IB DP Chemistry -Reactivity 1.2 Energy cycles in reactions - IB Style Questions For HL Paper 1A -FA 2025

IBDP Chemistry HL Paper 1A & 1B - All Chapters

Question 

Consider these equations:

\( \mathrm{S(s)+O_2(g)\rightarrow SO_2(g)} \qquad \Delta H^\circ=x\,\mathrm{kJ\,mol^{-1}} \)

\( \mathrm{2SO_2(g)+O_2(g)\rightarrow2SO_3(g)} \qquad \Delta H^\circ=y\,\mathrm{kJ\,mol^{-1}} \)

What is the value of \( \Delta H^\circ_{\mathrm{f}} \), in \( \mathrm{kJ\,mol^{-1}} \), for this reaction?

\( \mathrm{S(s)+\dfrac{3}{2}O_2(g)\rightarrow SO_3(g)} \)

(A) \(x+y\)
(B) \(x+\dfrac{1}{2}y\)
(C) \(x-\dfrac{1}{2}y\)
(D) \(x-y\)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{B}} \)

The first equation produces \(1\,\mathrm{mol}\) of \( \mathrm{SO_2} \) with enthalpy change \(x\).

The second equation produces \(2\,\mathrm{mol}\) of \( \mathrm{SO_3} \) with enthalpy change \(y\). Dividing the second equation by \(2\) gives:

\( \mathrm{SO_2(g)+\dfrac{1}{2}O_2(g)\rightarrow SO_3(g)} \)

The corresponding enthalpy change is:

\( \dfrac{1}{2}y \)

Adding this to the first equation gives the required reaction:

\( \mathrm{S(s)+\dfrac{3}{2}O_2(g)\rightarrow SO_3(g)} \)

Therefore:

\( \Delta H^\circ_{\mathrm{f}}=x+\dfrac{1}{2}y \)

Therefore, the correct answer is (B).

Question 

Which equation represents the enthalpy change of atomisation, \( \Delta H^\circ_{\mathrm{at}} \), of bromine?

(A) \( \mathrm{Br_2(l)\rightarrow Br_2(g)} \)
(B) \( \dfrac{1}{2}\mathrm{Br_2(g)\rightarrow Br(g)} \)
(C) \( \dfrac{1}{2}\mathrm{Br_2(l)\rightarrow Br(g)} \)
(D) \( \mathrm{Br_2(g)\rightarrow 2Br(g)} \)
▶️ Answer/Explanation

Correct Answer: \( \boxed{\mathrm{C}} \)

The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

Bromine exists as \( \mathrm{Br_2(l)} \) in its standard state. Therefore, the formation of one mole of gaseous bromine atoms is:

\( \dfrac{1}{2}\mathrm{Br_2(l)\rightarrow Br(g)} \)

The coefficient \( \dfrac{1}{2} \) is required because one mole of \( \mathrm{Br_2} \) contains two moles of bromine atoms.

Therefore, the correct answer is (C).

Question

Which of the following changes involve an increase in entropy?

(A) \( \text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g) \) and \( \text{Li}(g) \rightarrow \text{Li}^+(g) + e^- \)

(B) \( \text{Li}(s) \rightarrow \text{Li}(g) \) and \( \text{Li}^+(g) + \text{Cl}^-(g) \rightarrow \text{LiCl}(s) \)

(C) \( \text{Cl}(g) \rightarrow \tfrac{1}{2}\text{Cl}_2(g) \) and \( \text{Li}(s) \rightarrow \text{Li}(g) \)

(D) \( \tfrac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \) and \( \text{Li}(s) \rightarrow \text{Li}(g) \)

▶️ Answer/Explanation
Detailed solution

Entropy increases when:

  • a solid turns into a gas,
  • a liquid or gas breaks into more gas particles,
  • overall disorder increases.

Option D:

  • \( \tfrac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \): One molecule becomes two atoms → more disorder → entropy increases.
  • \( \text{Li}(s) \rightarrow \text{Li}(g) \): Solid → gas → large increase in entropy.

Both steps increase entropy, making option D the correct choice.

✅ Answer: (D)

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