Igcse Model Questions Vol1 2025 Chemistry Question 14

Practice objective / multiple choice question 14 from the 2025 Igcse Model Questions Vol1 Chemistry examination.

Igcse Model Questions Vol1 2025 Chemistry Objective / Multiple Choice Hard Difficulty
Question 14 IGCSE_MODEL_QUESTIONS_VOL1 • 2025 • CHEMISTRY • objective

Average bond enthalpies (kJ mol(^{-1})) are: C=C = 614, C-C = 348, C-H = 413, H-H = 436. Using these values, calculate the enthalpy change for the hydrogenation reaction \(\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)}\). The standard enthalpies of formation are: \(\Delta H_f^\circ(\mathrm{C_2H_4}) = +52\) kJ mol(^{-1}), \(\Delta H_f^\circ(\mathrm{C_2H_6}) = -84\) kJ mol(^{-1}), \(\Delta H_f^\circ(\mathrm{H_2}) = 0\). Using the formation data, calculate the enthalpy change for the same reaction. State both values and indicate which method predicts a more exothermic reaction, giving a brief explanation for any difference.

Answer Options

A. Bond method: -118 kJ/mol; Formation method: -130 kJ/mol; formation method more exothermic; difference due to experimental error.
B. Bond method: -130 kJ/mol; Formation method: -140 kJ/mol; formation method more exothermic; difference due to neglect of entropy.
C. Bond method: -124 kJ/mol; Formation method: -136 kJ/mol; bond method more exothermic; difference due to bond energy approximations.
Correct Answer D. Bond method: -124 kJ/mol; Formation method: -136 kJ/mol; formation method more exothermic; difference due to approximations in average bond enthalpies.
Correct Answer
D
Correct Option:
Bond method: -124 kJ/mol; Formation method: -136 kJ/mol; formation method more exothermic; difference due to approximations in average bond enthalpies.

About This Question

This is Igcse Model Questions Vol1 2025 Chemistry Question 14. It is one of the objective questions from the 2025 Igcse Model Questions Vol1 Chemistry examination.

Difficulty level: Hard .

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