Monitoring and Controlling Chemical Reactions (OCR)
A theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction.
It can be calculated using:
the balanced equation
the mass and relative formula mass of the limiting reactant
the relative formula mass of the product
Practice Question:
Lithium hydroxide is used to absorb exhaled carbon dioxide in spacecraft:
2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Calculate the maximum mass of water that can be made from an excess of carbon dioxide and 95.6 g of lithium hydroxide. (Relative atomic masses: H = 1.0, Li = 6.9, O = 16.0)
relative formula mass, Mr, of LiOH = 6.9 + 16.0 + 1.0 = 23.9
relative formula mass, Mr, of H2O = (2 × 1.0) + 16.0 = 18.0
Looking at the balanced equation:
sum of Mr for LiOH = (2 × 23.9) = 47.8
sum of Mr for H2O = 18.0
= 36.0g
An actual yield is the mass of a product actually obtained from the reaction. It is usually less than the theoretical yield. The reasons for this include:
incomplete reactions, in which some of the reactants do not react to form the product
practical losses during the experiment, such as during pouring or filtering
side reactions (unwanted reactions that compete with the desired reaction)
Calculating percentage yield
The percentage yield is calculated using this equation:
percentage yield = (actual yield)/(theoretical yield) * 100
Key fact
The percentage yield can vary from 100% (no product has been lost) to 0% (no product has been made)
Question:
Copper oxide reacts with sulfuric acid to make copper sulfate and water. In an experiment, 1.6 g of dry copper sulfate crystals are made. If the theoretical yield is 2.0 g, calculate the percentage yield of copper sulfate.
actual yield = 1.6 g
percentage yield = 1.6/2.0 × 100
percentage yield = 80%
No atoms are created or destroyed in a chemical reaction. However, the atoms in the reactants may not become the desired product. They may instead end up forming other products, which are regarded as waste products, also called by-products.
For example, hydrogen can be manufactured by reacting methane with steam:
methane + steam → hydrogen + carbon monoxide
CH4(g) + H2O(g) → 3H2(g) + CO(g)
In this reaction, carbon and oxygen atoms in the reactants do not form the useful product. Carbon monoxide is a waste gas.
The atom economy of a reaction is a measure of how many reactant atoms form a desired product.
Calculating atom economy
The atom economy of a reaction is calculated using this equation:
atom economy = (total M of the desired product)/(total M of all reactants) × 100
Key fact
The maximum atom economy possible for a reaction is 100%. This will be the case if there is only one product (the desired product) and no by-products.
Question:
Hydrogen can be manufactured by reacting methane with steam:
CH4(g) + H2O(g) → 3H2(g) + CO(g)
Calculate the atom economy for the reaction. (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
Mr of CH4 = 12.0 + (4 × 1.0) = 16.0
Mr of H2O = (2 × 1.0) + 16.0 = 18.0
total Mr of reactants = 16.0 + 18.0 = 34.0
Ar of H2 = (2 × 1.0) = 2.0
total Mr of desired product = 3 × 2.0 = 6.0 (there are three H2 in the balanced equation)
atom economy = 17.6% (to 3 significant figures)
It is important in this core practical to use appropriate apparatus to make and record a range of volume measurements accurately. This includes the safe use and handling of liquids, and monitoring chemical changes.
This outlines one way to carry out the practical. Eye protection must be worn.
Aims
To carry out an accurate titration using dilute hydrochloric acid, dilute sodium hydroxide solution, and phenolphthalein indicator.
Method
Use a pipette and pipette filler to add 25 cm3 of dilute sodium hydroxide solution to a clean conical flask.
Add a few drops of phenolphthalein indicator and put the conical flask on a white tile.
Fill the burette with dilute hydrochloric acid and note the starting volume.
Slowly add the acid from the burette to the conical flask, swirling to mix.
Stop adding the acid when the end-point is reached (when the colour first permanently changes from pink to colourless). Note the final volume reading.
Repeat steps 1 to 5 until you get concordant titres (see the Analysis).
Results
Record the results in a suitable table. The one here also shows some sample readings.
Run | Rough | 1 | 2 | 3 |
End reading (cm3) | 25.45 | 24.80 | 47.90 | 23.70 |
Start reading (cm3) | 0.00 | 1.00 | 23.80 | 0.00 |
Titre (cm3) | 25.45 | 23.80 ✓ | 24.10 | 23.70 ✓ |
Readings should be recorded to two decimal places, ending in 0 or 5 (where the liquid level is between two graduations on the burette). The titre is the volume added (the difference between the end and start readings).
Analysis
Tick (✓) at least two concordant titres. These are titres within 0.20 cm3 (or sometimes 0.10 cm3) of each other.
The concentration of a solution can be measured in g/dm3 or in mol/dm3.
Volume units
Volumes used in concentration calculations must be in dm3, not in cm3. It is useful to know that 1 dm3 = 1,000 cm3. This means:
divide by 1,000 to convert from cm3 to dm3
multiply by 1,000 to convert from dm3 to cm3
For example, 500 cm3 is 0.5 dm3 (500 ÷ 1,000). It is often easiest to convert from cm3 to dm3 before continuing with a concentration calculation.
The relative formula mass, Mr, of the solute is used to convert between mol/dm3 and g/dm3:
To convert from mol/dm3 to g/dm3, multiply by the Mr.
Example
Calculate the concentration of 0.100 mol/dm3 sodium hydroxide solution in g/dm3. (Relative formula mass: NaOH = 40.0)
concentration = 0.100 × 40.0
= 4.00 g/dm3
The rate of a reaction is a measure of how quickly a reactant is used up or a product is formed.
Collision theory
For a chemical reaction to happen:
reactant particles must collide with each other
the particles must have enough energy for them to react
A collision that produces a reaction is called a successful collision. The activation energy is the minimum amount of energy needed by particles for a collision to be successful. It is different for different reactions.
Measuring rates of reaction
There are different ways to determine the rate of a reaction. The method chosen usually depends on the reactants and products involved, and how easy it is to measure changes in them.
In addition, the rate of reaction determines how long a reaction is observed. Reactions can vary from being almost instant to taking years to complete. In the lab, reactions are usually followed over a few seconds or minutes.
The greater the rate or frequency of successful collisions, the greater the rate of reaction. If the concentration of a reacting solution or the pressure of a reacting gas is increased:
the reactant particles become more crowded
the frequency of collisions between reactant particles increases
the rate of reaction increases
Note that the mean energy of the particles does not change. However, since the frequency of collisions increases, the frequency of successful collisions also increases.
Graphs
The rates of two or more reactions can be compared using a graph of mass or volume of product formed against time. The graph shows this for two reactions.
Comparing reactions of different concentration and pressure
The gradient of the line is equal to the rate of reaction. The faster reaction at the higher concentration or pressure:
gives a steeper line
finishes sooner
Key fact
Make sure you answer questions in terms of frequency or rate of successful collisions, rather than the number of successful collisions. This is because, given enough time, even a slow reaction will have a large number of collisions.
Dividing lumps
For a given mass of a solid, large lumps have smaller surface area to volume ratios than smaller lumps or powders. If a large lump is divided or ground into a powder:
its total volume stays the same
the area of exposed surface increases
the surface area to volume ratio increases
Lumps vs powders
The greater the rate or frequency of successful collisions, the greater the rate of reaction. If the surface area to volume ratio of a reacting solid is increased:
more reactant particles are exposed at the surface
the frequency of successful collisions between reactant particles increases
the rate of reaction increases
Note that the mean energy of the particles does not change. However, since the frequency of collisions increases, the frequency of successful collisions also increases.
Graphs
The rates of two or more reactions can be compared using a graph of mass or volume of product formed against time. The graph shows this for two reactions.
Comparing reactions of different surface area
The gradient of the line is equal to the rate of reaction. The faster reaction with the powder:
gives a steeper line
finishes sooner
Key fact
Make sure you answer questions in terms of surface area to volume ratio, rather than just surface area. This is because the surface area also depends on the mass of solid reactant used.
In principle, all chemical reactions are reversible reactions. The products could be changed back into the original reactants using a suitable reaction. This is not obvious when a reaction ‘goes to completion’, a situation in which very little or no reactants are left. Examples of reactions that go to completion are:
complete combustion of a fuel
many precipitation reactions
reactions in which a product escapes, usually a gas
It is more obvious in reactions that do not go to completion that the reaction is reversible. The reaction mixture may contain reactants and products, and their proportions may be changed by altering the reaction conditions.
Two examples
Ammonium chloride
Ammonium chloride is a white solid. It breaks down when heated, forming ammonia and hydrogen chloride. When these two gases are cool enough, they react together to form ammonium chloride again. This reversible reaction can be modelled as:
ammonium chloride ⇌ ammonia + hydrogen chloride
NH4Cl(s) ⇌ NH3(g) + HCl(g)
The symbol ⇌ has two half arrowheads, one pointing in each direction. It is used in equations that model reversible reactions:
the forward reaction is the one that goes to the right
the backward reaction is the one that goes to the left
Copper sulfate
Blue copper sulfate is described as hydrated. The copper ions in its crystal lattice structure are surrounded by water molecules. This water is driven off when blue hydrated copper sulfate is heated, leaving white anhydrous copper sulfate. This reaction is reversible:
hydrated copper sulfate ⇌ anhydrous copper sulfate + water
CuSO4.5H2O(s) ⇌ CuSO4(s) + 5H2O(l)
Dynamic equilibrium
When a reversible reaction happens in a closed system, such as a stoppered flask, it reaches a dynamic equilibrium. At equilibrium:
the forward and backward reactions are still happening
the forward and backward reactions have an equal rate of reaction
the concentrations of all the reacting substances remain constant and do not change
The equilibrium position of a reversible reaction is a measure of the concentrations of the reacting substances at equilibrium. Using the Haber process, which makes ammonia, as an example:
nitrogen + hydrogen ⇌ ammonia
N2(g) + 3H2(g) ⇌ 2NH3(g)
The equilibrium position is:
to the left if the concentrations of N2 and H2 are greater than the concentration of NH3
to the right if the concentration of NH3 is greater than the concentrations of N2 and H2
The equilibrium position can be changed by changing the reaction conditions by:
changing the pressure
changing the concentration
changing the temperature
Changing the pressure
If the pressure is increased in a reaction involving gases, the equilibrium position moves in the direction of the fewest molecules of gas.
There are fewer molecules on the right hand side of the equation for the Haber process:
N2(g) + 3H2(g) ⇌ 2NH3(g)
1 + 3 = 4 molecules / 2 molecules
If the pressure is increased, the equilibrium position moves to the right.
Changing the concentration
If the concentration of a solute is increased in a reaction involving solutions, the equilibrium position moves in the direction away from this solute. For example, bismuth chloride reacts with water in a reversible reaction:
BiCl3(aq) + H2O(l) ⇌ BiOCl(s) + 2HCl(aq)
The concentration of hydrochloric acid can be increased by adding more hydrochloric acid. When this happens, the equilibrium position moves to the left, away from HCl(aq) in the equation.
Changing the temperature
In a reversible reaction, if the reaction is exothermic in one direction, it is endothermic in the other direction.
If the temperature is increased, the equilibrium position moves in the direction of the endothermic process. For example, sulfur dioxide reacts with oxygen in a reversible reaction:
2SO2(g) + O2(g) ⇌ 2SO3(g) (forward reaction is exothermic)
If the forward reaction is exothermic, the backward reaction must be endothermic. Therefore, if the temperature is increased, the equilibrium position moves to the left.
A theoretical yield is the maximum possible mass of a product that can be made in a chemical reaction.
It can be calculated using:
the balanced equation
the mass and relative formula mass of the limiting reactant
the relative formula mass of the product
Practice Question:
Lithium hydroxide is used to absorb exhaled carbon dioxide in spacecraft:
2LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Calculate the maximum mass of water that can be made from an excess of carbon dioxide and 95.6 g of lithium hydroxide. (Relative atomic masses: H = 1.0, Li = 6.9, O = 16.0)
relative formula mass, Mr, of LiOH = 6.9 + 16.0 + 1.0 = 23.9
relative formula mass, Mr, of H2O = (2 × 1.0) + 16.0 = 18.0
Looking at the balanced equation:
sum of Mr for LiOH = (2 × 23.9) = 47.8
sum of Mr for H2O = 18.0
= 36.0g
An actual yield is the mass of a product actually obtained from the reaction. It is usually less than the theoretical yield. The reasons for this include:
incomplete reactions, in which some of the reactants do not react to form the product
practical losses during the experiment, such as during pouring or filtering
side reactions (unwanted reactions that compete with the desired reaction)
Calculating percentage yield
The percentage yield is calculated using this equation:
percentage yield = (actual yield)/(theoretical yield) * 100
Key fact
The percentage yield can vary from 100% (no product has been lost) to 0% (no product has been made)
Question:
Copper oxide reacts with sulfuric acid to make copper sulfate and water. In an experiment, 1.6 g of dry copper sulfate crystals are made. If the theoretical yield is 2.0 g, calculate the percentage yield of copper sulfate.
actual yield = 1.6 g
percentage yield = 1.6/2.0 × 100
percentage yield = 80%
No atoms are created or destroyed in a chemical reaction. However, the atoms in the reactants may not become the desired product. They may instead end up forming other products, which are regarded as waste products, also called by-products.
For example, hydrogen can be manufactured by reacting methane with steam:
methane + steam → hydrogen + carbon monoxide
CH4(g) + H2O(g) → 3H2(g) + CO(g)
In this reaction, carbon and oxygen atoms in the reactants do not form the useful product. Carbon monoxide is a waste gas.
The atom economy of a reaction is a measure of how many reactant atoms form a desired product.
Calculating atom economy
The atom economy of a reaction is calculated using this equation:
atom economy = (total M of the desired product)/(total M of all reactants) × 100
Key fact
The maximum atom economy possible for a reaction is 100%. This will be the case if there is only one product (the desired product) and no by-products.
Question:
Hydrogen can be manufactured by reacting methane with steam:
CH4(g) + H2O(g) → 3H2(g) + CO(g)
Calculate the atom economy for the reaction. (Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
Mr of CH4 = 12.0 + (4 × 1.0) = 16.0
Mr of H2O = (2 × 1.0) + 16.0 = 18.0
total Mr of reactants = 16.0 + 18.0 = 34.0
Ar of H2 = (2 × 1.0) = 2.0
total Mr of desired product = 3 × 2.0 = 6.0 (there are three H2 in the balanced equation)
atom economy = 17.6% (to 3 significant figures)
It is important in this core practical to use appropriate apparatus to make and record a range of volume measurements accurately. This includes the safe use and handling of liquids, and monitoring chemical changes.
This outlines one way to carry out the practical. Eye protection must be worn.
Aims
To carry out an accurate titration using dilute hydrochloric acid, dilute sodium hydroxide solution, and phenolphthalein indicator.
Method
Use a pipette and pipette filler to add 25 cm3 of dilute sodium hydroxide solution to a clean conical flask.
Add a few drops of phenolphthalein indicator and put the conical flask on a white tile.
Fill the burette with dilute hydrochloric acid and note the starting volume.
Slowly add the acid from the burette to the conical flask, swirling to mix.
Stop adding the acid when the end-point is reached (when the colour first permanently changes from pink to colourless). Note the final volume reading.
Repeat steps 1 to 5 until you get concordant titres (see the Analysis).
Results
Record the results in a suitable table. The one here also shows some sample readings.
Run | Rough | 1 | 2 | 3 |
End reading (cm3) | 25.45 | 24.80 | 47.90 | 23.70 |
Start reading (cm3) | 0.00 | 1.00 | 23.80 | 0.00 |
Titre (cm3) | 25.45 | 23.80 ✓ | 24.10 | 23.70 ✓ |
Readings should be recorded to two decimal places, ending in 0 or 5 (where the liquid level is between two graduations on the burette). The titre is the volume added (the difference between the end and start readings).
Analysis
Tick (✓) at least two concordant titres. These are titres within 0.20 cm3 (or sometimes 0.10 cm3) of each other.
The concentration of a solution can be measured in g/dm3 or in mol/dm3.
Volume units
Volumes used in concentration calculations must be in dm3, not in cm3. It is useful to know that 1 dm3 = 1,000 cm3. This means:
divide by 1,000 to convert from cm3 to dm3
multiply by 1,000 to convert from dm3 to cm3
For example, 500 cm3 is 0.5 dm3 (500 ÷ 1,000). It is often easiest to convert from cm3 to dm3 before continuing with a concentration calculation.
The relative formula mass, Mr, of the solute is used to convert between mol/dm3 and g/dm3:
To convert from mol/dm3 to g/dm3, multiply by the Mr.
Example
Calculate the concentration of 0.100 mol/dm3 sodium hydroxide solution in g/dm3. (Relative formula mass: NaOH = 40.0)
concentration = 0.100 × 40.0
= 4.00 g/dm3
The rate of a reaction is a measure of how quickly a reactant is used up or a product is formed.
Collision theory
For a chemical reaction to happen:
reactant particles must collide with each other
the particles must have enough energy for them to react
A collision that produces a reaction is called a successful collision. The activation energy is the minimum amount of energy needed by particles for a collision to be successful. It is different for different reactions.
Measuring rates of reaction
There are different ways to determine the rate of a reaction. The method chosen usually depends on the reactants and products involved, and how easy it is to measure changes in them.
In addition, the rate of reaction determines how long a reaction is observed. Reactions can vary from being almost instant to taking years to complete. In the lab, reactions are usually followed over a few seconds or minutes.
The greater the rate or frequency of successful collisions, the greater the rate of reaction. If the concentration of a reacting solution or the pressure of a reacting gas is increased:
the reactant particles become more crowded
the frequency of collisions between reactant particles increases
the rate of reaction increases
Note that the mean energy of the particles does not change. However, since the frequency of collisions increases, the frequency of successful collisions also increases.
Graphs
The rates of two or more reactions can be compared using a graph of mass or volume of product formed against time. The graph shows this for two reactions.
Comparing reactions of different concentration and pressure
The gradient of the line is equal to the rate of reaction. The faster reaction at the higher concentration or pressure:
gives a steeper line
finishes sooner
Key fact
Make sure you answer questions in terms of frequency or rate of successful collisions, rather than the number of successful collisions. This is because, given enough time, even a slow reaction will have a large number of collisions.
Dividing lumps
For a given mass of a solid, large lumps have smaller surface area to volume ratios than smaller lumps or powders. If a large lump is divided or ground into a powder:
its total volume stays the same
the area of exposed surface increases
the surface area to volume ratio increases
Lumps vs powders
The greater the rate or frequency of successful collisions, the greater the rate of reaction. If the surface area to volume ratio of a reacting solid is increased:
more reactant particles are exposed at the surface
the frequency of successful collisions between reactant particles increases
the rate of reaction increases
Note that the mean energy of the particles does not change. However, since the frequency of collisions increases, the frequency of successful collisions also increases.
Graphs
The rates of two or more reactions can be compared using a graph of mass or volume of product formed against time. The graph shows this for two reactions.
Comparing reactions of different surface area
The gradient of the line is equal to the rate of reaction. The faster reaction with the powder:
gives a steeper line
finishes sooner
Key fact
Make sure you answer questions in terms of surface area to volume ratio, rather than just surface area. This is because the surface area also depends on the mass of solid reactant used.
In principle, all chemical reactions are reversible reactions. The products could be changed back into the original reactants using a suitable reaction. This is not obvious when a reaction ‘goes to completion’, a situation in which very little or no reactants are left. Examples of reactions that go to completion are:
complete combustion of a fuel
many precipitation reactions
reactions in which a product escapes, usually a gas
It is more obvious in reactions that do not go to completion that the reaction is reversible. The reaction mixture may contain reactants and products, and their proportions may be changed by altering the reaction conditions.
Two examples
Ammonium chloride
Ammonium chloride is a white solid. It breaks down when heated, forming ammonia and hydrogen chloride. When these two gases are cool enough, they react together to form ammonium chloride again. This reversible reaction can be modelled as:
ammonium chloride ⇌ ammonia + hydrogen chloride
NH4Cl(s) ⇌ NH3(g) + HCl(g)
The symbol ⇌ has two half arrowheads, one pointing in each direction. It is used in equations that model reversible reactions:
the forward reaction is the one that goes to the right
the backward reaction is the one that goes to the left
Copper sulfate
Blue copper sulfate is described as hydrated. The copper ions in its crystal lattice structure are surrounded by water molecules. This water is driven off when blue hydrated copper sulfate is heated, leaving white anhydrous copper sulfate. This reaction is reversible:
hydrated copper sulfate ⇌ anhydrous copper sulfate + water
CuSO4.5H2O(s) ⇌ CuSO4(s) + 5H2O(l)
Dynamic equilibrium
When a reversible reaction happens in a closed system, such as a stoppered flask, it reaches a dynamic equilibrium. At equilibrium:
the forward and backward reactions are still happening
the forward and backward reactions have an equal rate of reaction
the concentrations of all the reacting substances remain constant and do not change
The equilibrium position of a reversible reaction is a measure of the concentrations of the reacting substances at equilibrium. Using the Haber process, which makes ammonia, as an example:
nitrogen + hydrogen ⇌ ammonia
N2(g) + 3H2(g) ⇌ 2NH3(g)
The equilibrium position is:
to the left if the concentrations of N2 and H2 are greater than the concentration of NH3
to the right if the concentration of NH3 is greater than the concentrations of N2 and H2
The equilibrium position can be changed by changing the reaction conditions by:
changing the pressure
changing the concentration
changing the temperature
Changing the pressure
If the pressure is increased in a reaction involving gases, the equilibrium position moves in the direction of the fewest molecules of gas.
There are fewer molecules on the right hand side of the equation for the Haber process:
N2(g) + 3H2(g) ⇌ 2NH3(g)
1 + 3 = 4 molecules / 2 molecules
If the pressure is increased, the equilibrium position moves to the right.
Changing the concentration
If the concentration of a solute is increased in a reaction involving solutions, the equilibrium position moves in the direction away from this solute. For example, bismuth chloride reacts with water in a reversible reaction:
BiCl3(aq) + H2O(l) ⇌ BiOCl(s) + 2HCl(aq)
The concentration of hydrochloric acid can be increased by adding more hydrochloric acid. When this happens, the equilibrium position moves to the left, away from HCl(aq) in the equation.
Changing the temperature
In a reversible reaction, if the reaction is exothermic in one direction, it is endothermic in the other direction.
If the temperature is increased, the equilibrium position moves in the direction of the endothermic process. For example, sulfur dioxide reacts with oxygen in a reversible reaction:
2SO2(g) + O2(g) ⇌ 2SO3(g) (forward reaction is exothermic)
If the forward reaction is exothermic, the backward reaction must be endothermic. Therefore, if the temperature is increased, the equilibrium position moves to the left.
