This question tests the application of Hess’s law to determine the enthalpy change for the formation of ethane. To obtain the target $2\mathrm{C(s) + 3H_2(g) \rightarrow C_2H_6(g)}$, add the first reaction $2\mathrm{C(s) + 2H_2(g) \rightarrow C_2H_4(g)}$ with $\Delta H = +52\ \mathrm{kJ}$. Add the second reaction $\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)}$ with $\Delta H = -149\ \mathrm{kJ}$. The overall enthalpy change is $+52\ \mathrm{kJ} - 149\ \mathrm{kJ} = -97\ \mathrm{kJ}$, with $\mathrm{C_2H_4}$ canceling. A tempting distractor is $-149\ \mathrm{kJ}$, 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 $\Delta H$ values.

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.

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.

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.

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.

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.

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.

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.

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.

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.