Explanation​:
At equilibrium, Gibbs free energy is at a minimum.

Explanation:
Atomisation is defined as the formation of one mole of gaseous atoms.
Ionization is the process in which one mole of electrons is removed from one mole of gaseous atoms/ions.
Electron affinity is the addition of one mole of electrons to gaseous atoms/ions.

Explanation:
To answer this question, the student must first recognize that oxygen gas is reduced at the cathode, and the fuel (methanol) is oxidized at the anode.
Next, it is essential to select the answer choice where the amount of electrons produced in the oxidation is equivalent to the electrons used in the reduction.
Finally, the question states the reaction is under acidic conditions, and therefore, the correct answer is the one that includes H+ as opposed to OH−.

This question is categorized as hard because there are many parameters to consider in order to select the correct answer:

• Cathode = site of reduction
• Anode = site of oxidation
• Electrons must be equal in each half-reaction
• Acidic conditions require H+ to be present in half-reactions

Explanation​:
The reactants are at a lower energy level to the products and are therefore more stable. The products are at a higher energy level to the products meaning the enthalpy change is positive (endothermic reaction).

Explanation:​
Average bond enthalpies give data for species in the gaseous phase. In choice C, water is a liquid and so the enthalpy change for the condensation of water vapour would also be needed.

Explanation​:
ΔG=−RT ln(K).
If $\Delta G$ is positive, $ln(K)$ must be negative. $ln(K)$ is negative for values smaller than 1.
A positive $\Delta G$ value indicates that the reaction is not spontaneous. Therefore, the equilibrium concentrations of the reactants are larger than the ones of the products; the equilibrium constant is less than 1.

Explanation​:
The reaction is taking place under standard conditions, so it is fair to use standard enthalpies of formation to calculate $\Delta$Hr​. Even if the reaction takes place under non-standard conditions, the error is typically small unless a phase transition such as boiling or melting has taken place. In their standard states, carbon is a solid and bromine is a liquid, but it is not necessary for the elements to be gases as the enthalpy of vaporisation is accounted for in the measurement of ΔHfθ​.

Bond energy calculations do not account for the enthalpy changes of boiling, melting and sublimation, but the reactants and products in this reaction are gases so such errors are not relevant. The bond energies may, however, differ from the average values; for example, the C-H bond energies in the product may be affected by the neighbouring C-Br bond.

This question is only moderately difficult as the errors and assumptions of the two calculation methods are clearly stated in the syllabus.

Explanation​:
Gases have the highest entropy followed by liquids and then solids.
CO2​ has a higher entropy than O2 because it has more bonds, and therefore, more vibration modes.

Explanation:
Increased temperature increases the kinetic energy of the gas particles and therefore increases the disorder. Gases are more disordered than liquids and solids. The greater the pressure, the lower the disorder.

Explanation:​
Four $C-H$ bonds and two $O=O$ bonds are broken. Two $C=O$ and four $O-H$ bonds are formed.
ΔH =$\Sigma$ΔHBEθ​(bonds broken) - ΣΔHBEθ(bonds formed).

Explanation​:
Most combustion reactions are exothermic. D is the opposite of a combustion reaction

Explanation​:
The standard enthalpy of formation of a pure element in its standard state is zero; not for a compound.

Explanation​:
Crude oil is non-renewable since there is a finite amount of this resource available on Earth. Crude oil has a high energy density meaning that it stores considerable energy in a smaller quantity than other energy sources.
Combusting crude oil releases sulfur and nitrogen oxides (thus contributing to acid precipitation) and carbon dioxide. Importantly, carbon dioxide is always a product of hydrocarbon combustion.

The question difficulty is categorized as a medium because there are a number of facts that the student must remember to answer this question correctly, but there is nothing that the student must solve to arrive at the correct answer.

Explanation​:
When very little oxygen is present in the combustion of a hydrocarbon, the result is carbon soot (C solid) as opposed to carbon monoxide or carbon dioxide.

Sufficient oxygen $\to$ complete combustion resulting in CO2 and water;

Limited oxygen $\to$ incomplete combustion resulting in CO and water;

Very limited oxygen $\to$ incomplete combustion resulting in C soot and water.

For this question, the student must recognize that when there is limited oxygen, there are fewer oxygens in the products. This question is categorized as easy as the student needs only to recognize that the amount of oxygen in the combustion directly relates to the number of oxygens in the products.

Explanation:
An exothermic reaction is one in which heat energy is transferred from system to surroundings. Endothermic reaction is one in which heat energy is transferred from surroundings to system.

Explanation​:
ΔG= ΔH−TΔS, and reactions are spontaneous if $\Delta \mathrm{G}$ is negative.
In endothermic reactions, $\Delta \mathrm{H}$ is positive, and if entropy increases, $\Delta \mathrm{S}$ is also positive.
At high temperatures, $-$ resulting in a negative $\Delta \mathrm{G}$ value.

Explanation​:
Endothermic reactions absorb energy from the surroundings. This decreases the temperature of the surroundings and increases the energy in the bonds (the system).

Explanation​:
ΔG=ΔH−TΔS.
Negative $\Delta \mathrm{G}$ values result in spontaneous reactions.
Negative $\Delta \mathrm{H}$ and positive $\Delta \mathrm{S}$ values result in a negative $\Delta \mathrm{G}$ value at all temperatures.

Explanation​:
The reaction equation is N2​O4​ (l) $\to$ 2NO2​ (g).

Drawing an enthalpy cycle reveals that ΔHfΘ​ (N2​O4​) = 2ΔHfθ​ (NO2​) − ΔHrθ​=2×33.1−85.8=−19.6kJ mol−1.

The alternative options are obtained if the wrong signs are given to the enthalpy changes in the equation and/or ΔHfθ (NO2​) is not multiplied by 2.

This question is categorised as very hard, as the student must deduce the stoichiometry of the reaction, sum the enthalpy changes with the correct signs, and obtain the correct value for the unknown quantity, without the aid of a reaction equation or enthalpy cycle.

Explanation​:
The graph shows that the maximum change in temperature is 24.9−20.0=4.9°C. However, extrapolating the linear cooling trend to the time of addition gives a theoretical maximum change of 5.4$°$C. This is the value of $\Delta T$ that should be used to obtain the best estimate of $q$, as the extrapolation accounts for heat loss to the environment during the reaction.

The other distractors correspond to the temperature change at the latest recorded time point (option A) and the extrapolated temperature change at time = 0$s$ (option D), a time point which precedes the addition of the solid and is therefore irrelevant.

This question is moderately difficult as the student must interpret a graph and apply the heat loss correction with little guidance.

Explanation​:
Combining the first equation and the reverse of the second equation gives the equation for the enthalpy change in question. (+33.9 kJ)−(−56.5 kJ)=+90.4 kJ

Explanation:​
The products are at a lower energy level than the reactants meaning that the total average bond enthalpy of formation of bonds in the products is greater than the total average bond enthalpy of breaking of bonds in the reactants.

Explanation​:
The endothermic reaction is favoured when the temperature increases. The concentration of products increases at t=10 mins; therefore, the forward reaction must be endothermic.
Increasing the volume decreases the pressure and causes the side with more moles of gas to be favoured. The concentration of reactants increases at $t$; therefore, the reactants must have more moles of gas than the products.