R1.2 Exam Practice

Test your knowledge on Energy Cycles

Calculator and Data Booklet permitted.

Given:
C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹
H₂(g) + ½O₂(g) → H₂O(l) ΔH = −286 kJ mol⁻¹
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890 kJ mol⁻¹

(a) State Hess's Law. [1]

(b) Calculate ΔHf° for CH₄(g) using the data above. [2]

(c) Explain why the calculated value may differ from the experimental value. [2]

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(a) The enthalpy change for a reaction is independent of the route taken, provided initial and final conditions are the same [1]

(b) ΔHf° = [−394 + 2(−286)] − (−890) = −966 + 890 = −76 kJ mol⁻¹ [1] for method, [1] for answer

(c) Standard conditions may not be met [1]; incomplete combustion or heat loss in calorimetry [1]

For NaCl: ΔHf° = −411, ΔHat°(Na) = +107, IE₁(Na) = +496, ΔHat°(½Cl₂) = +122, EA(Cl) = −349 kJ mol⁻¹

(a) Define lattice enthalpy. [1]

(b) Calculate the lattice enthalpy of NaCl. [2]

(c) Predict whether MgO has a larger or smaller lattice enthalpy than NaCl. Justify. [1]

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(a) The enthalpy change when 1 mole of ionic compound is formed from gaseous ions under standard conditions [1]

(b) −411 = +107 + 496 + 122 + (−349) + LE → LE = −411 − 376 = −787 kJ mol⁻¹ [1][1]

(c) MgO has larger (more exothermic) lattice enthalpy: higher ionic charges (2+/2−) and smaller ions [1]

Show all working.

(a) Define standard enthalpy of formation. [1]

(b) Using ΔHf° values: CO₂ = −394, H₂O = −286, C₂H₆ = −85 kJ mol⁻¹, calculate ΔHc° for ethane. [2]

(c) State why ΔHf° for O₂(g) is zero. [1]

(d) State one limitation of using ΔHf° data from tables. [1]

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(a) Enthalpy change when 1 mole of compound is formed from its elements in their standard states [1]

(b) ΔHc° = [2(−394) + 3(−286)] − (−85) = −788 − 858 + 85 = −1561 kJ mol⁻¹ [1][1]

(c) O₂ is an element in its standard state [1]

(d) Values are measured at 298 K / may not apply at reaction temperature [1]

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