Physical (Year 2)

Enthalpy change when one mole of covalent bonds is broken in the gaseous state.

Enthalpy change when one mole of gaseous atoms is produced from an element in its standard state.

Enthalpy change when one mole of a solid is turned into a liquid.

Enthalpy change when one mole of a liquid is turned into a gas.

Enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants and products in standard states under standard conditions.Note: this definition is worth 2-3 marks.

Enthalpy change when one mole of a substance is formed from its constituent elements with all reactants and products in standard states under standard conditions.Note: this definition is worth 2-3 marks.

Enthalpy change when 1 mole of water is formed in a reaction between an acid and alkali under standard conditions.

Enthalpy change when 1 mol of a solid ionic compound is formed from its gaseous ionsNote: this definition is worth 2 marks.

Enthalpy change when one mole of a solid ionic compound is broken up into its constituent gaseous ions.Note: this definition is worth 2 marks.

Enthalpy change when one mole of gaseous atoms gain one mole of electrons to form one mole of gaseous 1− ions.

Enthalpy change when one mole of gaseous 1− gain one mole of electrons to form one mole of gaseous 2− ions.

Enthalpy change when one mole of gaseous ions becomes dissolved in water.

Enthalpy change when one mole of an ionic solid dissolves in a large volume of water to form an infinitely dilute solution.

1st electron affinity is exothermic (for many non-metals) as the electron being gained is attracted to the positive nucleus, whereas 2nd electron affinity is endothermic as the negative electron being gained is repelled by the negative ion.

Bond dissociation enthalpy. If you’re going from ½ O₂ -> O then you have to halve the value, as O₂ -> 2O is breaking 1 mole of covalent bonds.

It underestimates bond strength as a result of not accounting for covalent character, which makes the bonds holding the lattice together stronger.Note: Theoretical uses the perfect ionic model, which is inaccurate to real life. The experimental value is the ‘true’ value.

The cations being small and very positive, giving them a high charge density.

The anions being large and therefore easily polarised.

It reacts with water to form a solution of NaOH.Note: if a compound will only react with water, you cannot measure its enthalpy of solution.

Entropy will increase over time.

The entropy of a substance is zero at absolute zero and increases with temperature. This is because the higher the temperature, the faster the particles move so the greater the entropy.

Because the particles are stationary.

Use the equation ΔG = ΔH − TΔS and if ΔG ≤ 0 the reaction is feasible.

ΔH is the enthalpy change in kJmol-1.

T is the temperature in Kelvin.

ΔS is the entropy change in kJ K-1 mol-1.Note: remember that this means you’ll probably have to convert from J K-1 mol-1 to kJ K-1 mol-1 by dividing by 1000.

The gas is more disordered.Note: I included this just to show that the explanation is very simple.

The sum of the indices to which the concentrations are raised in the rate equation.Note: or just the ‘sum of the orders of the species involved’. Don’t say ‘reactants’ though as catalysts are also involved.

This is the point where the concentrations are known.

The higher the temperature, the higher the rate constant (k).

The rate constant also rises exponentially with temperature.

They are J mol⁻¹ in the equation. Therefore, if the question asks for the Ea in kJ mol⁻¹, you have to convert.Note: you don’t necessarily have to memorise this - just pay attention to the units of the other quantities in the question (e.g. the gas constant).

You should either write an equation showing their proportionality, e.g. X α Y² or X² α 1/Y (use proportional symbol instead of α obvs). OR you can just say ‘When X increases by a factor of 2, Y increases by a factor of 4.’Note: you could also just try and get across the idea that they have a square/ square root relationship another way - the key thing is that you acknowledge the square aspect of the relationship and don’t just say ‘inversely proportional’ or something.

You must remember to add s⁻¹ after the ‘mol dm⁻³’ part, as it is the constant for 'rate'.

1. Plot concentration against time then take tangents to calculate the rates at certain times.

2. Then plot a graph of log(rate) against log(concentration)

3. The gradient of this straight line tells you the order.

Alternatively:

2. After finding the rates at certain times, plot a graph of rates against concentration.

3. A linear line through the origin tells you it is order 1, a horizontal line tells you order 0, and an exponential line tells you order 2.

Temperature and Pressure.

Repeat the experiment with a catalyst while:

1. Keeping the same volume and concentration of the reactants.

2. Keeping the temperature the same.

They must first be quenched to stop the reaction.

This is done by extreme cooling or dilution.

It is a straight line through the origin.Note: mentioning ‘through the origin’ is key.

Draw a diagram if possible. Make sure to label the electrodes and solutions (with concentrations), show and label that the salt bridge connects the solutions, and show and label the two electrodes being connected by a wire with a ‘high resistance’ voltmeter in it.Note: I’d also add that you should indicate what reactants make the solutions, e.g. FeCl₂ for a solution of Fe²⁺ ions.Note 2: I’d also include a temp of 298K and explain how to make the salt bridge.

Its high resistance.Note: this is because it stops the reaction from occurring and keeps them still at equilibrium due to its very high resistance, which prevents electrons flowing through the external circuit. Therefore, to allow the cell reaction to go to completion (after you have finished measuring the EMF), you would remove the voltmeter as this allows electrons to flow through the circuit.

By containing mobile ions, meaning ions can move through it and complete the circuit.

It must allow ions to pass through it.Note: this is so the charges can remain balanced and the circuit can be complete.

They react to form a solid AgCl precipitate.Note: they would be fine with Fe²⁺ ions (even though Fe²⁺ ions are more reactive) as the ions would remain soluble.Note 2: mark schemes do normally allow you to just say they ‘react’ with Ag⁺ ions, but it’s important you understand the precipitate concept as saying they 'react with Fe²⁺ ions' is penalised. Also always remember to actually say which ions react/ form precipitates with which to gain the marks.

FeCl₂ and FeCl₃.Note: make sure that in any question where you’re asked to identify the substance inside of an electrode you give the full compounds (e.g. HCl for the SHE) not just the ions.

Non-standard conditions.Note: one of the concentrations being wrong could be a possible alternative answer, but I’d always go with non-standard conditions.

A list of the electrode potentials (E°) of half cells/ equations in numerical order.

‘The E° of the reactant of the reaction with the high E° is greater than the E° of the reactant in the reaction with a low E°’.Note: Making this comparison explains why the reaction would then occur between the reactant of the high E° reaction and the product of the low E° reaction, as it shows the product of the reaction with the lower E° is a better reducing agent than the product of the reaction with the high E°.Note 2: Alternatively, you could do a calculation for the EMF of a hypothetical cell formed from the two reactions and show that since it is positive the reaction would be spontaneous. This is the best approach in my opinion.

There is no effect. The E° of an equation is always the same, even if you have to double it to combine it with another equation, for example.

They have different ligands.

You must remember that Fe³⁺ can then form from further oxidation. This means that if you are reducing a substance with iron, the final product will most likely be Fe³⁺. This means you may have to combine 3 half equations to get the full equation for the redox reaction.

H₂O is included when it is an actual reactant or product in a reaction.Note: I have seen it left out when it’s a reactant, but it's always there when it’s a product of a reaction. Remember that it should go in the same state boundary as aqueous products/ reactants.Note: they do usually allow you to separate them on mark schemes.

You order them in the order they appear in the overall reaction, so the reactants on the left-hand side and the products on the right-hand side. This means that for the negative electrode side (the left-hand side, also the oxidation side), the products are closest to the salt bridge and the reactants are furthest away. For the positive electrode side (the reduction side), the reactants are closest to the salt bridge and the products are farthest away.Note: this logic can be used for any conventional cell representation question, it’s just that using ROOR is easier in most cases.

H+ should always be placed further away from the salt bridge than the other reactant/ product.

Example: H⁺ (aq) │CO₂ (g) ││O₂ (g) │H⁺ (aq)

Here H⁺ and CO₂ were both formed in the oxidation reaction, and CO₂ was placed closest to the salt bridge.

When O₂ and H+ were both reactants in the reduction reaction, O₂ was placed closest to the salt bridge.Note: this is in spite of H⁺ being the more oxidised species, so make sure you remember this. I do not know the reasoning behind this rule.

H₂ + ½ O₂ — H₂O.

H₂ + 2OH⁻ —> 2H₂O +2e-.

½ O₂ + H₂O +2e- —> 2OH⁻.

They have the same overall reaction.

The fuel (hydrogen) is supplied continuously.

The concentrations of the reactants must be kept constant by constantly adding reactants.Note: don’t say something like ‘the temperature must be kept the same’ when it’s in the context of specifically fuel cells.

A fuel cell converts more of the energy released in combustion into kinetic energy, as the internal combustion energy wastes more energy as heat.Note: always use this as your reason for why using hydrogen in fuel cells is better than using it in internal combustion engines, don’t mention waste products or anything like that.

Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻.

Li → Li⁺ + e⁻.

Rechargeable cells can be recharged because their electrode reactions can be reversed by applying a reverse potential.

It is the reverse of its discharging equation (its overall equation, which is reversible).Note: this is key knowledge - one question asked for the ‘overall equation that occurs when the cell is recharging’ and you had to work out the overall equation then reverse it to get the right answer.

The electricity needed often comes from the combustion of fuels.

There will be no effect, as surface area doesn’t affect electrode potential.Note: remember the conditions described for the SHE are the only things that affect EMF.

An acid that completely dissociates to form H⁺ ions.

Acid that partially dissociates when added to water.Note: do not say ‘it doesn’t fully dissociate’, as this could mean it doesn’t dissociate at all.

It is almost constant as it is very large in comparison to [H⁺] and [OH⁻].Note: saying it is ‘1’ was ignored.

It is composed of a weak acid and its conjugate salt, or a weak acid and a strong base.

For example, propanoic acid and sodium propanoate or sodium hydroxide.

Note: They have high concentrations of [HA] (from the weak acid) and [A⁻] (from the salt) relative to [H⁺]. Note 2: They naturally form in weak acid, strong base titrations.

Way 1: When the OH⁻ ions react with the H⁺ ions, the equilibrium for the dissociation of [HA] moves to the right to replace the H⁺ ions and (almost) maintain [H⁺].

Way 2: When the OH⁻ ions react with the HA (to form A⁻ + H₂O), the ratio of [HA] to [A⁻] remains almost constant since their concentrations are so high, meaning [H⁺] doesn’t decrease much.Note: Way 2 is the one that is used in buffer calculations, but I’d use both explanations in a theory question since both happen at the same time and explanation 1 is a bit easier to use.

The added H⁺ ions react with A⁻ to form HA. Since the [A⁻] and [HA] are very high, the ratio of [A⁻] to [HA] remains roughly constant, meaning pH doesn’t decrease much.

Add an excess of a weak acid to a strong base (e.g. NaOH), so the acid is partially neutralised and a mixture of the HA and A⁻ forms.

A weak base and the conjugate salt of that base.Note: the common example is NH₃ and NH₄Cl.Note 2: there is a high concentration of both NH₃ and NH₄⁺, although this isn’t as relevant as it is with acidic buffers.

The OH⁻ ions react with NH₄⁺ to form NH₃ + H₂O, meaning [OH⁻] does not increase much because the OH⁻ ions are used up.

The H⁺ ions react with NH₃ to form NH₄⁺, meaning [H⁺] does not increase much because the H⁺ ions are used up.

Acid: proton (H⁺) donor Base: proton (H⁺) acceptor.Note: fun fact - water is both a Brønsted–Lowry acid and base, as it can donate a proton to form OH⁻ ions or accept a proton to form H₃O⁺ ions.

Because [H⁺] = [OH⁻], which is the definition of ‘neutral’.Note: neutral isn’t defined as pH = 7, it’s defined as [H⁺] = [OH⁻].

It has no distinct colour change and turns a range of different colours during a titration.

The point where [H⁺] = [OH⁻]. It can be identified by the steep/ near vertical increase in pH. It is also therefore the point where all of the acid has reacted.

The pKa = pH.

So the Ka of the weak acid can be calculated by doing 10-pH.

It is also the point at which half the acid has reacted, which is why pH = pKa, as [A⁻] = [HA] so Ka = [H⁺].

You must ensure the indicator’s colour change falls within the range of the steep section of the curve.

They are weak acids, where HA is the dominant species at low pH and A⁻ is the dominant species at high pH, and HA and A⁻ are different colours.

Methyl orange:

Red at low pH and yellow at high pH.

Changes at approx. 3-4 pH.

Phenolphthalein:

Colourless at low pH and pink at high pH.

Changes at approx. 8-10 pH.

Because it can deliver variable volumes.

The indicator would change colour over a range of volumes of alkali added.

This is because there is a large change in pH for a small addition of OH⁻ near and at the end point, meaning small increments need to be added to avoid missing the end point.

They don’t give accurate readings after storage.

Maintain a constant temperature.Note: I’d describe a method for this, e.g. using an electric water bath or taking readings of the temperature at 30 second intervals.

Ineffective stirring of the mixture.Note: this would cause the OH⁻ ions to not spread out evenly through the solution.