3.2

Question 1

3.2.1 a) What is enthalpy and enthalpy change?

Answer 1

Enthalpy (H) is the thermal energy stored in a chemical system. An enthalpy change (in kJ/mol) is the heat energy transferred in a reaction at constant pressure - heat energy tends to be transferred between the system and the surroundings.
Enthalpy change is the difference between between the enthalpy of the products and the enthalpy of the reactants.
ΔH = H(products) - H(reactants)

Question 2

3.2.1 b) What is an exothermic enthalpy change and an endothermic enthalpy change?

Answer 2

In exothermic reactions, the enthalpy of the products is greater than that of the reactants, so heat is released to the surroundings (causing a rise in temperature), and the enthalpy change is negative. (In an energy profile diagram, the products have a lower energy than the reactants and are positioned below them. The arrow is going down (-ve).)
In endothermic reactions, the enthalpy of the reactants is greater than that of the products, so heat is absorbed from the surroundings (causing a fall in temperature), and the enthalpy change is positive. (In an energy profile diagram, the products have a higher energy than the reactants and are positioned above them. The arrow is going up (+ve).)

Question 3

3.2.1 c) What is activation energy?

Answer 3

During chemical reactions, the bonds in the reactants need to be broken by an input of energy. Activation energy is the minimum energy required for a reaction to take place (once an exothermic reaction begins, the activation energy is regenerated and the reaction becomes self-sustaining).

Question 4

3.2.1 d) i) What are standard conditions? Why are they used?

Answer 4

When carrying out an experiment, standard conditions are used - this allows comparisons to be made between different sets of data. They are set at 100kPa, 298K, with all solutions at a concentration of 1mol/dm3. All substances should also be in their standard physical state.

Question 5

3.2.1 d) ii) What is the enthalpy change of reaction, ∆rH?

Answer 5

The enthalpy change of reaction, ∆rH, is the enthalpy change associated with a given reaction, in the molar quantities shown in a balanced, stated equation.
This is calculated using the following equation:
ΔrH = H(products) - H(reactants)

Question 6

3.2.1 d) iii) What is the enthalpy change of formation, ∆fH?

Answer 6

The enthalpy change of formation, ∆fH, is the enthalpy change that takes place when 1 mole of a compound is formed from its constituent elements .

Question 7

3.2.1 d) iv) What is the enthalpy change of combustion, ∆cH?

Answer 7

The enthalpy change of combustion, ∆cH, is the enthalpy change that takes place when 1 mole of a compound is completely combusted (i.e. burned in oxygen).

Question 8

3.2.1 d) v) What is the enthalpy change of neutralisation, ∆neutH?

Answer 8

The enthalpy change of neutralisation, ∆neutH, is the enthalpy change that takes place when 1 mole of water is formed in a neutralisation reaction.

Question 9

3.2.1 e) Which equation allows you to calculate heat loss or gain?

Answer 9

q = mcΔt
q = heat exchanged with the surroundings (or the enthalpy change as long as the pressure is constant) - in J
m = mass (of the water in the calorimeter, or solution in the container) - in g {1cm3 of water = (approx.) 1g of water - since density of water is ~1.0 gcm–1}
c = specific heat capacity of water = 4.18 - in J/g/K
ΔT = change in temperature (of the water or solution) - in K
[Note: the mass is measured by identifying and weighing the materials changing the temperature - not the mass of the reactants]

Question 10

3.2.1 e) How can you calculate the molar enthalpy change?

Answer 10

The molar enthalpy change can be calculated by first working out the moles of the substance that released energy, and then dividing it by q {enthalpy change = q / n - in kJ/mol or J/mol}

Question 11

3.2.1 f) i) Explain what average bond enthalpy is. Why might it differ from an actual bond enthalpy?

Answer 11

The average bond enthalpy is the energy needed to break 1 mole of bonds in gaseous molecules (the energy needed to break a bond is the same amount of energy released when the bond is formed).
An actual bond enthalpy may differ from the average bond enthalpy, which is calculated from many different compounds (for example, the average O-H bond enthalpy is calculated over a range of molecules, such as the O-H bond in alcohols or the O-H bond in water - individually they are different). The larger the average bond enthalpy value, the stronger the bond.
[Note: energy is always required to break a bond, so bond enthalpies are always endothermic, with a positive enthalpy value]

Question 12

3.2.1 f) ii) Explain exothermic and endothermic reactions in terms of enthalpy changes associates with the breaking and making of bonds.

Answer 12

When reactant bonds are broken, energy is is absorbed from the surroundings - and so bond breaking is endothermic, accompanied by a positive change in enthalpy.
When product bonds are formed, energy is released into the surroundings - and so bond forming is exothermic, accompanied by a negative change in enthalpy.
The enthalpy change for a reaction is the overall effect of these two changes.
In an endothermic reaction, reactant bonds require more energy to break than product bonds require to form - more energy is therefore absorbed from the surroundings, and the enthalpy change is positive.
In an exothermic reaction, product bonds require more energy to form than product bonds require to break - more energy is therefore released into the surroundings, and the enthalpy change is negative.

Question 13

3.2.1 f) How can enthalpy changes be calculated using average bond enthalpies?

Answer 13

The enthalpy change for a reaction can be calculated from average bond enthalpy data using the following expression:
ΔrH = ∑(bond enthalpies in reactants) - ∑(bond enthalpies in products)
[Note: this is different from the following equation: ΔH = H(products) - H(reactants) - learn the difference]

When calculating the enthalpy change of reaction, remember to first multiply each bond enthalpy by the number of bonds in 1 molecule, and then by the number of molecules indicated by the balanced equation (i.e. the number of moles for this particular molecule}. Remember that all bond enthalpy values are positive.

If the enthalpy change is negative, then the reaction is exothermic. If the enthalpy change is positive, then the reaction is endothermic.

Question 14

3.2.1 g) What is Hess’ law?

Answer 14

Hess’ law states that the enthalpy change in a chemical reaction is independent of the route is takes (i.e. if a reaction can take place by two routes - and the starting and finishing conditions are the same - the total enthalpy change is the same for each route)

Question 15

3.2.1 g) i) Explain how you can calculate the enthalpy change of reaction from the enthalpy changes of combustion?

Answer 15

You can work out the enthalpy change of reaction from enthalpy changes of combustion - create an enthalpy cycle, writing the enthalpy change you want to find as a balanced, horizontal equation, with two arrows pointing down from the reaction equation to the combustion products (CO2 + H2O).
ΔrH = ∑( ΔcH of reactants) - ∑( ΔcH of products)
{You should include all signs when inputting each individual enthalpy. Remember to multiply each enthalpy by the number of molecules indicated in the balanced equation, i.e. the number of moles for this particular molecule}

Question 16

3.2.1 g) ii) Explain how you can calculate the enthalpy change of reaction from the enthalpy changes of formation?

Answer 16

You can work out the enthalpy change of reaction from the enthalpy changes of formation - create an enthalpy cycle, writing the enthalpy change you want to find as a balanced, horizontal equation, with two arrows pointing up from the elements to the reaction equation.
ΔrH = ∑( ΔfH of products) - ∑( ΔfH of reactants)
{You should include all signs when inputting each individual enthalpy. Remember to multiply each enthalpy by the number of molecules indicated in the balanced equation, i.e. the number of moles for this particular molecule}

Question 17

3.2.1 h) Explain a procedure used to calculate enthalpy of a (non-combustion) reaction.

Answer 17

A simple polystyrene calorimeter of low heat capacity can be used for any non–combustion reaction that will happen spontaneously at room temperature - these reactions should involve at least one solution, even if its water - it doesn’t matter if the reaction is exothermic or endothermic. For example, the mixture could be a salt and water (with a heat change on dissolving) or an acid and an alkali solution (with a heat change on neutralisation) - ensure that the mass of the solution is measured. When the reaction takes place, record the temperature change of the solution, and use the following equations to calculate ΔH:
q=mcΔt and ΔH = q/n.

Limitations:
While the polystyrene cup is insulated, it may still absorb/release heat (depending on an exothermic/endothermic reaction) and a small amount of heat will still be lost to or gained from the surroundings. Heat loss can be reduced by using a polystyrene lid, however.
It is often assumed that the heat capacity of water is the same as the solution, which provides a margin of error when calculating the enthalpy change.

Question 18

3.2.1 h) Explain a procedure used to calculate enthalpy of combustion of a flammable liquid.

Answer 18

Liquid fuels can be burnt using small spirit burners:
- Using a measuring cylinder, measure out a known volume of water into a beaker. Record the initial temperature of the water.
- Weigh the spirit burner, and then place it under the beaker.
- Light the burner, and stir the water with the thermometer whilst the fuel burns.
- After a few minutes, extinguish the flames and immediately record the temperature rise.
- Re-weigh the spirit burner.
Use the following equations to calculate ΔH:
q=mcΔt and ΔH = q/n.

Question 19

3.2.2 a) Which two conditions are required for particles to collide successfully?

Answer 19

A reaction will only take place between two particles if they collide with the correct orientation, with the activation energy (the minimum amount of kinetic energy required to break the bonds and start the reaction).

Question 20

3.2.2 a) What effect does increasing the concentration of reactants have on the rate of reaction?

Answer 20

If you increase the concentration of reactants, the particles will be closer together (with more molecules in the same volume), resulting in more frequent collisions, and increasing the chance that two molecules successfully collide (in the right direction, with sufficient energy) - so the rate of reaction will increase.

Question 21

3.2.2 a) What effect does increasing the pressure of the reactants have on the rate of reaction?

Answer 21

If you increase the pressure of the reactants, the particles will be closer together (with the same number of molecules occupying a smaller volume), resulting in more frequent collisions, and increasing the chance that two molecules successfully collide (in the right direction, with sufficient energy) - so the rate of reaction will increase.

Question 22

3.2.2 a) What is the rate of reaction? How can it be calculated?

Answer 22

The rate of reaction is defined as the change in concentration of a reactant or a product in a given time (units: mol/dm3/s). It can be calculated by dividing the amount of reactant used up, or the amount of product formed, by time.

Question 23

3.2.2 b) How can you use experimental date to calculate rate of reaction?

Answer 23

The rate of reaction can be calculated by monitoring changes in physical quantities, with values (for a reactant used up or a product formed) plotted (on the y-axis) against time (on the x-axis). The rate of reaction is equal to the gradient of the graph, calculated by dividing the change in y with the change in x.

Question 24

3.2.2 c) What is the role of a catalyst?

Answer 24

The role of a catalyst is

i) in increasing the reaction rate without being used up by the overall reaction
ii) in allowing a reaction to process via a different route with lower activation energy.

Question 25

3.2.2 d) i) Explain what homogeneous catalysts are, and how they work.

Answer 25

Homogeneous catalysts are in the same physical state as the reactants.
They work by combining with the reactants to form an intermediate species, which further reacts to form the products and reform the catalyst.

Question 26

3.2.2 d) i) Explain what heterogeneous catalysts are, and how they work.

Answer 26

Heterogeneous catalysts are in a different phase (such as different physical states) from the reactants - different phases also include a reaction between immiscible liquids (which don’t mix) and a catalyst in a different liquid layer from the reactants.
The reaction takes place on the surface of the heterogeneous catalyst. Therefore increasing the surface area of the catalyst increases the number of molecules that can react at the same times, which increases the rate of reaction.

Question 27

3.2.2 d) ii) Evaluate both the economic importance and the benefits to the environment that catalysts have.

Answer 27

By employing catalysts, lower temperatures can be used, reducing energy demand from combustion of fossil fuels. This results in lower production costs, more product in a shorter amount of time, and a reduction in CO2 emissions. Catalysts can also reduce waste by allowing a different reaction to be used with a better atom economy. However, the benefits to the environment of improved sustainability should be weighed against the toxicity of some catalysts, and the economic importance that catalysts have can be impacted by how certain catalysts are often specific to only one reaction.

Question 28

3.2.2 e) Briefly outline common techniques and procedures to investigate reaction rates.

Answer 28

The rate of reaction can be calculated by monitoring changes in physical quantities during a reaction, such as:

• using a gas syringe to measure the volume of (gaseous) product given off, over time
• monitoring the change in a reactants mass, over time (usually the other reactant is a liquid and the product is a gas - as it forms, the mass decreases)
• recording the time it takes for a particular change to occur, such as a precipitate forming
• using a calorimeter to monitor the changes in concentration of coloured reactants or products, over time
• monitoring changes in pressure, or changes in conductivity, over time

Question 29

3.2.2 f) Give a qualitative explanation of the Boltzmann distribution.

Answer 29

The Boltzmann distribution is a frequency distribution of particles in a system over various possible energy states - it shows the number of molecules in a system against their kinetic energy (at a particular temperature). The area under the graph is equal to the number of molecules in the sample - this doesn’t change with conditions. No molecules have zero energy, so the graph begins at the origin (0,0). While some particles have low energies (moving relatively slowly), the majority of particles have moderate energies, with only a few particles having or surpassing the activation energy needed to react. There is no maximum energy for a molecule, so the curve gets close to, but doesn’t touch or cross, the energy (x) axis.

Question 30

3.2.2 i) Explain, using the Boltzmann distribution, the effect that temperature change can have on the rate of reaction.

Answer 30

As the temperature of a reaction increases, the particles will, on average, have greater kinetic energy, moving faster, and resulting in more frequent collision. A higher proportion of molecule will have energy that surpasses the activation energy - this means that more collisions are likely to be successful. At a higher temperature, the Boltzmann distribution therefore flattens and shifts to the right (the number of molecules doesn’t change, however, and the area under the graph remains the same). So with an increase in temperature, more successful collisions occur in a certain length of time, increasing the rate of reaction.

Question 31

3.2.2 g) ii) Explain, using the Boltzmann distribution, the effect that catalytic behaviour can have on the rate of reaction.

Answer 31

Catalysts don’t affect the distribution of energy within molecules, but instead lower the activation energy - while the shape of a Boltzmann distribution won’t change, a greater proportion of the molecules will exceed the activation energy. This results in more successful collision within a certain length of time, thereby increasing the rate of reaction.

Question 32

3.2.3 a) What is dynamic equilibrium?

Answer 32

(As the reactants get used up in a reversible reaction, the forward reaction slows down, and as more product forms, the reverse reaction speeds up - after some time, both reactions go at the same rate.)
Dynamic equilibrium exists in a closed system, when the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products do not change.

Question 33

3.2.3 b) What is le Chatelier’s principle?

Answer 33

le Chatelier’s principle states that when a system in dynamic equilibrium is subjected to change, the position of equilibrium will shift to minimise the change.

Question 34

3.2.3 b) Employ le Chatelier’s principle to explain the effect that a change in concentration has on the position of equilibrium.

Answer 34

Increasing the concentration of a reactant results in the position of equilibrium shifting to the right to counteract this change (by removing the increased reactant): this favours the forward reaction, which produces more products.
Increasing the concentration of a product results in the position of equilibrium shifting to the left to counteract this change (by removing the increased product): this favours the reverse reaction, which produces more reactants.

Question 35

3.2.3 b) Employ le Chatelier’s principle to explain the effect that a change in pressure has on the position of equilibrium.

Answer 35

Increasing the pressure of the system causes the position of equilibrium to counteract this change by shifting to side with fewer gas molecules, thus decreasing the pressure.
Decreasing the pressure of the system causes the position of equilibrium to counteract this change by shifting to side with more gas molecules, thus increasing the pressure.
[If the position of equilibrium shifts to the right, it favours the forward reaction, and more product is formed. If the position of equilibrium shifts to the left, it favours the reverse reaction, and more reactant is formed.]

Question 36

3.2.3 b) Employ le Chatelier’s principle to explain the effect that a change in temperature has on the position of equilibrium.

Answer 36

Increasing the temperature (or adding heat to the surroundings) causes the position of equilibrium to counteract this change by shifting in the endothermic direction, thereby absorbing heat and decreasing the temperature.
Decreasing the temperature (or removing heat from the surroundings) causes the position of equilibrium to counteract this change by shifting in the exothermic direction, thereby releasing heat and increasing the temperature.
[If the position of equilibrium shifts to the right, it favours the forward reaction, and more product is formed. If the position of equilibrium shifts to the left, it favours the reverse reaction, and more reactant is formed.]

Question 37

3.2.3 c) What effect does a catalyst have on a reversible reaction in equilibrium?

Answer 37

A catalyst increases the rate of both the forward and reverse reactions in an equilibrium by the same amount, resulting in an unchanged position of equilibrium (they therefore have no effect of the yield, but do allow equilibrium to be reached faster).

Question 38

3.2.3 e) Use the Haber process to consider how there must be a compromise between chemical equilibrium and reaction rate in deciding operational conditions.
[N2(g) + 3H2(g) ⇄ 2NH3(g), exothermic forward reaction]

Answer 38

By increasing the pressure, the position of equilibrium will shift to the right to counteract this change, favouring the forward reaction where there are fewer gas molecules (thereby decreasing the pressure) - this produces a higher equilibrium yield. High pressures also increase the concentration of the gases, and therefore increase the rate of reaction. However, they require a lot of energy (which isn’t cost effective or sustainable to the environment), and create a possible hazard (should there be a chemical leak, for example).
By decreasing the temperature, the position the position of equilibrium will shift to the right to counteract this change, favouring the forward reaction which is exothermic (thereby increasing the temperature) - this produces a higher equilibrium yield. However, lower temperature would result in a slower rate of reaction. Therefore a compromise must be met between equilibrium yield, the rate of reaction, and cost effectiveness.

Question 39

3.2.3 f) What is the equilibrium constant, and how can it be calculated?

Answer 39

When a homoegeneous reaction (where all the reactants and products are in the same physical states) reaches dynamic equilibrium, Kc (the equilibrium constant) gives a measure of where the equilibrium lies, whether it’s more to the right or the left.
aA + bB ⇄ cC + dD
Kc = [C]c[D]d / [A]a[B]b

Question 40

3.2.3 g) How can you estimate the position of equilibrium from the magnitude of Kc?

Answer 40

A Kc value of 1 would indicate the position of equilibrium to lie halfway between reactants and products, producing equal concentrations of both.
A Kc value above 1, shows that the position of equilibrium lies to the right, favouring the forward reaction, and producing more product.
A Kc value below 1, shows that the position of equilibrium lies to the left, favouring the reverse reaction, and producing more reactants.