Question Detail

CHM 101 CHM 101 Past Questions Objective Question
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Question

Given the standard enthalpy of formation for the following in kJ/mol: ΔHf°(CaCO₃) = -1207; ΔHf°(CaO) = -636 and ΔHf°(CO₂) = -394. Calculate the enthalpy change for the reaction: CaCO₃ → CaO + CO₂.
Options
A 177 kJ/mol
Correct Answer
B 2237 kJ/mol
C -2237 kJ/mol
D -177 kJ/mol
Correct Answer

Option A is the correct answer.

Detailed Explanation

ΔH°rxn = Σ ΔHf°(products) - Σ ΔHf°(reactants) = [(-636) + (-394)] - (-1207) = (-1030) + 1207 = 177 kJ/mol.

Hint

Use ΔH°rxn = Σ ΔHf°(products) - Σ ΔHf°(reactants).

Question Info

Course: CHM 101
Topic: CHM 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1045 of 1872
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