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

Given enthalpy of combustion data:
Substance\(\Delta H_c^\circ\) (kJ mol−1)
C2H6(g)-1561
C2H4(g)-1411
H2(g)-286
Which expression represents the enthalpy change for this process?
\[C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)\]
(A) -1561 + 286 – 1411
(B) -1411 – 286 + 1561
(C) -1411 + 286 + 1561
(D) -1561 – 286 – 1411
▶️ Answer/Explanation
Detailed solution

Using combustion data in a Hess’s law cycle:
Target: C2H4 + H2 → C2H6
Path: combust reactants and products:
C2H4 + H2 → 2CO2 + 3H2O (ΔH = -1411 – 286)
Reverse: C2H6 → 2CO2 + 3H2O (ΔH = +1561)
Add: ΔH = (-1411 – 286) + 1561

✅ Answer: (B)

Question

Which is the correct combination of \(\Delta H^\ominus\) and \(\Delta S^\ominus\) for a reaction that is only spontaneous at high temperature?
Option\(\Delta H^\ominus\)\(\Delta S^\ominus\)
A\(+\)\(+\)
B\(+\)\(-\)
C\(-\)\(+\)
D\(-\)\(-\)
▶️ Answer/Explanation
Detailed solution

The spontaneity of a reaction is determined by the Gibbs free energy change \(\Delta G\), given by \(\Delta G = \Delta H – T\Delta S\).

A reaction is spontaneous when \(\Delta G < 0\).

  • If \(\Delta H^\ominus > 0\) (endothermic) and \(\Delta S^\ominus > 0\) (increase in entropy), then \(\Delta G = \Delta H – T\Delta S\) can become negative only when the temperature \(T\) is sufficiently large, because the term \(T\Delta S\) must outweigh the positive \(\Delta H\).
  • Thus, a reaction with \(\Delta H^\ominus > 0\) and \(\Delta S^\ominus > 0\) is spontaneous only at high temperature.
  • The other sign combinations either make the reaction never spontaneous, always spontaneous, or spontaneous only at low temperature.

✅ Answer: (A)

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)

More DP Chemistry HL Paper 1 Questions..
Scroll to Top