*Question*

What is the correct combination of \(ΔH^{\theta }\) and \(ΔS^{\theta }\) for a reaction which is only spontaneous at high temperature?

**▶️Answer/Explanation**

**Markscheme: A**

The spontaneity of a reaction is determined by the Gibbs free energy change (\(ΔG\)), which is related to enthalpy change (\(ΔH\)) and entropy change (\(ΔS\)) through the equation:

\[ΔG = ΔH – TΔS\]

For a reaction to be spontaneous at any temperature, \(ΔG\) must be negative. At high temperatures, the \(TΔS\) term becomes more significant, and the reaction becomes spontaneous if \(ΔH\) is less positive and/or \(ΔS\) is more positive.

Therefore, for a reaction to be only spontaneous at high temperatures, it suggests that the \(TΔS\) term dominates over the \(ΔH\) term at high temperatures, making \(ΔG\) negative. This could happen if \(ΔS\) is significantly positive and \(ΔH\) is not too positive.

So, a possible combination for a reaction to be only spontaneous at high temperatures is a relatively positive \(ΔH\) and a significantly positive \(ΔS\), where \(TΔS\) outweighs \(ΔH\) at higher temperatures.

*Question*

Which of these changes involve an increase in entropy?

A. \(Cl(g) + e^- \rightarrow CL^-(g)\) and \(Li(g) \rightarrow Li^+(g) + e^-\)

B. \(Li(s) \rightarrow Li(g)\) and \(Li^+(g) + Cl^-(g) \rightarrow LiCl(s)\)

C. \(Cl(g) \rightarrow \frac{1}{2} Cl_2 (g)\) and \(Li(s) \rightarrow Li(g)\)

D. \(\frac{1}{2} Cl_2(g) \rightarrow Cl (g)\) and \(Li(s) \rightarrow Li(g)\)

**▶️Answer/Explanation**

**Markscheme: D**

Entropy is a measure of the randomness or disorder of a system. An increase in the number of particles or an increase in the randomness of their arrangement generally corresponds to an increase in entropy.

Let’s analyze each change:

A. \(Cl(g) + e^- \rightarrow Cl^-(g)\) and \(Li(g) \rightarrow Li^+(g) + e^-\)

The first process involves the addition of an electron to a chlorine atom, resulting in an increase in the number of particles. This process increases entropy.

The second process involves the removal of an electron from a lithium atom, again leading to an increase in the number of particles. This process also increases entropy.

B. \(Li(s) \rightarrow Li(g)\) and \(Li^+(g) + Cl^-(g) \rightarrow LiCl(s)\)

The first process involves the transformation from a solid to a gas, which generally results in an increase in entropy.

The second process involves the formation of a solid ionic compound from gaseous ions, leading to a decrease in entropy.

C. \(Cl(g) \rightarrow \frac{1}{2} Cl_2 (g)\) and \(Li(s) \rightarrow Li(g)\)

The first process involves the combination of individual chlorine atoms into diatomic molecules, resulting in a decrease in the number of particles. This process decreases entropy.

The second process involves the transformation from a solid to a gas, which generally results in an increase in entropy.

D. \(\frac{1}{2} Cl_2(g) \rightarrow Cl (g)\) and \(Li(s) \rightarrow Li(g)\)

The first process involves the dissociation of diatomic chlorine molecules into individual atoms, resulting in an increase in the number of particles. This process increases entropy.

The second process involves the transformation from a solid to a gas, which generally results in an increase in entropy.

Based on the analysis, the changes that involve an increase in entropy are:

A. \(Cl(g) + e^- \rightarrow Cl^-(g)\) and \(Li(g) \rightarrow Li^+(g) + e^-\)

D. \(\frac{1}{2} Cl_2(g) \rightarrow Cl (g)\) and \(Li(s) \rightarrow Li(g)\)**

So, the correct answer is A and D.

**Question**

Consider the following two equations.

\({\text{2Ca(s)}} + {{\text{O}}_2}{\text{(g)}} \to {\text{2CaO(s)}}\) \(\Delta {H^\Theta } = + x{\text{ kJ}}\)

\({\text{Ca(s)}} + {\text{0.5}}{{\text{O}}_2}{\text{(g)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}} \to {\text{CaC}}{{\text{O}}_3}{\text{(s)}}\) \(\Delta {H^\Theta } = + y{\text{ kJ}}\)

What is \(\Delta {H^\Theta }\), in kJ, for the following reaction?

\({\text{CaO(s)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}} \to {\text{CaC}}{{\text{O}}_3}{\text{(s)}}\)

A. \(y – 0.5x\)

B. \(y – x\)

C. \(0.5 – y\)

D. \(x – y\)

**Answer/Explanation**

## Markscheme

A

## Examiners report

This was a common question with standard level where there was concern about the use of algebraic notation rather than actual numerical data. Algebraic notation has been used since November 2010 so candidates should be familiar with this type of question.

While one comment in HL agreed with this sentiment, the other said it was “good to use pronumerals”. In the event, it was the fifth easiest question; nearly 91% of candidates gave the correct answer and less than 6% gave B.

**Question**

_{2}(g) → C

_{2}H

_{2}(g)

B. 2 × (−394) + (−572) − (−2602)

C. 2 × (−394) + \(\frac{1}{2}\) (−572) + \(\frac{1}{2}\) (−2602)

D. 2 × (−394) + (−572) + (−2602)

**Answer/Explanation**

## Markscheme

A

**Question**

Consider the following reactions.

\[\begin{array}{*{20}{l}} {{{\text{N}}_2}({\text{g)}} + {{\text{O}}_2}{\text{(g)}} \to {\text{2NO(g)}}}&{\Delta {H^\Theta } = + 180{\text{ kJ}}} \\ {2{\text{N}}{{\text{O}}_2}({\text{g)}} \to {\text{2NO(g)}} + {{\text{O}}_2}{\text{(g)}}}&{\Delta {H^\Theta } = + 112{\text{ kJ}}} \end{array}\]

What is the \({\Delta {H^\Theta }}\) value, in kJ, for the following reaction?

\[{{\text{N}}_2}({\text{g)}} + {\text{2}}{{\text{O}}_2}{\text{(g)}} \to {\text{2N}}{{\text{O}}_2}{\text{(g)}}\]

A. \( – 1 \times ( + 180) + – 1 \times ( + 112)\)

B. \( – 1 \times ( + 180) + 1 \times ( + 112)\)

C. \(1 \times ( + 180) + – 1 \times ( + 112)\)

D. \(1 \times ( + 180) + 1 \times ( + 112)\)

**Answer/Explanation**

## Markscheme

C

## Examiners report

Two respondents stated that there was too much mathematics required to answer this question. However, candidates simply had to use Hess‟s law and were not required to determine the numerical value of the final answer. In fact, the question was the second easiest question on the paper and 82.41% of candidates got the correct answer C.