Hess's Law and Enthalpies of Formation: Quiz

Question 1EasyWhich substance has a standard enthalpy of formation of 0 kJ/mol?

H₂O(l)CO₂(g)N₂(g)O₃(g)

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Answer: N₂(g)

N₂(g) is an element in its standard state. O₃ is not the standard state of oxygen (O₂ is), so its ΔHf° is not zero.

Question 2MediumA → B has ΔH = −50 kJ. What is ΔH for 2B → 2A?

−100 kJ+100 kJ+50 kJ−50 kJ

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Answer: +100 kJ

Reversing changes the sign (+50 kJ), and doubling multiplies by 2: +100 kJ.

Question 3MediumN₂ + O₂ → 2NO, ΔH = +182.6 kJ; 2NO + O₂ → 2NO₂, ΔH = −116.2 kJ. What is ΔH for N₂ + 2O₂ → 2NO₂?

+298.8 kJ+66.4 kJ−66.4 kJ+33.2 kJ

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Answer: +66.4 kJ

The two equations add directly (2NO cancels): +182.6 kJ + (−116.2 kJ) = +66.4 kJ.

Question 4EasyWhat is the formula for ΔH° from enthalpies of formation?

Σ ΔHf°(reactants) − Σ ΔHf°(products)Σ nΔHf°(products) − Σ nΔHf°(reactants)Σ ΔHf°(products) + Σ ΔHf°(reactants)Σ nΔHf°(products) × Σ nΔHf°(reactants)

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Answer: Σ nΔHf°(products) − Σ nΔHf°(reactants)

Products minus reactants, each multiplied by its coefficient n.

Question 5MediumCalculate ΔH° for C₂H₄(g) + H₂(g) → C₂H₆(g). (ΔHf°: C₂H₄ +52.4 kJ/mol, C₂H₆ −84.0 kJ/mol)

−31.6 kJ+136.4 kJ−136.4 kJ−84.0 kJ

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Answer: −136.4 kJ

ΔH° = (−84.0 kJ) − (+52.4 kJ + 0 kJ) = −136.4 kJ. H₂ is an element, so its ΔHf° is 0.

Question 6MediumWhy is Hess's law useful for C(s) + ½O₂(g) → CO(g)?

The reaction is too fast to measureSome CO₂ always forms, so the heat cannot be measured cleanlyCO is an elementThe reaction has no enthalpy change

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Answer: Some CO₂ always forms, so the heat cannot be measured cleanly

Hess's law gets ΔH from two reactions that can be measured: the combustions of C and of CO.

Question 7HardΔH° for CH₄ + 2O₂ → CO₂ + 2H₂O(l) is −890.5 kJ. Which ΔHf° values does the calculation need?

Only CO₂ and H₂OCH₄, CO₂ and H₂O (O₂ is zero)CH₄ and O₂ onlyNone: ΔH is measured, not calculated

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Answer: CH₄, CO₂ and H₂O (O₂ is zero)

[−393.5 kJ + 2(−285.8 kJ)] − [−74.6 kJ + 2(0 kJ)] = −890.5 kJ. O₂ is an element, so it contributes 0.