Explanation:
Born-Haber cycles are used for ionic compounds only.

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

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:
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:
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:
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 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:
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:
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:
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:
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.

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:
Combining the first equation and the reverse of the second equation gives the equation for the enthalpy change in question.
ΔH∘=−393.5−(−283.0)=−110.5

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

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:
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=−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 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) →2Ag2O(s)

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