Chemistry Module 3.2

Question 1

What happens to bonds in reactants when during a chemical reaction?

Answer 1

When chemical reactions happen, some bonds are broken and some bonds are made. Usually this’ll cause a change in energy.

Question 2

Define enthalpy change.

Answer 2

Enthalpy change, ∆H (delta H), is the heat energy transferred in a reaction at a constant pressure. The units of ∆H are kJ mol^-1.

Question 3

How should one write the symbol for enthalpy when the measurements were made under standard conditions and the elements were at their standard states?

Answer 3

One should write ‘∆H°’ to show that the measurement s were maded under standard conditions and that the elements were in their standard states. Standard conditions are 100kPa of pressure and temperature of 298K.

Question 4

What do exothermic reactions do? Give two examples.

Answer 4

Exothermic reactions give out energy. ∆H is negative.

In exothermic reactions, the temperature often goes up.

Oxidation is usually exothermic. Here are two examples:

• The combustion of a fuel like methane:
  
  • CH_{4 (g)} + 2O_{2 (g)} → CO_{2 (g)} + 2H₂O _(l)
  • This reaction has an ∆_rHº = -890kJ mol^-1 exothermic.
• The oxidation of carbohydrates, such as glucose in respiration.

Question 5

What do endothermic reactions do? Give two examples.

Answer 5

Exothermic reactions absorb energy ∆H is positive.

In these reactions, temperature falls.

• The thermal decomposition of calcium carbonate is endothermic:
  
  • CaCO_{3 (s)} → CaO _(s) + CO_{2 (g)}
  • ∆_rHº = +178kJ mol^-1 endothermic.
• The main reactions of photosynthesis are also endothermic - sunlight supplies the energy.

Question 6

What do enthalpy profile diagrams show?

Answer 6

Enthalpy profile diagrams show you how the enthalpy (energy) changes during reactions.

Question 7

What is the activation energy?

Answer 7

The activation energy ,E_a, is the minimum amount of energy needed to begin breaking reactant bonds and start a chemical reaction.

Question 8

What can you tell from a substance by its enthalpy?

Answer 8

The less enthalpy a substance has, the more stable it is.

Question 9

Why must one specify if the change in enthalpy was found in standard conditions?

Answer 9

This is because changes in enthalpy are affected by temperature and pressure - using standard conditions means that everyone can know exactly what the enthalpy change is describing.

Question 10

List the four different types of enthalpy change.

Answer 10

1. Standard enthalpy change of reaction, ∆_rHº, is the enthalpy change when the reaction occurs in molar quantities shown in the chemical equation, under standard conditions.
2. Standard enthalpy change of formation, ∆_fH°, is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states, under standard conditions.
3. Standard enthalpy change of combustion, ∆_cH°, is the enthalpy change when 1 mole of a substance is completely burned in oxygen, under standard conditions.
4. Standard enthalpy change of netralisation, ∆_neutH°, is the enthalpy change when an acid and alkali react together, under standard conditions, to form 1 mole of water.

Question 11

Is breaking a bond between a molecule an endothermic or exothermic process?

Answer 11

You need energy to break bonds, so breaking bonds is endothermic.

Question 12

Is forming a bond in a molecule an endothermic or exothermic process?

Answer 12

Energy is released when bonds are formed, so this is exothermic.

Question 13

What is enthalpy change for a reaction?

Answer 13

The enthalpy change for a reaction is the overall effect of these two changes. If you need more energy to break the bonds than is released when the bonds are made, ∆H is positive and the reaction is endothermic. If you need less ∆H is negative and the reaction is exothermic.

Question 14

Explain how ionic bonding works.

Answer 14

In ionic bonding, positive and negative ions are attracted to each other.

Question 15

Explain how covalent bonding works.

Answer 15

In covalent molecules, the positive nuclei are attracted to the negative charge of the shared electrons in a covalent bond.

Question 16

What do you need to break the attraction between atoms in molecules?

Answer 16

You need energy to break this attraction - stronger bonds take more energy to break. The amount of energy you need per mole is the bond dissociation enthalpy.

Question 17

What do Bond dissociation enthalpies always involve?

Answer 17

Bond dissociation enthalpies always involve bond breaking in gaseous compounds - this makes comparisons fair.

Question 18

What is the average bond enthalpy?

Answer 18

The energy needed to break one mole of bonds in the gas phase, averaged over many different compounds.

Question 19

How does one find the average bond enthalpy of a molecule?

Answer 19

Find the bond enthalpies of all the bonds in the molecule and divide them by how many there are.

Question 20

Give a method to find out enthalpy changes in the lab.

Answer 20

To measure the enthalpy change for a reaction you only need to know two things -

• the number of moles of the reactant
• the change in temperature

How you go about the experiment depends on what type of reaction it is.

• To find the enthalpy of combustion of a flammable liquid, you burn it.
• As the fuel burns, it heats the water. You can work out the heat absorbed by the water if you know the mass of the water, the change in temperature of the water (∆T), and the specific heat capacity of water (4.18 J/g/K).
• Ideally all the heat given out by the fuel as it burns would be absorbed by the water - allowing you to work out the enthalpy change of combsution. In practice though, you always lose some heat (as you heat the appuratus and the surroundings).

Similar methods to this (also known as calorimetry) can also be used to calculate an enthalpy change for a reaction that happens in solution, such as neutralisation or displacement. For a neutralisation reaction, combine known quantities of acid and alkali in an insulated container, and measure the temperature change. The heat given out can be calculated using the formula q=mc∆T.

Question 21

Define each symbol in the formula:

q = mc∆T

Answer 21

q = heat loss or gained (in joules). This is the same as the enthalpy change if the pressure is constant.

m = mass of water in the calorimeter or solution in the insulated container (in grams).

c = specific heat capacity of water (4.18 J/g/K)

∆T = the change in temperature of the water or solution in (K)

Question 22

Give the formula to find the change in enthalpy.

Answer 22

^-q/_n = ∆H

∆H = change in enthalpy

q = heat lost or gained in joules

n = number of moles

Question 23

What does Hess’ law state?

Answer 23

Hess’ law states that the total enthalpy change of a reaction is always the same, no matter which route is taken.

∆_rH + sum of ∆_fH (reactants) = sum of ∆_fH (products)

Question 24

What is required of the particles in order for a reaction to take place?

Answer 24

• The particles must collide in the right direction. They need to be facing each other the right way.
• They collide with at least a certain minimum amount of kinetic energy.

Question 25

What is the minimum amount of kinetic energy particles need to react also called?

Answer 25

The minimum amount of kinetic energy particles need to react is called the activation energy. The particles need this much energy to break the bonds to start the reaction.

Reactions with a low activation energy often happen pretty easily. But reactions with high activation energies don’t. You need to give the particles extra energy by heating them.

Question 26

What happens when you increase the temperature of particles?

Answer 26

If you increase the temperature, the particles will (on average) have more kinetic energy and will move faster.

So a greater proportion of molecules will have at least the activation energy and be able to react. This changes the shape of the Boltzmann distribution curve - it pushes over to the right.

Because the molecules are moving around faster, they’ll collide more often. This is another reason why increasing the temperature makes a reaction faster.

Question 27

What is the effect of increasing the concentration of reactants on the rate of reaction?

Answer 27

If you increase the concentration of reactants in a solution, the particles will be closer together, on average. If they’re closer, they’ll collide more frequently. If there are more collisions, they’ll have more chances to react.

Question 28

What is the effect of increasing the pressure of the reactants on the rate of reaction?

Answer 28

If any of your reactants are gases, increasing the pressure will increase the rate of reaction. It’s similar to increasing the concentration of a solution - at higher pressures, the particles will be closer together, increasing the chance of successful collisions.

Question 29

How does the presence of a catalyst affect the rate of reaction?

Answer 29

Catalysts are very useful. They lower the activation energy by providing a different way for the bonds to be broken and remade. If the activation energy’s lower, more particles will have enough energy to react.

Question 30

What does a catalyst do?

Answer 30

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. This catalyst is chemically unchanged at the end of the reaction.

Question 31

What’s a benefit of using a catalyst?

Answer 31

Catalysts don’t get used up in reactions, so you only need a small amount of catalyst to catalyse a huge amount of reactant. They do take part in reactions but they’re remade at the end.

Question 32

What is a drawback of using a catalyst in a reaction?

Answer 32

Catalysts are very specific about which reactions they catalyse. Many will only work on a single reaction.

An example of this is iron. It’s used in the Haber process to make ammonia.

Fe(s)

N_2(g) + 3H_2(g) ⇔ 2NH_3(g)

Question 33

What is a heterogeneous catalyst?

Answer 33

A heterogeneous catalyst is one in a different phase from the reactants - i.e. in a different physical state. For example, in the Haber process, gases are passed over a solid iron catalyst.

The reaction happens on the surface of the heterogeneous catalyst. So increasing the surface area of the catalyst increases the number of molecules that can react at the same time, increasing the rate of reaction.

Question 34

What is a homogeneous catalyst?

Answer 34

Homogeneous catalysts are in the same physical state as the reactants. Usually a homogeneous catalyst is an aqueous catalyst for a reaction between two aqeous solutions.

A homogeneous catalyst works by forming an intermediate species. The reactants combine with the catalyst to make an intermediate species, which then reacts to form the products and reform the catalyst.

Question 35

Give an examples of how catalysts can be good for industry.

Answer 35

Iron is used as a catalyst in ammonia production. If it wasn’t for the catalyst, the temperature would have to be raised loads to make the reaction happen quick enough. Not only would this be bad for fuel consumption levels but it’d also reduce the amount of ammonia produced.

Using a catalyst can change the properties of a product to make it more useful e.g. poly(ethene).

When polyethene is made without a catalyst, it’s less dense and less rigid. When polyethene is made with a catalyst it’s more dense, more rigid and a higher melting point.

Question 36

How are catalysts good for the environment?

Answer 36

• Lower temperaturtes and oressures can be used so energy is saved and less carbon dioxide is released and fossil fuels are preserved. Catalysts can also reduce waste by allowing a different reaction to be used with a better atom economy.
• Catalytic converters on cars are made from alloys of platinum, palladium and rhodium. They reduce the pollution released into the atmosphere by speeding up the reaction, 2CO + 2NO → 2CO₂ + N_​2

Question 37

Define the term ‘rate of reaction’ and give the formula for it.

Answer 37

The rate of reaction is the rate at which a product is formed or a reactant is used up.

Rate of reaction = ^{amount of reactant used or product formed}/_time

Question 38

How would one investigate reaction rates using the change in mass?

Answer 38

1. When the product is a gas, its formation can be measured using a mass balance.
2. The amount of product formed is the mass disappearing from the container.
3. When the reaction starts, you should start a stop clock or timer. Then take mass measurements at regular time intervals.
4. Make a table with a column for ‘time’ and a column for ‘mass’ and fill it in as the reaction goes on.
5. You’ll know the reaction is finished when the reading on the mass balance stops decreasing.
6. This method is very accurate and easy to use but does release gas into the room, which could be dangerous if the gas is toxic or flammable.

Question 39

How would one investigate the reaction rate with the volume of gas given off?

Answer 39

1. You can use a gas syringe to emasure the volume of product formed.
2. The experiment is carried out the same way as above but you measure the volume of gas in the syringe rather than the mass from the balance.
3. This method is accurate but vigorous reactions can blow the plunger out of the syringe.

Question 40

How could one work out a reaction rate from the gradient of a graph?

Answer 40

1. Draw a line of best fit through the data points.
2. Pick two points on the line that are easy to read.
3. Then draw a vertical line down from one point and a horizontal line across from the other to make a triangle.
4. Calculate the change in x and y of the triangle.
5. Divide the change in y by the change in x to find the gradient.
6. Now you have a reaction rate.

Question 41

How could one work out the gradient from a curved graph?

Answer 41

1. Find the point on the curve that you need to look at. I.e. the question may ask you for the rate of reaction at 3 minutes, so find 3 on the axis and go up the curve from there.
2. Place a ruler at that point so that it’s touching the curve. Position the ruler so that you can see the whole curve.
3. Adjust the ruler until the space between the ruler and the curve is equal on both sides of the point
4. Draw a line across the ruler to make a tangent. Extend the line across the graph - it’ll help make your gradient calculation easier as you’ll have more points to choose from.
5. Calculate the gradient of the tangent to find the rate:

gradient = change in y/change in x

gradient = reaction rate

Question 42

What does it mean for a chemical reaction to be reversible?

Answer 42

If a chemical reaction is reversible, it means they go both ways. To show a reaction’s reversible, you stick in a ⇌ symbol.

H_2(g) + I_2(g) ⇌ 2HI_(g)

This reaction can go in either direction -

forwards H_2(g) + I_2(g) ⇌ 2HI_(g)

or backlwards -

2HI_(g) ⇌ I_{2(g) +} H_2(g)

Question 43

How can a reversible reaction reach dynamic equilibrium?

Answer 43

As the reactants get used up, the forwards reaction slows down - and as more product is formed, the reverse reaction speeds up.

After a while, the forwards reaction will be going at the exact same rate as the backwards reaction, so the amounts of reactants and products won’t be changing anymore - it’ll seem like nothing’s happening.

This is called dynamic equilibrium. At equilibrium the concentrations of reactants and products stays constant.

A dynamic equilibrium can only happen in a closed system. This just means nothing can get in or out.

Question 44

What happens if you change the concentration, pressure or temperature of a reversible reaction?

Answer 44

If you change the concentration, pressure or temperature of a reversible reaction, you tend to alter the position of equillibrium. This means you’ll end up with different amounts of reactants and products at equilbrium.

If the position of the equilibrium moves to the left, you’ll get more reactants:

H_2(g) + I_{<strong>2(g)</strong>} ⇌ 2HI_(g)

If the position of the equilibrium moves to the right, you’ll get more products:

H_2(g) + I_2(g) ⇌ 2HI_(g)

Question 45

What does Le Chatelier’s principle tell you?

Answer 45

Le Chatelier’s principle tells you how the position of the equilibrium changes if a condition changes:

If there’s a change in concentration, pressure or temperature, the equilibrium will move to help counteract this change.

In essence, if you raise the temperature, the position of the equilibrium will shift in an attempt to cool things down.

And, if you raise the pressure or concentration, the position of the equilibrium will shift to reduce it again.

Question 46

What are the rules for using Le Chatelier’s principle in terms of concentration?

Answer 46

Concentration

2SO_2(g) + O_2(g) → 2SO₃

1. If you increase the concentration of a reactant (SO₂ or O₂), the equilibrium tries to get rid of the extra reactant. It does this by making more product (SO₃).
2. If you increase the concentration of the product (SO₃), the equilibrium tries to get rid of the extra product. This makes the reverse reaction go faster. So the equilibrium shifts to the left.
3. Decreasing the concentrations has an opposite effect.

Question 47

What are the rules for using Le Chatelier’s principle in terms of pressure?

Answer 47

Pressure (this only affects gases)

1. Increasing the pressure shifts equilibrium to the side with fewer gas molecules. This reduces the pressure.
2. Decreasing the pressure shifts the equilibrium to the side with more gas molecules. This raises the pressure again.

_{There are 3 moles on the left, but only 2 right. So, increasing the pressure shifts the equilibrium to the right.}

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

Question 48

What are the rules for using Le Chatelier’s principle in terms of temperature?

Answer 48

Temperature

1. Increasing the temperature means adding heat. The equilibrium shifts in the endothermic (positive ∆H) direction to absorb this heat.
2. Decreasing the temperature removes heat. The equilibrium shfts in the exothermic (negative ∆H) direction to try to replace the heat.
3. If the forward reaction’s endothermic, the reverse reaction will be exothermic, and vise versa.

_{This reaction’s exothermic in the forward direction (∆H = -197kJ mol^-1). If you increas the temperature, the equilibrium shifts to the left to absorb the extra heat.}

_Exothermic →

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

_Endothermic ←

Question 49

What effect does a catalyst have on the postion of equilibrium?

Answer 49

Catalysts have no effect on the position of equilibrium. They speed up the forward and reverse reactions by the same amount. They can’t increase yield - but they do mean equilibrium is reached much faster.

Question 50

How can Ethanol be produced from ethene and steam?

Answer 50

1. The industrial production of ethanol is a good example of why Le Chatelier’s principle is important in real life.
2. Ethanol is produced through a reversible reaction between ethene and steam.
3. The reaction is carried out at a pressure of 60-70 atomspheres and a temperature of 300ºC, with a phosphoric (V) acid catalyst.

C₂H_4(g) + H₂O_(g) ⇌ C₂H₅OH_(g) ∆H = -46kJmol^-1

Question 51

How are the conditions chosen to make ethanol a compromise?

Answer 51

1. Because it’s an exothermic reaction, the lower temperatures favour the forward reaction. This means more ethene and steam are converted to ethanol at lower temperatures - you get a better yield.
2. But lower temperatures mean a slower rate of reaction. so the 300ºC is a compromise between maximum yield and a faster rate of reaction.
3. Higher pressures favour the forward reaction, so a pressure of 60-70 atmospheres is used - high pressures moves the reaction to the side with fewer molecules of gas. Increasing the pressure also increases the rate of reaction.
4. However, high pressures are expensive to produce. You need stronger pipes and containers to withstand high pressure. In this case, increasing the pressure would lead to side reactions occuring.
5. So the 60-70 atmospheres is a compromise between maximum yield and expense. In the end it all comes down to minimising costs.

Question 52

What is K_c?

Answer 52

K_c is the equilibrium constant.

Question 53

When can one work out the equilibrium constant?

Answer 53

When you have a homogeneous reaction (where all the reactants are in the same physical state) that’s reached equilibrium constant, K_c, using the concentrations of the products and reactants at equilibrium. K_c gives you an idea of how far left or right the equilibrium is.

Question 54

Give the expression of the equilibrium constant.

Answer 54

aA + bB ⇌ dD + eE,

K_c = [D]^d [E]^e / [A]^a [B]^b

The products would be the numerator, the square brackets mean concentration in mol dm^-3.

The lower case letters are the number of moles of each substance.

So for the reaction H₂ + I₂ ⇌ 2HI, K_c = [HI]² / [H₂] [I₂]

Question 55

What are the units for the K_c?

Answer 55

There are no units for K_c

Question 56

What happens to the K_c if there’s a change in temperature?

Answer 56

A change in temperature can change the point of equilibrium, so K_c varies with temperature.

Question 57

How can you predict the position of equilibrium using the value of K_c?

Answer 57

1. The larger the value of K_c, the more products there are at equilibrium, so the further the equilibrium lies to the right.
2. The smaller the value of K_c the more reactants there are at equilibrium, so the further the equilibrium lies to the left.

Question 58

How can you investigate the equilibrium position with changing temperature?

Answer 58

In a closed system, the brown gas NO₂ exists in equilibrium with the colourless gas N₂O_4. This reversible reaction can be used to invesigate the effects of changing temperature on equilibrium position.

1. Place two sealed tubes containing the equilibrium mixture in two baths, one in a warm water bath and one in a cool water bath, and observe the colour of the mixtures.
2. The tube in the warm water bath will change to a darker brown colour as the endothermic reaction speeds up to absorb the extra heat, pushing equilibrium to the left.
3. The tube in the cool water bath will lose colour as the exothermic reaction speeds up to try and replace the lost heat, pushing equilibrium to the right.

Question 59

How can you investigate the equilibrium position with changing concnentration?

Answer 59

Mixing iron (III) nitrate (yellow) and potassium thiocynate (colourless) results in a reversible reaction where the product is iron (III) thiocynate (blood red).

Fe³⁺_(aq) + 3SCN^-_(aq) ⇌ Fe(SCN)_3(aq)

Yellow + Colourless ⇌ Blood red

Add equal amounts of the equilibrium mixture to four test tubes.

1. Test tube 1 is the ‘control’ and nothing is added to it. It keeps its initial red colour.
2. Add some iron (III) nitrate to test tube 2. The mixture turns a deep red collour.
3. Add some potassium thiocynate to test tube 3. The mixture turns a deep red colour.
4. Add some iron (III) thiocynate to test tube 4. The mixture turns a yellow colour.

By adding more reactants, the forward reaction speeds up to produce more product as can be seen in the deep red colour in test tubes 1 and 2, so the equilibrium moves to the right.

By adding more product, the reverse reaction speeds up to produce more reactant as can be seen in the yellow colour in test tube 4, so the equilibrium moves to the left.