AP Chemistry Quiz: Hesss Law
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Hesss LawQuestion 1 of 20

Use Hess's law to determine ΔH\Delta H for: 2C(s)+3H2(g)→C2H6(g)\mathrm{2C(s) + 3H_2(g) \rightarrow C_2H_6(g)}. The following reactions are given:

  1. 2C(s)+2H2(g)→C2H4(g)\mathrm{2C(s) + 2H_2(g) \rightarrow C_2H_4(g)} ΔH=+52 kJ\Delta H = +52\ \mathrm{kJ}

  2. C2H4(g)+H2(g)→C2H6(g)\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)} ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}

ΔH=−201 kJ\Delta H = -201\ \mathrm{kJ}
ΔH=+201 kJ\Delta H = +201\ \mathrm{kJ}
ΔH=−97 kJ\Delta H = -97\ \mathrm{kJ}
ΔH=+97 kJ\Delta H = +97\ \mathrm{kJ}
ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}
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AP Chemistry Quiz

AP Chemistry Quiz: Hesss Law

Practice Hesss Law in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Hesss Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Use Hess's law to determine ΔH\Delta H for: 2C(s)+3H2(g)→C2H6(g)\mathrm{2C(s) + 3H_2(g) \rightarrow C_2H_6(g)}. The following reactions are given:

  1. 2C(s)+2H2(g)→C2H4(g)\mathrm{2C(s) + 2H_2(g) \rightarrow C_2H_4(g)} ΔH=+52 kJ\Delta H = +52\ \mathrm{kJ}

  2. C2H4(g)+H2(g)→C2H6(g)\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)} ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}

  1. ΔH=−201 kJ\Delta H = -201\ \mathrm{kJ}
  2. ΔH=+201 kJ\Delta H = +201\ \mathrm{kJ}
  3. ΔH=−97 kJ\Delta H = -97\ \mathrm{kJ} (correct answer)
  4. ΔH=+97 kJ\Delta H = +97\ \mathrm{kJ}
  5. ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to determine the enthalpy change for the formation of ethane. To obtain the target 2C(s)+3H2(g)→C2H6(g)2\mathrm{C(s) + 3H_2(g) \rightarrow C_2H_6(g)}, add the first reaction 2C(s)+2H2(g)→C2H4(g)2\mathrm{C(s) + 2H_2(g) \rightarrow C_2H_4(g)} with ΔH=+52 kJ\Delta H = +52\ \mathrm{kJ}. Add the second reaction C2H4(g)+H2(g)→C2H6(g)\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)} with ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}. The overall enthalpy change is +52 kJ−149 kJ=−97 kJ+52\ \mathrm{kJ} - 149\ \mathrm{kJ} = -97\ \mathrm{kJ}, with C2H4\mathrm{C_2H_4} canceling. A tempting distractor is −149 kJ-149\ \mathrm{kJ}, from using only the second reaction, misconceiving it as the complete formation from elements. A transferable strategy is to chain stepwise reactions that build from elements to the final product, summing their ΔH\Delta H values.

Question 2

Use Hess's law to find ΔH\Delta H for: CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}. The following reactions are given:

  1. C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} ΔH=−394 kJ\Delta H = -394\ \mathrm{kJ}

  2. C(s)+12O2(g)→CO(g)\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)} ΔH=−111 kJ\Delta H = -111\ \mathrm{kJ}

  1. ΔH=+283 kJ\Delta H = +283\ \mathrm{kJ}
  2. ΔH=−505 kJ\Delta H = -505\ \mathrm{kJ}
  3. ΔH=−283 kJ\Delta H = -283\ \mathrm{kJ} (correct answer)
  4. ΔH=+505 kJ\Delta H = +505\ \mathrm{kJ}
  5. ΔH=−111 kJ\Delta H = -111\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to calculate the enthalpy change for the oxidation of CO to CO_2. To obtain the target CO(g) + ½O_2(g) → CO_2(g), reverse the second reaction to get CO(g) → C(s) + ½O_2(g) with ΔH = +111 kJ. Add the first reaction C(s) + O_2(g) → CO_2(g) with ΔH = -394 kJ. The overall enthalpy change is +111 kJ - 394 kJ = -283 kJ, with C and ½O_2 canceling. A tempting distractor is -111 kJ, from using the second reaction without reversing, misconceiving the reaction direction. A transferable strategy is to reverse reactions as necessary to align reactants and products with the target.

Question 3

Use Hess's law to determine ΔH\Delta H for: 2NO(g)+O2(g)→2NO2(g)\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}. The following reactions are provided:

  1. N2(g)+O2(g)→2NO(g)\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)} ΔH=+180 kJ\Delta H = +180\ \mathrm{kJ}

  2. N2(g)+2O2(g)→2NO2(g)\mathrm{N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)} ΔH=+66 kJ\Delta H = +66\ \mathrm{kJ}

  1. ΔH=+114 kJ\Delta H = +114\ \mathrm{kJ}
  2. ΔH=−114 kJ\Delta H = -114\ \mathrm{kJ} (correct answer)
  3. ΔH=+246 kJ\Delta H = +246\ \mathrm{kJ}
  4. ΔH=−246 kJ\Delta H = -246\ \mathrm{kJ}
  5. ΔH=+66 kJ\Delta H = +66\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to find the enthalpy change for the oxidation of NO to NO_2. To obtain the target 2NO(g) + O_2(g) → 2NO_2(g), reverse the first reaction to get 2NO(g) → N_2(g) + O_2(g) with ΔH = -180 kJ. Add the second reaction N_2(g) + 2O_2(g) → 2NO_2(g) with ΔH = +66 kJ. The overall enthalpy change is -180 kJ + 66 kJ = -114 kJ, with N_2 and O_2 canceling appropriately. A tempting distractor is +66 kJ, resulting from using only the second reaction without adjustment, misconceiving it as the target. A transferable strategy is to manipulate reactions by reversing and adding to match the target equation exactly.

Question 4

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2SO2(g)+O2(g)→2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}

Given:

  1. S(s)+O2(g)→SO2(g)\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)} ΔH=−297 kJ\Delta H = -297\ \text{kJ}
  2. S(s)+32O2(g)→SO3(g)\mathrm{S(s) + \tfrac{3}{2}O_2(g) \rightarrow SO_3(g)} ΔH=−396 kJ\Delta H = -396\ \text{kJ}
  1. ΔH=−198 kJ\Delta H = -198\ \text{kJ} (correct answer)
  2. ΔH=+198 kJ\Delta H = +198\ \text{kJ}
  3. ΔH=−99 kJ\Delta H = -99\ \text{kJ}
  4. ΔH=+99 kJ\Delta H = +99\ \text{kJ}
  5. ΔH=−693 kJ\Delta H = -693\ \text{kJ}

Explanation: This question tests the application of Hess's law, emphasizing the manipulation of formation reactions. For 2SO₂(g) + O₂(g) → 2SO₃(g), reverse two copies of the first reaction to 2SO₂(g) → 2S(s) + 2O₂(g) with ΔH = +594 kJ, then add two copies of the second reaction 2S(s) + 3O₂(g) → 2SO₃(g) with ΔH = -792 kJ. This cancels 2S(s) and 2O₂(g), leaving the target with net O₂(g) on left and ΔH = +594 kJ - 792 kJ = -198 kJ. The exothermic nature aligns with sulfur trioxide formation. A tempting distractor is -99 kJ, resulting from the misconception of not doubling the reactions to balance stoichiometry. Always scale reactions appropriately in Hess's law to ensure coefficients match the target equation.

Question 5

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2NO(g)+O2(g)→2NO2(g)\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}

Given:

  1. N2(g)+O2(g)→2NO(g)\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)} ΔH=+180 kJ\Delta H = +180\ \text{kJ}
  2. N2(g)+2O2(g)→2NO2(g)\mathrm{N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)} ΔH=+66 kJ\Delta H = +66\ \text{kJ}
  1. ΔH=+114 kJ\Delta H = +114\ \text{kJ}
  2. ΔH=−114 kJ\Delta H = -114\ \text{kJ} (correct answer)
  3. ΔH=−246 kJ\Delta H = -246\ \text{kJ}
  4. ΔH=+246 kJ\Delta H = +246\ \text{kJ}
  5. ΔH=+66 kJ\Delta H = +66\ \text{kJ}

Explanation: This question tests the application of Hess's law with nitrogen oxides. For 2NO(g) + O₂(g) → 2NO₂(g), reverse the first reaction to 2NO(g) → N₂(g) + O₂(g) with ΔH = -180 kJ, then add the second N₂(g) + 2O₂(g) → 2NO₂(g) with ΔH = +66 kJ. This cancels N₂(g) and O₂(g), resulting in the target with ΔH = -180 kJ + 66 kJ = -114 kJ. The exothermic shift favors NO₂ formation. A tempting distractor is +114 kJ, due to forgetting to reverse the sign of the first ΔH. A key strategy in Hess's law is to reverse signs when reversing reactions and double-check cancellations.

Question 6

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: N2(g)+2H2(g)→N2H4(l)\mathrm{N_2(g) + 2H_2(g) \rightarrow N_2H_4(l)}

Given:

  1. N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)} ΔH=−92 kJ\Delta H = -92\ \text{kJ}
  2. N2H4(l)+H2(g)→2NH3(g)\mathrm{N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)} ΔH=−47 kJ\Delta H = -47\ \text{kJ}
  1. ΔH=−139 kJ\Delta H = -139\ \text{kJ}
  2. ΔH=−45 kJ\Delta H = -45\ \text{kJ} (correct answer)
  3. ΔH=+45 kJ\Delta H = +45\ \text{kJ}
  4. ΔH=+139 kJ\Delta H = +139\ \text{kJ}
  5. ΔH=−92 kJ\Delta H = -92\ \text{kJ}

Explanation: This question tests the application of Hess's law, which states that the total enthalpy change for a reaction is the same regardless of the pathway taken. To find ΔH for N₂(g) + 2H₂(g) → N₂H₄(l), reverse the second given reaction to get 2NH₃(g) → N₂H₄(l) + H₂(g) with ΔH = +47 kJ, then add it to the first reaction N₂(g) + 3H₂(g) → 2NH₃(g) with ΔH = -92 kJ. This combination cancels out the 2NH₃(g) and one H₂(g), resulting in the target reaction with ΔH = -92 kJ + 47 kJ = -45 kJ. The negative value indicates an exothermic reaction, consistent with the formation of hydrazine. A tempting distractor is +45 kJ, which arises from the misconception of adding the enthalpies without reversing the sign for the reversed reaction. When applying Hess's law, always adjust the sign of ΔH when reversing a reaction and ensure intermediates cancel out properly.

Question 7

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2H2O2(l)→2H2O(l)+O2(g)\mathrm{2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)}

Given:

  1. 2H2(g)+O2(g)→2H2O(l)\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)} ΔH=−572 kJ\Delta H = -572\ \text{kJ}
  2. 2H2(g)+2O2(g)→2H2O2(l)\mathrm{2H_2(g) + 2O_2(g) \rightarrow 2H_2O_2(l)} ΔH=−376 kJ\Delta H = -376\ \text{kJ}
  1. ΔH=−948 kJ\Delta H = -948\ \text{kJ}
  2. ΔH=+196 kJ\Delta H = +196\ \text{kJ}
  3. ΔH=−196 kJ\Delta H = -196\ \text{kJ} (correct answer)
  4. ΔH=−572 kJ\Delta H = -572\ \text{kJ}
  5. ΔH=+948 kJ\Delta H = +948\ \text{kJ}

Explanation: This question tests the application of Hess's law for peroxide decomposition. For 2H₂O₂(l) → 2H₂O(l) + O₂(g), reverse the second reaction to 2H₂O₂(l) → 2H₂(g) + 2O₂(g) with ΔH = +376 kJ, then add the first 2H₂(g) + O₂(g) → 2H₂O(l) with ΔH = -572 kJ. This cancels 2H₂(g) and O₂(g) (net O₂ on right from extra), ΔH = +376 kJ - 572 kJ = -196 kJ. The exothermic value shows instability of peroxide. A tempting distractor is -572 kJ, from ignoring the peroxide formation enthalpy. Use Hess's law to combine formation and decomposition paths for accurate ΔH.

Question 8

Use Hess's law to determine ΔH\Delta H for the reaction: C(s)+12O2(g)→CO(g)\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)} given: (1) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} ΔH=−394 kJ mol−1\Delta H = -394\ \mathrm{kJ\,mol^{-1}}; (2) CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)} ΔH=−283 kJ mol−1\Delta H = -283\ \mathrm{kJ\,mol^{-1}}.

  1. ΔH=−677 kJ mol−1\Delta H = -677\ \mathrm{kJ\,mol^{-1}}
  2. ΔH=+111 kJ mol−1\Delta H = +111\ \mathrm{kJ\,mol^{-1}}
  3. ΔH=−111 kJ mol−1\Delta H = -111\ \mathrm{kJ\,mol^{-1}} (correct answer)
  4. ΔH=+677 kJ mol−1\Delta H = +677\ \mathrm{kJ\,mol^{-1}}
  5. ΔH=−394 kJ mol−1\Delta H = -394\ \mathrm{kJ\,mol^{-1}}

Explanation: This question tests the skill of applying Hess's law to calculate enthalpy changes for reactions. To find ΔH for C(s) + ½O₂(g) → CO(g), we need to manipulate the given equations so they add up to our target equation. Starting with equation (1) C(s) + O₂(g) → CO₂(g) with ΔH = -394 kJ/mol, and equation (2) CO(g) + ½O₂(g) → CO₂(g) with ΔH = -283 kJ/mol, we need to reverse equation (2) to get CO₂(g) → CO(g) + ½O₂(g) with ΔH = +283 kJ/mol. Adding this reversed equation to equation (1) gives us C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g), which simplifies to C(s) + ½O₂(g) → CO(g) with ΔH = -394 + 283 = -111 kJ/mol. A common error is forgetting to change the sign when reversing a reaction, which would incorrectly give -677 kJ/mol (option A). When using Hess's law, always check that your manipulated equations add up to give exactly the target equation, and remember to change the sign of ΔH when reversing a reaction.

Question 9

Use Hess's law to determine ΔH\Delta H for: 2Al(s)+Fe2O3(s)→Al2O3(s)+2Fe(s)\mathrm{2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(s)} given: (1) 2Al(s)+32O2(g)→Al2O3(s)\mathrm{2Al(s) + \tfrac{3}{2}O_2(g) \rightarrow Al_2O_3(s)} ΔH=−1676 kJ mol−1\Delta H = -1676\ \mathrm{kJ\,mol^{-1}}; (2) 2Fe(s)+32O2(g)→Fe2O3(s)\mathrm{2Fe(s) + \tfrac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)} ΔH=−824 kJ mol−1\Delta H = -824\ \mathrm{kJ\,mol^{-1}}.

  1. ΔH=−2500 kJ mol−1\Delta H = -2500\ \mathrm{kJ\,mol^{-1}}
  2. ΔH=+852 kJ mol−1\Delta H = +852\ \mathrm{kJ\,mol^{-1}}
  3. ΔH=−852 kJ mol−1\Delta H = -852\ \mathrm{kJ\,mol^{-1}} (correct answer)
  4. ΔH=+2500 kJ mol−1\Delta H = +2500\ \mathrm{kJ\,mol^{-1}}
  5. ΔH=−424 kJ mol−1\Delta H = -424\ \mathrm{kJ\,mol^{-1}}

Explanation: This question tests the skill of using Hess's law for thermite reactions. To find ΔH for 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s), we use equation (1) 2Al(s) + 3/2O₂(g) → Al₂O₃(s) with ΔH = -1676 kJ/mol and equation (2) 2Fe(s) + 3/2O₂(g) → Fe₂O₃(s) with ΔH = -824 kJ/mol. We keep equation (1) as is and reverse equation (2) to get Fe₂O₃(s) → 2Fe(s) + 3/2O₂(g) with ΔH = +824 kJ/mol. Adding these equations cancels the oxygen gas: 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s) with ΔH = -1676 + 824 = -852 kJ/mol. A common mistake is adding the enthalpies without reversing equation (2), which would give -1676 + (-824) = -2500 kJ/mol (option A), failing to recognize that Fe₂O₃ must be a reactant, not a product. The key insight for Hess's law is identifying which equations need to be reversed based on where each compound appears in the target equation.

Question 10

Use Hess's law to determine ΔH\Delta H for: CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)} given: (1) Ca(s)+C(s)+32O2(g)→CaCO3(s)\mathrm{Ca(s) + C(s) + \tfrac{3}{2}O_2(g) \rightarrow CaCO_3(s)} ΔH=−1207 kJ mol−1\Delta H = -1207\ \mathrm{kJ\,mol^{-1}}; (2) Ca(s)+12O2(g)→CaO(s)\mathrm{Ca(s) + \tfrac{1}{2}O_2(g) \rightarrow CaO(s)} ΔH=−635 kJ mol−1\Delta H = -635\ \mathrm{kJ\,mol^{-1}}; (3) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} ΔH=−394 kJ mol−1\Delta H = -394\ \mathrm{kJ\,mol^{-1}}.​

  1. ΔH=−178 kJ mol−1\Delta H = -178\ \mathrm{kJ\,mol^{-1}}
  2. ΔH=+178 kJ mol−1\Delta H = +178\ \mathrm{kJ\,mol^{-1}} (correct answer)
  3. ΔH=−246 kJ mol−1\Delta H = -246\ \mathrm{kJ\,mol^{-1}}
  4. ΔH=+246 kJ mol−1\Delta H = +246\ \mathrm{kJ\,mol^{-1}}
  5. ΔH=+2236 kJ mol−1\Delta H = +2236\ \mathrm{kJ\,mol^{-1}}

Explanation: This question tests the skill of applying Hess's law to determine enthalpy changes for decomposition reactions. To find ΔH for CaCO₃(s) → CaO(s) + CO₂(g), we need to manipulate the three given equations. We have (1) Ca(s) + C(s) + 3/2O₂(g) → CaCO₃(s) with ΔH = -1207 kJ/mol, (2) Ca(s) + 1/2O₂(g) → CaO(s) with ΔH = -635 kJ/mol, and (3) C(s) + O₂(g) → CO₂(g) with ΔH = -394 kJ/mol. To get CaCO₃ as a reactant, we reverse equation (1): CaCO₃(s) → Ca(s) + C(s) + 3/2O₂(g) with ΔH = +1207 kJ/mol. Adding equations (2) and (3) to this reversed equation gives us CaCO₃(s) → CaO(s) + CO₂(g) with ΔH = +1207 + (-635) + (-394) = +178 kJ/mol. A common error is forgetting to reverse equation (1), which would lead to an incorrect negative value like -178 kJ/mol (option A). The key strategy is to identify which compounds need to be on which side of the equation and systematically reverse equations as needed to achieve the target reaction.

Question 11

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2H2O2(l)→2H2O(l)+O2(g)\mathrm{2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)}

Given:

  1. 2H2(g)+O2(g)→2H2O(l)\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)} ΔH=−572 kJ\Delta H = -572\ \mathrm{kJ}
  2. 2H2(g)+2O2(g)→2H2O2(l)\mathrm{2H_2(g) + 2O_2(g) \rightarrow 2H_2O_2(l)} ΔH=−376 kJ\Delta H = -376\ \mathrm{kJ}

What is ΔH\Delta H for the target reaction?​​

  1. ΔH=−948 kJ\Delta H = -948\ \mathrm{kJ}
  2. ΔH=+196 kJ\Delta H = +196\ \mathrm{kJ}
  3. ΔH=−196 kJ\Delta H = -196\ \mathrm{kJ} (correct answer)
  4. ΔH=+948 kJ\Delta H = +948\ \mathrm{kJ}
  5. ΔH=−98 kJ\Delta H = -98\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to calculate the enthalpy change for a target reaction using given thermochemical equations. To find ΔH for 2H₂O₂(l) → 2H₂O(l) + O₂(g), we need to manipulate the given equations. We keep equation (1) as is: 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = -572 kJ. We reverse equation (2): 2H₂O₂(l) → 2H₂(g) + 2O₂(g), ΔH = +376 kJ. Adding these equations gives: 2H₂(g) + O₂(g) + 2H₂O₂(l) → 2H₂O(l) + 2H₂(g) + 2O₂(g). After canceling 2H₂(g) from both sides and simplifying, we get: 2H₂O₂(l) → 2H₂O(l) + O₂(g), with ΔH = -572 + 376 = -196 kJ. A common misconception is to add the ΔH values without reversing equation (2), which would give -948 kJ (choice A). When using Hess's law, always ensure that reactants and products appear on the correct sides of your manipulated equations.

Question 12

Use Hess's law to find ΔH\Delta H for: CaO(s)+CO2(g)→CaCO3(s)\mathrm{CaO(s) + CO_2(g) \rightarrow CaCO_3(s)}. The following reactions are given:

  1. CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)} ΔH=+178 kJ\Delta H = +178\ \mathrm{kJ}

  2. CaO(s)+H2O(l)→Ca(OH)2(s)\mathrm{CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(s)} ΔH=−65 kJ\Delta H = -65\ \mathrm{kJ}

  3. CaCO3(s)+H2O(l)→Ca(OH)2(s)+CO2(g)\mathrm{CaCO_3(s) + H_2O(l) \rightarrow Ca(OH)_2(s) + CO_2(g)} ΔH=+113 kJ\Delta H = +113\ \mathrm{kJ}

  1. ΔH=+65 kJ\Delta H = +65\ \mathrm{kJ}
  2. ΔH=−113 kJ\Delta H = -113\ \mathrm{kJ}
  3. ΔH=−178 kJ\Delta H = -178\ \mathrm{kJ} (correct answer)
  4. ΔH=+178 kJ\Delta H = +178\ \mathrm{kJ}
  5. ΔH=−65 kJ\Delta H = -65\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to determine the enthalpy change for the formation of CaCO_3. The target CaO(s) + CO_2(g) → CaCO_3(s) is the reverse of the first reaction, giving ΔH = -178 kJ. Alternatively, reverse the third reaction to get Ca(OH)_2(s) + CO_2(g) → CaCO_3(s) + H_2O(l) with ΔH = -113 kJ, then add the second CaO(s) + H_2O(l) → Ca(OH)_2(s) with ΔH = -65 kJ. The overall enthalpy change is -113 kJ - 65 kJ = -178 kJ, canceling Ca(OH)_2 and H_2O. A tempting distractor is +178 kJ, from not reversing the first reaction, misconceiving the reaction direction. A transferable strategy is to identify when the target is the reverse of a given reaction and simply negate the ΔH value.

Question 13

Use Hess's law to determine ΔH\Delta H for: 2SO2(g)+O2(g)→2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}. The following reactions are given:

  1. S(s)+O2(g)→SO2(g)\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)} ΔH=−297 kJ\Delta H = -297\ \mathrm{kJ}

  2. 2S(s)+3O2(g)→2SO3(g)\mathrm{2S(s) + 3O_2(g) \rightarrow 2SO_3(g)} ΔH=−792 kJ\Delta H = -792\ \mathrm{kJ}

  1. ΔH=+495 kJ\Delta H = +495\ \mathrm{kJ}
  2. ΔH=−198 kJ\Delta H = -198\ \mathrm{kJ} (correct answer)
  3. ΔH=+198 kJ\Delta H = +198\ \mathrm{kJ}
  4. ΔH=−495 kJ\Delta H = -495\ \mathrm{kJ}
  5. ΔH=−1086 kJ\Delta H = -1086\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to calculate the enthalpy change for the oxidation of SO_2 to SO_3. To obtain the target 2SO_2(g) + O_2(g) → 2SO_3(g), reverse the first reaction twice to get 2SO_2(g) → 2S(s) + 2O_2(g) with ΔH = +594 kJ (2 × +297 kJ). Add the second reaction 2S(s) + 3O_2(g) → 2SO_3(g) with ΔH = -792 kJ. The overall enthalpy change is +594 kJ - 792 kJ = -198 kJ, canceling S and O_2 appropriately. A tempting distractor is -495 kJ, resulting from not multiplying the first reaction by two, misconceiving the balancing of coefficients. A transferable strategy is to scale reactions appropriately to ensure all intermediate species cancel out.

Question 14

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: H2(g)+Cl2(g)→2HCl(g)\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)}

Given:

  1. H2(g)+Cl2(g)→2HCl(aq)\mathrm{H_2(g) + Cl_2(g) \rightarrow 2HCl(aq)} ΔH=−167 kJ\Delta H = -167\ \text{kJ}
  2. HCl(g)→HCl(aq)\mathrm{HCl(g) \rightarrow HCl(aq)} ΔH=−74 kJ\Delta H = -74\ \text{kJ}
  1. ΔH=+19 kJ\Delta H = +19\ \text{kJ}
  2. ΔH=−315 kJ\Delta H = -315\ \text{kJ}
  3. ΔH=−19 kJ\Delta H = -19\ \text{kJ} (correct answer)
  4. ΔH=+315 kJ\Delta H = +315\ \text{kJ}
  5. ΔH=−167 kJ\Delta H = -167\ \text{kJ}

Explanation: This question tests the application of Hess's law, focusing on phase changes in reactions. For H₂(g) + Cl₂(g) → 2HCl(g), reverse two copies of the second reaction to 2HCl(aq) → 2HCl(g) with ΔH = +148 kJ, then add the first reaction H₂(g) + Cl₂(g) → 2HCl(aq) with ΔH = -167 kJ. This cancels 2HCl(aq), resulting in the target with ΔH = -167 kJ + 148 kJ = -19 kJ. The slightly exothermic value reflects the gas-phase formation. A tempting distractor is -167 kJ, arising from ignoring the phase change and not including the dissolution enthalpy. In Hess's law problems involving different states, incorporate all necessary steps to match the target's phases.

Question 15

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2NO2(g)→2NO(g)+O2(g)\mathrm{2NO_2(g) \rightarrow 2NO(g) + O_2(g)}

Given:

  1. N2(g)+O2(g)→2NO(g)\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)} ΔH=+180 kJ\Delta H = +180\ \mathrm{kJ}
  2. N2(g)+2O2(g)→2NO2(g)\mathrm{N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)} ΔH=+66 kJ\Delta H = +66\ \mathrm{kJ}

What is ΔH\Delta H for the target reaction?​​

  1. ΔH=−114 kJ\Delta H = -114\ \mathrm{kJ}
  2. ΔH=+114 kJ\Delta H = +114\ \mathrm{kJ} (correct answer)
  3. ΔH=−246 kJ\Delta H = -246\ \mathrm{kJ}
  4. ΔH=+246 kJ\Delta H = +246\ \mathrm{kJ}
  5. ΔH=+90 kJ\Delta H = +90\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to calculate the enthalpy change for a target reaction using given thermochemical equations. To find ΔH for 2NO₂(g) → 2NO(g) + O₂(g), we need to manipulate the given equations. We keep equation (1) as is: N₂(g) + O₂(g) → 2NO(g), ΔH = +180 kJ. We reverse equation (2): 2NO₂(g) → N₂(g) + 2O₂(g), ΔH = -66 kJ. Adding these equations gives: N₂(g) + O₂(g) + 2NO₂(g) → 2NO(g) + N₂(g) + 2O₂(g). After canceling N₂(g) from both sides and simplifying, we get: 2NO₂(g) → 2NO(g) + O₂(g), with ΔH = +180 + (-66) = +114 kJ. A common misconception is to subtract the values incorrectly or reverse the wrong equation, potentially getting -114 kJ (choice A). When applying Hess's law, double-check your arithmetic and ensure you've correctly identified which equations to reverse.

Question 16

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2SO2(g)+O2(g)→2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}

Given:

  1. S(s)+O2(g)→SO2(g)\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)} ΔH=−297 kJ\Delta H = -297\ \mathrm{kJ}
  2. 2S(s)+3O2(g)→2SO3(g)\mathrm{2S(s) + 3O_2(g) \rightarrow 2SO_3(g)} ΔH=−792 kJ\Delta H = -792\ \mathrm{kJ}

What is ΔH\Delta H for the target reaction?​​

  1. ΔH=−198 kJ\Delta H = -198\ \mathrm{kJ} (correct answer)
  2. ΔH=+198 kJ\Delta H = +198\ \mathrm{kJ}
  3. ΔH=−495 kJ\Delta H = -495\ \mathrm{kJ}
  4. ΔH=+495 kJ\Delta H = +495\ \mathrm{kJ}
  5. ΔH=−1089 kJ\Delta H = -1089\ \mathrm{kJ}

Explanation: This question tests the application of Hess's law to calculate the enthalpy change for a target reaction using given thermochemical equations. To find ΔH for 2SO₂(g) + O₂(g) → 2SO₃(g), we need to manipulate the given equations appropriately. We can multiply equation (1) by 2 to get: 2S(s) + 2O₂(g) → 2SO₂(g), ΔH = -594 kJ. We keep equation (2) as is: 2S(s) + 3O₂(g) → 2SO₃(g), ΔH = -792 kJ. Now we reverse the first manipulated equation to get: 2SO₂(g) → 2S(s) + 2O₂(g), ΔH = +594 kJ. Adding this to equation (2) gives us the target reaction, and the ΔH values sum to +594 + (-792) = -198 kJ. A common misconception is to subtract the ΔH values incorrectly or forget to multiply equation (1) by 2, which might lead to +495 kJ (choice D). When applying Hess's law, carefully track coefficients when multiplying equations and remember to reverse the sign when reversing reactions.

Question 17

Use Hess's law to determine ΔH\Delta H for: 2SO2(g)+O2(g)→2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)} given: (1) S(s)+O2(g)→SO2(g)\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)} ΔH=−297 kJ mol−1\Delta H = -297\ \mathrm{kJ\,mol^{-1}}; (2) 2S(s)+3O2(g)→2SO3(g)\mathrm{2S(s) + 3O_2(g) \rightarrow 2SO_3(g)} ΔH=−792 kJ mol−1\Delta H = -792\ \mathrm{kJ\,mol^{-1}}.

  1. ΔH=−198 kJ mol−1\Delta H = -198\ \mathrm{kJ\,mol^{-1}} (correct answer)
  2. ΔH=+198 kJ mol−1\Delta H = +198\ \mathrm{kJ\,mol^{-1}}
  3. ΔH=−495 kJ mol−1\Delta H = -495\ \mathrm{kJ\,mol^{-1}}
  4. ΔH=+495 kJ mol−1\Delta H = +495\ \mathrm{kJ\,mol^{-1}}
  5. ΔH=−792 kJ mol−1\Delta H = -792\ \mathrm{kJ\,mol^{-1}}

Explanation: This question tests the skill of using Hess's law to determine enthalpy changes. To find ΔH for 2SO₂(g) + O₂(g) → 2SO₃(g), we use equation (1) S(s) + O₂(g) → SO₂(g) with ΔH = -297 kJ/mol and equation (2) 2S(s) + 3O₂(g) → 2SO₃(g) with ΔH = -792 kJ/mol. We need to eliminate solid sulfur from our equations. Multiplying equation (1) by 2 gives 2S(s) + 2O₂(g) → 2SO₂(g) with ΔH = -594 kJ/mol. Reversing this gives 2SO₂(g) → 2S(s) + 2O₂(g) with ΔH = +594 kJ/mol. Adding equation (2) to this reversed equation gives 2SO₂(g) + 2S(s) + 3O₂(g) → 2S(s) + 2O₂(g) + 2SO₃(g), which simplifies to 2SO₂(g) + O₂(g) → 2SO₃(g) with ΔH = +594 + (-792) = -198 kJ/mol. A common error is adding the equations without proper manipulation, which might give -495 kJ/mol (option C) by incorrectly subtracting 297 from 792. When using Hess's law, always ensure that intermediate species (like S(s) here) cancel out completely.

Question 18

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: C(s)+H2O(g)→CO(g)+H2(g)\mathrm{C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)}

Given:

  1. C(s)+12O2(g)→CO(g)\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)} ΔH=−111 kJ\Delta H = -111\ \text{kJ}
  2. H2(g)+12O2(g)→H2O(g)\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(g)} ΔH=−242 kJ\Delta H = -242\ \text{kJ}
  1. ΔH=−353 kJ\Delta H = -353\ \text{kJ}
  2. ΔH=+353 kJ\Delta H = +353\ \text{kJ}
  3. ΔH=−131 kJ\Delta H = -131\ \text{kJ}
  4. ΔH=+131 kJ\Delta H = +131\ \text{kJ} (correct answer)
  5. ΔH=−111 kJ\Delta H = -111\ \text{kJ}

Explanation: This question tests the application of Hess's law, which allows calculation of ΔH by combining given reactions. For the target C(s) + H₂O(g) → CO(g) + H₂(g), reverse the second reaction to H₂O(g) → H₂(g) + ½O₂(g) with ΔH = +242 kJ, then add it to the first reaction C(s) + ½O₂(g) → CO(g) with ΔH = -111 kJ. This cancels the ½O₂(g), yielding the target with ΔH = -111 kJ + 242 kJ = +131 kJ. The positive value shows the reaction is endothermic, as expected for the water-gas reaction. A tempting distractor is -131 kJ, stemming from the misconception of subtracting instead of adding the reversed ΔH. To use Hess's law effectively, manipulate reactions to match the target and sum the adjusted ΔH values accordingly.

Question 19

Use Hess's law to determine ΔH\Delta H for the target reaction:

Target: 2Al(s)+Fe2O3(s)→Al2O3(s)+2Fe(s)\mathrm{2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(s)}

Given:

  1. 2Al(s)+32O2(g)→Al2O3(s)\mathrm{2Al(s) + \tfrac{3}{2}O_2(g) \rightarrow Al_2O_3(s)} ΔH=−1676 kJ\Delta H = -1676\ \text{kJ}
  2. 2Fe(s)+32O2(g)→Fe2O3(s)\mathrm{2Fe(s) + \tfrac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)} ΔH=−824 kJ\Delta H = -824\ \text{kJ}
  1. ΔH=−2500 kJ\Delta H = -2500\ \text{kJ}
  2. ΔH=−852 kJ\Delta H = -852\ \text{kJ} (correct answer)
  3. ΔH=+852 kJ\Delta H = +852\ \text{kJ}
  4. ΔH=+2500 kJ\Delta H = +2500\ \text{kJ}
  5. ΔH=−1676 kJ\Delta H = -1676\ \text{kJ}

Explanation: This question tests the application of Hess's law for thermite reaction. For 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s), reverse the second to Fe₂O₃(s) → 2Fe(s) + ³⁄₂O₂(g) with ΔH = +824 kJ, add the first 2Al(s) + ³⁄₂O₂(g) → Al₂O₃(s) with ΔH = -1676 kJ. Cancels O₂, ΔH = +824 -1676 = -852 kJ. Exothermic, as known. Tempting +852 kJ from sign error. Strategy: reverse signs correctly.

Question 20

Use Hess's law to determine ΔH\Delta H for the target reaction:\n\nTarget: 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)\mathrm{4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)}\n\nGiven:\n1) N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)} ΔH=−92 kJ\Delta H = -92\ \text{kJ}\n2) N2(g)+O2(g)→2NO(g)\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)} ΔH=+180 kJ\Delta H = +180\ \text{kJ}\n3) 2H2(g)+O2(g)→2H2O(g)\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} ΔH=−484 kJ\Delta H = -484\ \text{kJ}

  1. ΔH=−1816 kJ\Delta H = -1816\ \text{kJ}
  2. ΔH=−1368 kJ\Delta H = -1368\ \text{kJ}
  3. ΔH=+1368 kJ\Delta H = +1368\ \text{kJ}
  4. ΔH=−908 kJ\Delta H = -908\ \text{kJ} (correct answer)
  5. ΔH=+908 kJ\Delta H = +908\ \text{kJ}

Explanation: This question tests the application of Hess's law for the Ostwald process. For 4\mathrm{NH_3(g) + 5O_2(g) \rightarrow 4\mathrm{NO(g) + 6H_2O(g)}, reverse two copies of the first to 2NH3(g)→N2(g)+3H2(g)2\mathrm{NH_3(g) \rightarrow N_2(g) + 3H_2(g)} with ΔH=+92 kJ\Delta H = +92\ \text{kJ}, but for four, 4NH3(g)→2N2(g)+6H2(g)4\mathrm{NH_3(g) \rightarrow 2N_2(g) + 6H_2(g)} with ΔH=+184 kJ\Delta H = +184\ \text{kJ}, add two copies of the second 2N2(g)+2O2(g)→4NO(g)2\mathrm{N_2(g) + 2O_2(g) \rightarrow 4NO(g)} with ΔH=+360 kJ\Delta H = +360\ \text{kJ}, and three copies of the third 6H2(g)+3O2(g)→6H2O(g)6\mathrm{H_2(g) + 3O_2(g) \rightarrow 6H_2O(g)} with ΔH=−1452 kJ\Delta H = -1452\ \text{kJ}. This cancels N₂ and H₂, leaving 5O25\mathrm{O_2} with ΔH=+184+360−1452=−908 kJ\Delta H = +184 + 360 - 1452 = -908\ \text{kJ}. The exothermic value drives the reaction. A tempting distractor is -1816 kJ, from not scaling properly. Scale reactions in Hess's law to match target stoichiometry.