g (2 × 1) g 12 g g g g Total mass of reactants Mass of desired product = = 48 g = 4 g This reaction route has a very low atom economy and is an inefficient method of producing hydrogen. % AE = 4 / 48 × 100 = 8.3%

2 = [(6 × 12) + (12 × 1)] = [(6 × 12) + (12 × 1)] = 84 g = 84 g
Calculate the percentage atom economy for the reaction below. C6H C6H12 Total mass of reactants Mass of desired product = [(6 × 12) + (12 × 1)] = [(6 × 12) + (12 × 1)] = 84 g = 84 g % AE = (84 / 84)× 100 = 100% This reaction route has a very high atom economy as all reactant atoms are incorporated into the desired product.

3 Hydrazine (N2H4) is used for rocket fuel. Calculate the atom economy for hydrazine production. Total mass of reactants Mass of desired product = = g = 32 g 1 mol 18 g 2 mol 34 g 1 mol 74.5 g 1 mol 32 g 1 mol 58 g % AE = (32 / 108.5) x 100 = 30% This reaction route has an atom economy of 30%. The remaining 70% is waste product (NaCl and H2O).

Atom Economy I understand what is meant by “atom economy”.into the desired product. I can calculate the atom economy of a reaction using the correct formula.tal mass of I understand why some reactions that have a high percentage yield may have a low atom economy. I can relate percentage yield and atom economy to different routes taken in manufacturing products.

NH4Cl(s) + NaOH(s) NH3(g) + NaCl(aq) + H2O(l)
Question Ammonia can be made in two ways: Method 1: NH4Cl(s) NaOH(s) NH3(g) NaCl(aq) H2O(l) In an industrial unit it was found that 1000kg of ammonium chloride made 150kg of ammonia. Method 2: N2(g) + 3H2(g)  2NH3(g) In an industrial unit it was found that 1000kg of nitrogen made 150kg of ammonia. With reference to both % yield and atom economy comment on which process is the most efficient. 18/11/2018

% Yield: NH4Cl(s) + NaOH(s)  NH3(g) + NaCl(aq) + H2O(l) 1mole 1 mole
1mole mole 53.5g  g 1000kg  kg % Yield = (150/ 318) x 100 = 47 % N2(g) + 3H2(g)  2NH3(g) 1mole mole 28g  g 1000kg  kg % Yield = (150/ 1214) x 100 = 12 % 18/11/2018

% Atom Economy: Method 2 is actually much more efficient as all of the atoms from the reactants are converted into the desired product and there is no waste produced at all. 18/11/2018

1. CH3OH + CH3COOH  CH3COOCH3 + H2O methanol ethanoic acid methylethanoate In preparation, 60g of methylethanoate is made from 20g of methanol. a. Calculate the % yield. b. Calculate the atom economy for this reaction. 2. Nitrous oxide NO can be made by the catalytic oxidation of ammonia: 2NH3(g) + 2.5O2(g)  2NO(g) + 3H2O(g) a. What is the % yield of nitrous oxide if 340g of ammonia gas produces 500g of nitrous oxide? b. Calculate the atom economy for this reaction. 77% 92.5% 83% 2.6% 18/11/2018

18/11/2018 Molar Volume I can calculate the volume of a gas from the number of moles and vice versa. I can calculate the volumes of reactant and product gases from the number of moles of each reactant and product.

Volume and Concentration
0.5moll-1 5cm3 0.83moll-1 18/11/2018

Calculations based on equations
N2(g) + 2O2(g)  2NO2(g) What mass of nitrogen would be required to produce 460kg of nitrogen dioxide? Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) In a blast furnace 6 x 106 kg of iron are produced every day, what mass of iron oxide is required to make this? 3CuO(s) + 2NH3(g)  3Cu(s) + N2(g) + 3H2O(g) What mass of copper can be produced using 4 x 105 kg of copper oxide in this reaction? 140kg 8.6x106kg 3.2x106kg 18/11/2018

Molar volume When we are thinking about the mole we are usually concerned with the mass of one mole. When we think about gases, however, it is more appropriate to think about the volume they occupy. Molar volume is simply the volume of one mole of a gas at a set temperature and pressure. How could we find out the volume of one mole of a gas?

Molar Volume by Experiment
We can carry out an experiment to find the volume of 1 mole of a gas. We will need to record the following measurements: Mass of flask + gas = _________ g Mass of empty flask = _________ g Mass of gas = _________ g Volume of flask = _________ L

Calculation

Errors The Molar volume of nitrogen is given as 22.4L at stp (standard temperature and pressure) Obviously our experiment has some flaws…

Calculate the molar volume of the gases below
Mass of flask + oxygen = g Mass of empty flask = g Mass of oxygen = Volume of flask = 575 ml Mass of flask + hydrogen = g Mass of empty flask = g Mass of hydrogen = Volume of flask = ml

Different gases 1 mole H2  22.2 L 1 mole CH4  22.5 L
If the experiment was repeated using different gases the volume of 1 mole of each gas works out to be roughly the same. All of the following molar volumes were measured at stp: 1 mole H2  22.2 L 1 mole CH4  22.5 L 1 mole O2  22.4 L

Why? It may seem surprising that the molar volume is the same for all gases, even at the same temperature and pressure. In a gas the molecules have much more kinetic energy and are relatively far apart so the volume of the gas does not depend on the sizes of the particles. In a gas the at room temperature and pressure the molecules only occupy about 0.1% of the volume of a gas. The rest is empty space!

“Standard Temperature & Pressure”
The volume of a gas will change under different temperatures and pressures. It is therefore very important when we talk about the volume of a gas to state the temperature and pressure of the gas. STP – standard temperature (0OC) and 1atm

Measuring Molar gas volume
Expt: 3.2 Finding out the molar volume by experiment. Method 2

Calculate the molar gas volume: ____ g magnesium has produced ____cm3 hydrogen ____ g ↔ ________ cm3 24 g ↔ _________ The volume of one mole of H2 gas is ________ Litres

Calculations involving molar gas volume
At room temperature and pressure the molar volume of a gas is 24 litres mol-1. Example 1. Calculate the volume of moles of oxygen. 1 mole  24 litres 0.025 moles  x 24 / 1 = litres Example 2. Calculate the number of moles in 72 litres of hydrogen. 1 mole  litres  litres 72 x 1 / 24 = 3 moles

Calculations involving mass and volume
Worked example 1.The equation below shows the reaction between calcium carbonate and hydrochloric acid. CaCO3(s) + 2HCl(aq)  CaCl2(aq) CO2(g) H2O(l) 20g of calcium carbonate reacts with excess hydrochloric acid. Calculate the volume of carbon dioxide gas formed. (Take the molar volume to be 23.0 litre mol-1) Balanced equation CaCO3(s) + 2HCl(aq)  CaCl2 (aq) CO2(g) +H2O(l) Underline 1 mol  mol Show mole ratio Change moles into required units 100 g  litres 20 g  /100 x 23.0 litre 4.6 litres Cross multiply

Calculations involving molar gas volume
What is the mass of steam in 180 cm3 of the gas, when the molar volume is 24 litres mol-1? 0.135 g

Calculations for you to try:
Calculate the number of moles in 0.36 litres of argon (molar gas volume = 24 litres mol-1 ). Calculate the volume of 0.04 moles of CO2. (molar gas volume = 24 litres mol-1 ). Under certain conditions oxygen has a density of 1.44 g l-1. Calculate the molar volume of oxygen under these conditions. A gas has a density of 2.74 g l-1 and a molar volume of 23.4 litre mol-1. Calculate the molecular mass of the gas. 0.015moles 0.96L 22.22 litres 64.1 g

18/11/2018

Density and Molar Volume
REMEMBER: Molar volume is the volume occupied by one mole of a gas. Worked example 1. In an experiment the density of carbon dioxide was measured and found to be 1.85 g l-1. Calculate the molar volume of carbon dioxide. 1.85 g  1 litre So 1 mole, g  44/1.85 x 1 = litres Worked example 2. A gas has a molar volume of 24 litres and a density of 1.25 g l -1. Calculate the mass of 1 mole of the gas. 1 litre  g So 1 mole, litre  24/1 x = 30 g

Calculations for you to try.
Under certain conditions oxygen has a density of 1.44 gl-1. Calculate the molar volume of oxygen under these conditions. 1.44g occupies 1 litre. 1 mole of O2 = 32g So 32g occupies x 1 / litres = litres A gas has a density of 2.74 gl-1 and a molar volume of 23.4 l mol-1. Calculate the molecular mass of the gas. 1 litre weighs g 23.4 litres weighs x 2.74 / 1 = g

Use the data below to calculate the molecular mass of gas X.
Mass of empty bottle = g Mass of bottle filled with gas X = g Capacity of bottle = 500 cm3 Molar volume of gas X = 23.8 litres Volume of gas in litres = 500/ = 0.5 litres Mass of gas X in bottle = – g = g 0.5 litres of gas has a mass of 0.336g So 1mole, 23.8 litres has a mass of 23.8/0.5 x g = g

Reactions involving just gases
If a reaction just involves gases then it is possible to just compare the reaction ratios and volumes without calculating the actual numbers of moles. Since one mole of any gas occupies the same volume under the same conditions of temperature and pressure we can use the balanced equation to calculate the volume of gases. 2NO(g) H2(g)  N2(g) H2O(g) 2 moles 2 moles mole moles 2 volumes 2 volumes volumes volumes 18/11/2018

Calculate the volume of nitrogen dioxide gas produced when 100cm3 of nitrogen is sparked in excess oxygen N2 (g) + 2O2 (g)  2NO2 (g) 1mole moles 1 volume 2 volumes 100 cm cm3 200cm3 of NO2 107

What volume of C02, is produced if 100 cm3 of O2 is used to completely to burn some CH4 gas?
CH4 (g) O2 (g)  CO2 (g) H20 (l) 50cm3 of CO2 108

What is the volume and composition of the resultant gas mixture when 10 litres of sulfur dioxide reacts with 50 litres of oxygen gas to form sulfur trioxide? 2SO2(g) + O2(g)  2SO3(g) Mole ratio: 2 : 1 : 2 Start: (what you start with) Used/made: (use ratio to work it out) End: (what you end up with) Resultant gas composition and volume = 45 litres of un-reacted O2(g) and 10 litres of SO3(g) 18/11/2018

50 cm3 of C0, is burned with 20cm3 of oxygen.
Which gas is in excess What is the volume and composition of the resulting gas mixture CO (g) + ½ O2 (g)  CO2 (g) CO in excess 10cm3 CO + 40cm3 CO2 = 50cm3 of gas 110

Calculate the volume and the composition of the resulting gas mixture.
10 cm3 of propane gas is mixed with 75cm3 of O2 and the mixture exploded. Calculate the volume and the composition of the resulting gas mixture. b) What will be the change in volume when the resulting gas mixture is shaken with dilute sodium hydroxide solution. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H20 (l) 25cm3 of O2 + 30cm3 of CO2 = 55cm3 of gas Volume will decrease by 30cm3 (CO2) removed 111

Nitrogen monoxide reacts with oxygen to form nitrogen dioxide
Nitrogen monoxide reacts with oxygen to form nitrogen dioxide. What is the volume composition of the gases present when 40cm3 of nitrogen monoxide reacts with 100 cm3 of oxygen? = 40cm3 of NO2 (g) + 80cm3 of unreacted O2 40cm3 of propane is burned in 250cm3 of oxygen. Calculate the volume and composition of the resulting gas mixture. (All measurements are made at room temperature and pressure). = 120cm3 of CO2 (g) + 50cm3 of unreacted O2 200 litres of carbon monoxide is reacted with 500 litres of hydrogen to form gaseous methanol, CH3OH. Calculate the volume and composition of the resulting gas mixture. (All measurements carried out at 200oC) = 200 litres of CH3OH(g) litres of H2 left unreacted

NH W 7

NH W 7 gfm = molar volume 64g = L 88.5L 118g =

2014 W14

W 7 = gfm = molar volume 222g = L 1g = 0.973L

Molar Volume I can calculate the volume of a gas from the number of moles and vice versa. Give the units for molar volume. I can calculate the volumes of reactant and product gases from the number of moles of each reactant and product. Dynamic Equilibrium

Dynamic Equilibrium 18/11/2018
I understand what is meant by “dynamic equilibrium”.. I know what happens to reaction rates at equilibrium. I know what happens to concentrations of products and reactants at equilibrium.

Molar Volumes 450cm3 O2 & 400 cm3 CO2 50 cm3 x = 3, y = 6 18/11/2018

Reversible Reactions

Reversible Reaction 2 Experiment 1.
The conversion of hydrated cobalt chloride into anhydrous cobalt chloride is an example of a reversible reaction, i.e. the reaction can go forwards or backwards. Experiment 2. When heated, ammonium chloride decomposes: NH4Cl(s)  NH3(g) HCl(g) On cooling, the ammonia gas and hydrogen chloride gas react to reform ammonium chloride. NH3(g) HCl(g)  NH4Cl(s) The thermal decomposition of ammonium chloride is another example of a reversible reaction.

• What colour is hydrated cobalt(II) chloride
• What colour is hydrated cobalt(II) chloride? • When it is heated, what colour change takes place? • What is the blue substance formed? • Write an equation for this reaction. • When water is added to anhydrous cobalt(II) chloride, what colour change takes place? • What substance has been reformed? • What is meant by a reversible reaction? • Describe what is observed when ammonium chloride is heated. • Explain why this happens. • Write equations for the forward and reverse reactions. 18/11/2018

What does the Arrow tell you?
REACTANTS PRODUCTS

REACTANTS PRODUCTS The RATE of the forward reaction is EQUAL to the RATE of the reverse reaction

REACTANTS PRODUCTS The RATE of the forward reaction is LESS than the RATE of the reverse reaction

REACTANTS PRODUCTS The RATE of the forward reaction is GREATER than the RATE of the reverse reaction

Most chemical reactions can be considered to be REVERSIBLE (where the products are able to change back into the reactants). How easy the reaction is to reverse will depend on the Activation Energy (EA) needed. In reversible reactions, a 100% yield is never going to be possible, as the reverse reaction begins as soon as product has been made.

Equilibrium Equilibrium is a relationship between the reactants and products. It is talked about as having a ‘position’, and this ‘position’ can move to the (i) LEFT = Reactant side (the concentration of reactant increases) Or (ii) RIGHT = Product side (the concentration of product increases)

Rate of the FORWARD REACTION Rate of the REVERSE REACTION =
Dynamic Equilibrium The state of equilibrium is reached when the: Rate of the FORWARD REACTION Rate of the REVERSE REACTION = It is known as a dynamic equilibrium as the reactions don’t stop.

Dynamic Equilibrium Graph 1 Reaction Forward reaction rate equilibrium
Time Forward reaction Reverse reaction Graph 1

Concentrations of REACTANTS and PRODUCTS are CONSTANT
Dynamic Equilibrium At equilibrium: Concentrations of REACTANTS and PRODUCTS are CONSTANT These two definitions must BOTH be met for an equilibrium to exist. Not necessarily EQUAL

Dynamic Equilibrium

Closed System An equilibrium can only ever exist in a CLOSED
SYSTEM (where the reactants and products cannot escape!) This would happen inside a reaction vessel / tank with a closed lid.

Equilibrium on a Graph Notice that both graphs finish at the same concentrations – it doesn’t matter if you start with the Reactants or the Products (the final ratio is the same for each).

18/11/2018 Dynamic Equilibrium I understand what is meant by “dynamic equilibrium”. I know what happens to reaction rates at equilibrium. I know what happens to concentrations of products and reactants at equilibrium. Le Chatelier’s Principle

Le Chatelier’s Principle
18/11/2018 Le Chatelier’s Principle I can state Le Chatelier’s Principle I can explain using Le Chatelier’s Principle, the likely effect on the equilibrium position of changing:- Pressure Concentration of reactants or product Temperature Catalyst I can relate the effects of temperature, pressure, concentration of reactants and products to industrial processes including the Haber Process.

Factors That Change Equilibrium
The position of equilibrium can be changed by altering the conditions of the reaction. Conditions that will have an effect are: (i) Temperature of the reaction mixture (ii) Pressure of the reaction vessel (iii) Concentration of reactants/products

Enthalpy – Reminder! Remember that: EXOTHERMIC = NEGATIVE enthalpy
Temperature INCREASES ENDOTHERMIC = POSITIVE enthalpy Temperature DECREASES

Temperature & Equilibrium
For any equilibrium questions about temperature, the question will tell you whether the FORWARD reaction is endo- or exothermic. The negative value for ∆H tells us that the forward reaction (left to right) is EXOTHERMIC. (So the REVERSE reaction will be ENDOTHERMIC). CO(g) H2(g) CH3OH(g) ∆H = - 91kJmol-1

Increasing Temperature
If the temperature is increased, the equilibrium will move to decrease the temperature. So it will move in the ENDOTHERMIC direction.

Decreasing Temperature
If the reaction mixture is cooled down, the equilibrium will move to heat it back up. The equilibrium will move in the EXOTHERMIC direction.

Example 2NO2(g) N2O4(aq) ∆H = -ve
What would happen to the equilibrium position if the temperature was increased? 2NO2(g) N2O4(aq) ∆H = -ve The forward reaction is EXOTHERMIC (gives out heat). If an equilibrium is heated, it tries to cool down, so it moves in the ENDOTHERMIC direction. So this equilibrium will move LEFT.

Example 2NO2(g) N2O4(aq) ∆H = -ve
What would happen to the equilibrium position if the temperature was decreased? 2NO2(g) N2O4(aq) ∆H = -ve The forward reaction is EXOTHERMIC (gives out heat). If an equilibrium is cooled it FAVOURS this reaction and so it moves in the EXOTHERMIC direction. So this equilibrium will move RIGHT.

Pressure & Equilibrium
A change in pressure will ONLY affect equilibria that contain GASES. The number of moles of gas directly relates to the pressure in the equilibrium. More moles of gas = More gas particles = More pressure

Increasing Pressure If the pressure is increased, the equilibrium will
move to decrease the pressure. It will move in the direction with LEAST MOLES OF GAS.

Decreasing Pressure If the pressure is decreased, the equilibrium will
move to try and increase the pressure. It will move in the direction with MOST MOLES OF GAS.

RECAP DECREASE INCREASE TEMPERATURE PRESSURE EXOTHERMIC ENDOTHERMIC
EXO / ENDO PRESSURE MOST LESS LESS / MOST Moles of gas

Example The forward reaction is EXOTHERMIC (gives out heat).
What would you do to the temperature to obtain more product? ∆H= -157 kJ mol-1 The forward reaction is EXOTHERMIC (gives out heat). We want the equilibrium to move to the RIGHT. If an equilibrium is cooled, it tries to heat up, so it moves in the EXOTHERMIC direction.

Example The forward reaction is ENDOTHERMIC (takes in heat).
What would you do to the temperature to obtain more reactants? ∆H= -157 kJ mol-1 The forward reaction is ENDOTHERMIC (takes in heat). We want the equilibrium to move to the LEFT. If an equilibrium is HEATED, it tries to cool down, so it moves in the ENDOTHERMIC direction.

Example There are 2 moles of gas on the left, and 1 mole of gas
What would we do to the pressure to obtain more product?? ∆H= -157 kJ mol-1 There are 2 moles of gas on the left, and 1 mole of gas on the right. WE want the equilibrium to shift to the right Less moles= less particles = less pressure An increase in pressure will move this equilibrium RIGHT.

Example There are 2 moles of gas on the left, and 1 mole of gas
What would we do to the pressure to obtain more Reactant?? ∆H= -157 kJ mol-1 There are 2 moles of gas on the left, and 1 mole of gas on the right. WE want the equilibrium to shift to the LEFT More moles= more particles = more pressure A decrease in pressure will move this equilibrium LEFT.

Starter N2(g) + 3H2(g) 2NH3(g) ∆H = -ve
The reaction below has reached equilibrium. N2(g) H2(g) 2NH3(g) ∆H = -ve Changes to the equilibrium can increase the volume of NH3 produced. List as many changes as you can, that will move the equilibrium to the RIGHT!

Catalysts Recap The use of a catalyst will speed up chemical reactions
Remain unchanged/ not used up in reaction Lowers the activation Energy for reactions Heterogeneous/ Homogeneous

Catalysts & Equilibrium
The use of a catalyst will speed up BOTH the forward and reverse reactions, but it will have no effect on the concentrations of reactant and product Catalysts have NO EFFECT on the position of equilibrium, but their use allow it to be achieved sooner.

Le Chatelier’s Principle
18/11/2018 I can state Le Chatelier’s Principle I can explain using Le Chatelier’s Principle, the likely effect on the equilibrium position of changing:- Pressure Concentration of reactants or product Temperature Catalyst I can relate the effects of temperature, pressure, concentration of reactants and products to industrial processes including the Haber Process. Percentage Yield

Getting the Most From Reactants

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