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:
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:
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:
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:
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 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:
If a chemical reaction leads to an increase in enthalpy, the heat energy is absorbed from the surroundings.

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 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​:
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​:
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.

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:
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​:
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​:
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:​
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:
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​:
When a metal is combusted, a tremendous amount of energy is released in the form of heat or light. Since the energy is released, the reaction is exothermic, and the enthalpy is negative.
Additionally, the product of the combustion of the metal is the synthesis of the metal oxide.

This question is a medium difficulty where the student must know that combustions are exothermic and correspond to a negative enthalpy change. Further, the student must recognize that in the product compound, magnesium oxide, the magnesium has a +2 charge, and the oxygen has a −2 charge, resulting in MgO.

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