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:​
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:
As temperature increases, the equilibrium shifts to the left hand side (more blue) to oppose the change.
The backward reaction is therefore endothermic and absorbs energy, whereas the forward reaction is exothermic.
As equilibrium shifts to the left, Kc​ decreases as the concentration of reactants increases.

Explanation:
Many products can be made from the same reactants by more than one route. Regardless of the route by which a chemical reaction proceeds, the enthalpy change will always be the same provided that the initial and final states of the system are identical.

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​:
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​:
The silver in silver oxide has a $+1$ charge, resulting in Ag2O as the product. When arriving at the sum of the coefficients, it is important to remember to add a 1 for the O2​. Hence, the sum of the coefficients is 4+1+2=7

4Ag(s)+1O2​(g) →2Ag2​O(s)

This question is categorized as medium as the student must ensure the silver's correct charge and add a 1 for the O2.

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.

C6H12O6+yeast →2CH3CH2OH+2CO2​

To eliminate the incorrect answers, students must recognize that carbon fixation is the process where inorganic carbon (such as CO2COX2​) is converted to organic compounds during photosynthesis.
Combustion of hydrocarbons would be the process that would take place when the biofuel is combusted.

This question is categorized as easy since the student needs only to recall that ethanol for biofuel is produced via the fermentation of plant material.

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:​
Energy is absorbed when breaking bonds (endothermic). The amount of energy required is dependent on the strength of the bond.
Statement III is incorrect because average bond enthalpies are averaged over a range of compounds.

Explanation:
Coal is an impure form of fossil fuel and therefore contains many impurities that will also combust. This will result in many byproducts from the impurities combusting, so more oxygen is required to combust the hydrocarbons completely.
Propane, natural gas, and crude oil are all cleaner fuels (have fewer impurities). So, they are more likely to undergo complete combustion with a comparable amount of oxygen per mole of fuel.

Explanation:
In an electrochemical fuel cell, the fuel (methanol, for example) is oxidized at the anode, while oxygen gas is reduced at the cathode to form water. As is the case for any electrochemical cell, the electrons flow through an external wire from the anode to the cathode.

This question is categorized as medium because it requires the knowledge that oxygen gas is reduced at the cathode and that the product is water. The CO2​ would be a product for oxidizing a fuel such as methanol.

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:
D is a thermal decomposition reaction and so requires energy from the surroundings through heating in order to take place. A, B, and C are all combustion reactions and so exothermic.

Explanation:
The full, balanced equation divided into the balanced half-reactions where the hydrogen gas is oxidized, and the oxygen gas is reduced, is:

Anode half-reaction: 2H2(g) →4H^+(aq)+4 e^−

Cathode half-reaction: O2(g)+4H+(aq)+4e^− →2H2O(l)

The number of electrons required from these half-reactions can be identified as 4.

This question is categorized as hard. The student must be able to separate the hydrogen and oxygen gases to their correct electrodes and determine which is oxidized and which is reduced.
The easiest way to start this is to recognize that if the hydrogen were to be reduced, then it become negatively charged, which is not as stable as the cationic form (H+). Therefore, the hydrogen gas must be oxidized and the oxygen gas reduced.
From that point, the student must correctly place the water in the oxygen’s half-reaction.

Another strategy that is effective is to determine the oxidation state changes for the elements in each half-reaction. Oxidation will occur at the anode, and so the half-reaction must have an increased change in oxidation state. At the cathode, reduction will be occurring and so the oxidation state will decrease.

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 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.

Explanation​:
By applying Hess’s law, regardless of the route by which a chemical reaction proceeds, the enthalpy change will always have the same, provided that the initial and final states of the system are identical.

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).