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AP Chemistry

AP Chemistry Help: Hesss Law

Review real example questions for Hesss Law in AP Chemistry.

Question 1

Use Hess’s law to determine ΔH\Delta HΔH for: 2C(s)+3H2(g)→C2H6(g)\mathrm{2C(s) + 3H_2(g) \rightarrow C_2H_6(g)}2C(s)+3H2​(g)→C2​H6​(g). The following reactions are given:

  1. 2C(s)+2H2(g)→C2H4(g)\mathrm{2C(s) + 2H_2(g) \rightarrow C_2H_4(g)}2C(s)+2H2​(g)→C2​H4​(g) ΔH=+52 kJ\Delta H = +52\ \mathrm{kJ}ΔH=+52 kJ

  2. C2H4(g)+H2(g)→C2H6(g)\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)}C2​H4​(g)+H2​(g)→C2​H6​(g) ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}ΔH=−149 kJ

  1. ΔH=−201 kJ\Delta H = -201\ \mathrm{kJ}ΔH=−201 kJ
  2. ΔH=+201 kJ\Delta H = +201\ \mathrm{kJ}ΔH=+201 kJ
  3. ΔH=−97 kJ\Delta H = -97\ \mathrm{kJ}ΔH=−97 kJ
  4. ΔH=+97 kJ\Delta H = +97\ \mathrm{kJ}ΔH=+97 kJ
  5. ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}ΔH=−149 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)}2C(s)+3H2​(g)→C2​H6​(g), add the first reaction 2C(s)+2H2(g)→C2H4(g)2\mathrm{C(s) + 2H_2(g) \rightarrow C_2H_4(g)}2C(s)+2H2​(g)→C2​H4​(g) with ΔH=+52 kJ\Delta H = +52\ \mathrm{kJ}ΔH=+52 kJ. Add the second reaction C2H4(g)+H2(g)→C2H6(g)\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)}C2​H4​(g)+H2​(g)→C2​H6​(g) with ΔH=−149 kJ\Delta H = -149\ \mathrm{kJ}ΔH=−149 kJ. The overall enthalpy change is +52 kJ−149 kJ=−97 kJ+52\ \mathrm{kJ} - 149\ \mathrm{kJ} = -97\ \mathrm{kJ}+52 kJ−149 kJ=−97 kJ, with C2H4\mathrm{C_2H_4}C2​H4​ canceling. A tempting distractor is −149 kJ-149\ \mathrm{kJ}−149 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 HΔH values.

Question 2

Use Hess’s law to find ΔH\Delta HΔH for: CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}CO(g)+21​O2​(g)→CO2​(g). The following reactions are given:

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

  2. C(s)+12O2(g)→CO(g)\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)}C(s)+21​O2​(g)→CO(g) ΔH=−111 kJ\Delta H = -111\ \mathrm{kJ}ΔH=−111 kJ

  1. ΔH=+283 kJ\Delta H = +283\ \mathrm{kJ}ΔH=+283 kJ
  2. ΔH=−505 kJ\Delta H = -505\ \mathrm{kJ}ΔH=−505 kJ
  3. ΔH=−283 kJ\Delta H = -283\ \mathrm{kJ}ΔH=−283 kJ
  4. ΔH=+505 kJ\Delta H = +505\ \mathrm{kJ}ΔH=+505 kJ
  5. ΔH=−111 kJ\Delta H = -111\ \mathrm{kJ}ΔH=−111 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 HΔH for: 2NO(g)+O2(g)→2NO2(g)\mathrm{2NO(g) + O_2(g) \rightarrow 2NO_2(g)}2NO(g)+O2​(g)→2NO2​(g). The following reactions are provided:

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

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

  1. ΔH=+114 kJ\Delta H = +114\ \mathrm{kJ}ΔH=+114 kJ
  2. ΔH=−114 kJ\Delta H = -114\ \mathrm{kJ}ΔH=−114 kJ
  3. ΔH=+246 kJ\Delta H = +246\ \mathrm{kJ}ΔH=+246 kJ
  4. ΔH=−246 kJ\Delta H = -246\ \mathrm{kJ}ΔH=−246 kJ
  5. ΔH=+66 kJ\Delta H = +66\ \mathrm{kJ}ΔH=+66 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 HΔH for the target reaction:

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

Given:

  1. S(s)+O2(g)→SO2(g)\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)}S(s)+O2​(g)→SO2​(g) ΔH=−297 kJ\Delta H = -297\ \text{kJ}ΔH=−297 kJ
  2. S(s)+32O2(g)→SO3(g)\mathrm{S(s) + \tfrac{3}{2}O_2(g) \rightarrow SO_3(g)}S(s)+23​O2​(g)→SO3​(g) ΔH=−396 kJ\Delta H = -396\ \text{kJ}ΔH=−396 kJ
  1. ΔH=−198 kJ\Delta H = -198\ \text{kJ}ΔH=−198 kJ
  2. ΔH=+198 kJ\Delta H = +198\ \text{kJ}ΔH=+198 kJ
  3. ΔH=−99 kJ\Delta H = -99\ \text{kJ}ΔH=−99 kJ
  4. ΔH=+99 kJ\Delta H = +99\ \text{kJ}ΔH=+99 kJ
  5. ΔH=−693 kJ\Delta H = -693\ \text{kJ}ΔH=−693 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 HΔH for the target reaction:

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

Given:

  1. N2(g)+O2(g)→2NO(g)\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)}N2​(g)+O2​(g)→2NO(g) ΔH=+180 kJ\Delta H = +180\ \text{kJ}ΔH=+180 kJ
  2. N2(g)+2O2(g)→2NO2(g)\mathrm{N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)}N2​(g)+2O2​(g)→2NO2​(g) ΔH=+66 kJ\Delta H = +66\ \text{kJ}ΔH=+66 kJ
  1. ΔH=+114 kJ\Delta H = +114\ \text{kJ}ΔH=+114 kJ
  2. ΔH=−114 kJ\Delta H = -114\ \text{kJ}ΔH=−114 kJ
  3. ΔH=−246 kJ\Delta H = -246\ \text{kJ}ΔH=−246 kJ
  4. ΔH=+246 kJ\Delta H = +246\ \text{kJ}ΔH=+246 kJ
  5. ΔH=+66 kJ\Delta H = +66\ \text{kJ}ΔH=+66 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 HΔH for the target reaction:

Target: N2(g)+2H2(g)→N2H4(l)\mathrm{N_2(g) + 2H_2(g) \rightarrow N_2H_4(l)}N2​(g)+2H2​(g)→N2​H4​(l)

Given:

  1. N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}N2​(g)+3H2​(g)→2NH3​(g) ΔH=−92 kJ\Delta H = -92\ \text{kJ}ΔH=−92 kJ
  2. N2H4(l)+H2(g)→2NH3(g)\mathrm{N_2H_4(l) + H_2(g) \rightarrow 2NH_3(g)}N2​H4​(l)+H2​(g)→2NH3​(g) ΔH=−47 kJ\Delta H = -47\ \text{kJ}ΔH=−47 kJ
  1. ΔH=−139 kJ\Delta H = -139\ \text{kJ}ΔH=−139 kJ
  2. ΔH=−45 kJ\Delta H = -45\ \text{kJ}ΔH=−45 kJ
  3. ΔH=+45 kJ\Delta H = +45\ \text{kJ}ΔH=+45 kJ
  4. ΔH=+139 kJ\Delta H = +139\ \text{kJ}ΔH=+139 kJ
  5. ΔH=−92 kJ\Delta H = -92\ \text{kJ}ΔH=−92 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 HΔH for the target reaction:

Target: 2H2O2(l)→2H2O(l)+O2(g)\mathrm{2H_2O_2(l) \rightarrow 2H_2O(l) + O_2(g)}2H2​O2​(l)→2H2​O(l)+O2​(g)

Given:

  1. 2H2(g)+O2(g)→2H2O(l)\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)}2H2​(g)+O2​(g)→2H2​O(l) ΔH=−572 kJ\Delta H = -572\ \text{kJ}ΔH=−572 kJ
  2. 2H2(g)+2O2(g)→2H2O2(l)\mathrm{2H_2(g) + 2O_2(g) \rightarrow 2H_2O_2(l)}2H2​(g)+2O2​(g)→2H2​O2​(l) ΔH=−376 kJ\Delta H = -376\ \text{kJ}ΔH=−376 kJ
  1. ΔH=−948 kJ\Delta H = -948\ \text{kJ}ΔH=−948 kJ
  2. ΔH=+196 kJ\Delta H = +196\ \text{kJ}ΔH=+196 kJ
  3. ΔH=−196 kJ\Delta H = -196\ \text{kJ}ΔH=−196 kJ
  4. ΔH=−572 kJ\Delta H = -572\ \text{kJ}ΔH=−572 kJ
  5. ΔH=+948 kJ\Delta H = +948\ \text{kJ}ΔH=+948 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 HΔH for the reaction: C(s)+12O2(g)→CO(g)\mathrm{C(s) + \tfrac{1}{2}O_2(g) \rightarrow CO(g)}C(s)+21​O2​(g)→CO(g) given: (1) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}C(s)+O2​(g)→CO2​(g) ΔH=−394 kJ mol−1\Delta H = -394\ \mathrm{kJ\,mol^{-1}}ΔH=−394 kJmol−1; (2) CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)}CO(g)+21​O2​(g)→CO2​(g) ΔH=−283 kJ mol−1\Delta H = -283\ \mathrm{kJ\,mol^{-1}}ΔH=−283 kJmol−1.

  1. ΔH=−677 kJ mol−1\Delta H = -677\ \mathrm{kJ\,mol^{-1}}ΔH=−677 kJmol−1
  2. ΔH=+111 kJ mol−1\Delta H = +111\ \mathrm{kJ\,mol^{-1}}ΔH=+111 kJmol−1
  3. ΔH=−111 kJ mol−1\Delta H = -111\ \mathrm{kJ\,mol^{-1}}ΔH=−111 kJmol−1
  4. ΔH=+677 kJ mol−1\Delta H = +677\ \mathrm{kJ\,mol^{-1}}ΔH=+677 kJmol−1
  5. ΔH=−394 kJ mol−1\Delta H = -394\ \mathrm{kJ\,mol^{-1}}ΔH=−394 kJmol−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 HΔH for: 2Al(s)+Fe2O3(s)→Al2O3(s)+2Fe(s)\mathrm{2Al(s) + Fe_2O_3(s) \rightarrow Al_2O_3(s) + 2Fe(s)}2Al(s)+Fe2​O3​(s)→Al2​O3​(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)}2Al(s)+23​O2​(g)→Al2​O3​(s) ΔH=−1676 kJ mol−1\Delta H = -1676\ \mathrm{kJ\,mol^{-1}}ΔH=−1676 kJmol−1; (2) 2Fe(s)+32O2(g)→Fe2O3(s)\mathrm{2Fe(s) + \tfrac{3}{2}O_2(g) \rightarrow Fe_2O_3(s)}2Fe(s)+23​O2​(g)→Fe2​O3​(s) ΔH=−824 kJ mol−1\Delta H = -824\ \mathrm{kJ\,mol^{-1}}ΔH=−824 kJmol−1.

  1. ΔH=−2500 kJ mol−1\Delta H = -2500\ \mathrm{kJ\,mol^{-1}}ΔH=−2500 kJmol−1
  2. ΔH=+852 kJ mol−1\Delta H = +852\ \mathrm{kJ\,mol^{-1}}ΔH=+852 kJmol−1
  3. ΔH=−852 kJ mol−1\Delta H = -852\ \mathrm{kJ\,mol^{-1}}ΔH=−852 kJmol−1
  4. ΔH=+2500 kJ mol−1\Delta H = +2500\ \mathrm{kJ\,mol^{-1}}ΔH=+2500 kJmol−1
  5. ΔH=−424 kJ mol−1\Delta H = -424\ \mathrm{kJ\,mol^{-1}}ΔH=−424 kJmol−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 HΔH for: CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}CaCO3​(s)→CaO(s)+CO2​(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)}Ca(s)+C(s)+23​O2​(g)→CaCO3​(s) ΔH=−1207 kJ mol−1\Delta H = -1207\ \mathrm{kJ\,mol^{-1}}ΔH=−1207 kJmol−1; (2) Ca(s)+12O2(g)→CaO(s)\mathrm{Ca(s) + \tfrac{1}{2}O_2(g) \rightarrow CaO(s)}Ca(s)+21​O2​(g)→CaO(s) ΔH=−635 kJ mol−1\Delta H = -635\ \mathrm{kJ\,mol^{-1}}ΔH=−635 kJmol−1; (3) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)}C(s)+O2​(g)→CO2​(g) ΔH=−394 kJ mol−1\Delta H = -394\ \mathrm{kJ\,mol^{-1}}ΔH=−394 kJmol−1.​

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