Explanation​:
$\mathrm{G}$ becomes more negative as a spontaneous reaction proceeds until it reaches its minimum. At this point, there will be no net change, and the system has reached equilibrium. $\Delta \mathrm{G}$ will equal zero but $\mathrm{G}$ will not.

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

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

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

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:​
Condensation (gas to liquid) requires making bonds and therefore, releases energy. Sublimation (solid to gas) requires breaking bonds and therefore absorbs energy.

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