IB DP Chemistry -Reactivity 1.1 Measuring enthalpy changes - IB Style Questions For HL Paper 1A -FA 2025

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

Question

Which equation correctly shows both enthalpy of formation, \(\Delta H_f^\ominus\), of a compound and enthalpy of combustion, \(\Delta H_c^\ominus\), of its elements?
(A) \(CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)\)
(B) \( \frac{1}{2}N_2(g) + \frac{1}{2}H_2(g) \rightarrow NH_3(g)\)
(C) \(2S(s) + 3O_2(g) \rightarrow 2SO_2(g)\)
(D) \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)\)
▶️ Answer/Explanation
Detailed solution

Formation reaction: elements in standard states → 1 mole of compound

Combustion of elements: elements + O2 → oxides

For H2O(l):

  • Formation: H2(g) + ½O2(g) → H2O(l)
  • Combustion of H2: H2(g) + ½O2(g) → H2O(l)
  • These are identical

✅ Answer: (D)

Question

What is the enthalpy change for the reaction shown below? (Use the data given in the table.)
CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)
ReactionΔH° (kJ mol⁻¹)
2C(graphite) + O₂(g) → 2CO(g)−222
C(graphite) + 2H₂(g) → CH₄(g)−74
2H₂(g) + O₂(g) → 2H₂O(g)−484
A. −74 − 242 + 111
B. +74 + 242 + 111
C. −74 + 242 − 111
D. +74 + 484 − 222
▶️ Answer/Explanation

Correct answer: B

• Reverse the CH₄ formation reaction → sign becomes +74 kJ mol⁻¹.
• Reverse part of the H₂O formation reaction to match 1 mol H₂O → +242 kJ mol⁻¹.
• Use half of the CO formation reaction accordingly → +111 kJ mol⁻¹.

Adding: 74 + 242 + 111 gives the correct enthalpy expression.

Question

What is the percentage error if the enthalpy of combustion of a substance is determined experimentally to be \( -2100\,\text{kJ mol}^{-1} \), while the literature value is \( -3500\,\text{kJ mol}^{-1} \)?
(A) \(80\%\)
(B) \(60\%\)
(C) \(40\%\)
(D) \(20\%\)
▶️ Answer/Explanation
Detailed solution

The formula for percentage error is:
\( \text{Percentage Error} = \left| \dfrac{\text{Experimental Value} – \text{Literature Value}}{\text{Literature Value}} \right| \times 100 \)

Experimental value \(= -2100\,\text{kJ mol}^{-1}\)
Literature value \(= -3500\,\text{kJ mol}^{-1}\)

\( \text{Percentage Error} = \left| \dfrac{-2100 – (-3500)}{-3500} \right| \times 100 \)

\( = \left| \dfrac{1400}{3500} \right| \times 100 \)

\( = \dfrac{2}{5} \times 100 = 40\% \)

✅ Answer: (C)

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