This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. For water vaporization: ΔH°rxn = [1(-242)] - [1(-286)] = -242 - (-286) = -242 + 286 = +44 kJ/mol. The positive value confirms that vaporization is endothermic, requiring energy input. A common mistake would be to calculate -44 kJ/mol (choice B) by subtracting in the wrong order, confusing the direction of the phase change. Remember that ΔH°rxn = products minus reactants, and vaporization always requires energy input, so ΔH must be positive.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. ΔH_rxn° involves summing coefficient-adjusted ΔH_f° for products and subtracting that for reactants. Products 2 NO2 at +33 kJ/mol total +66 kJ/mol; reactants 2 NO at +90 kJ/mol (+180 kJ/mol) plus O2 (0), so +66 - +180 = -114 kJ/mol, exothermic. This fits NO to NO2 conversion. A tempting distractor is +114 kJ/mol, from the misconception of subtracting products from reactants. Consistently apply the products-minus-reactants rule and verify with known reaction energetics to ensure accuracy.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. Using Hess's law, $\Delta H_{\text{rxn}}^\circ$ = $\Sigma$ (coefficients × $\Delta H_f^\circ$ products) - $\Sigma$ (coefficients × $\Delta H_f^\circ$ reactants). For this, products 2 HCl at -92 kJ/mol sum to -184 kJ/mol, reactants H2 and Cl2 both 0, so -184 - 0 = -184 kJ/mol, exothermic. This reflects the energy release in forming HCl bonds. A tempting distractor is -92 kJ/mol, due to the misconception of omitting the coefficient 2 for HCl. A key strategy is to list all terms with coefficients before summing to ensure no multiplication errors in enthalpy calculations.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. ΔH_rxn° equals the sum of ΔH_f° for products minus that for reactants, with coefficients applied. For this decomposition, products CaO (-635 kJ/mol) and CO2 (-394 kJ/mol) sum to -1029 kJ/mol, and reactant CaCO3 is -1207 kJ/mol, so -1029 - (-1207) = +178 kJ/mol, indicating endothermic. This matches the known endothermic nature of limestone decomposition. A tempting distractor is -178 kJ/mol, stemming from the misconception of reversing the subtraction order. To compute ΔH_rxn° reliably, consistently subtract reactants' enthalpy sum from products' and verify the sign aligns with reaction type.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. The standard enthalpy of reaction, $\Delta H_{\text{rxn}}^\circ$, is found by subtracting the sum of the standard enthalpies of formation of the reactants, each multiplied by their stoichiometric coefficients, from that of the products. For this reaction, the products' enthalpies sum to $-394 \text{ kJ/mol}$ for CO2 plus 2 times $-286 \text{ kJ/mol}$ for H2O(l), equaling $-966 \text{ kJ/mol}$, while the reactants sum to $-75 \text{ kJ/mol}$ for CH4 and 0 for O2. Subtracting gives $-966 - (-75) = -891 \text{ kJ/mol}$, indicating an exothermic reaction. A tempting distractor is $-605 \text{ kJ/mol}$, which arises from the misconception of forgetting to multiply the enthalpy of formation of H2O by its coefficient of 2. Always remember to multiply each enthalpy of formation by the stoichiometric coefficient and ensure the states of matter match the given data when calculating $\Delta H_{\text{rxn}}^\circ$.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. ΔH_rxn° is computed as products' enthalpy sum minus reactants', with stoichiometric multipliers. Product C2H6 is -85 kJ/mol; reactants C2H4 (+52 kJ/mol) and H2 (0), so -85 - 52 = -137 kJ/mol, exothermic hydrogenation. This aligns with alkene to alkane conversion releasing energy. A tempting distractor is +137 kJ/mol, from the misconception of subtracting products from reactants. Remember to always use the formula products minus reactants and check the sign against reaction thermodynamics for verification.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. The standard enthalpy of reaction, ΔH_rxn°, is calculated as the sum of the products' formation enthalpies minus the sum of the reactants', each scaled by coefficients. Here, the products are 2 NH3 with ΔH_f° of -46 kJ/mol each, summing to -92 kJ/mol, while reactants N2 and H2 both have 0 kJ/mol. Thus, ΔH_rxn° = -92 - 0 = -92 kJ/mol, showing the synthesis of ammonia is exothermic. A tempting distractor is -46 kJ/mol, resulting from the misconception of not multiplying the enthalpy of NH3 by 2. Always double-check that you've applied the stoichiometric coefficients correctly to avoid undercounting multi-mole products in enthalpy calculations.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. For reverse reactions like this decomposition, ΔH_rxn° = Σ ΔH_f° products - Σ ΔH_f° reactants, including fractional coefficients. Products H2 (0) and 0.5 O2 (0) sum to 0; reactant H2O(g) is -242 kJ/mol, so 0 - (-242) = +242 kJ/mol, endothermic. This is the enthalpy of vaporization inverted but for gas phase. A tempting distractor is -242 kJ/mol, arising from the misconception of not reversing the sign for decomposition. To avoid errors, recognize that decomposition ΔH is the negative of formation ΔH for the compound, and apply coefficients precisely.

This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. ΔH_rxn° is the sum of products' ΔH_f° times coefficients minus that of reactants. Products 2 NaOH at -426 kJ/mol total -852 kJ/mol; reactants Na2O (-414 kJ/mol) and H2O (-286 kJ/mol) sum to -700 kJ/mol, so -852 - (-700) = -152 kJ/mol, exothermic. This reflects the hydration energy release. A tempting distractor is +152 kJ/mol, from the misconception of inverting the subtraction. To compute correctly, write out each term explicitly and use a calculator for summation to prevent sign or arithmetic errors.

This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. Using $\Delta H^\circ_{\text{rxn}} = \Sigma(\Delta H^\circ_f \text{ products}) - \Sigma(\Delta H^\circ_f \text{ reactants})$, we calculate the enthalpy change for decomposition. For products: CaO(s) has $\Delta H^\circ_f = -635$ kJ/mol and CO$_2$(g) has $\Delta H^\circ_f = -394$ kJ/mol, giving a total of -1029 kJ/mol. For reactants: CaCO$3$(s) has $\Delta H^\circ_f = -1207$ kJ/mol. Therefore, $\Delta H^\circ{\text{rxn}} = -1029 - (-1207) = +178$ kJ/mol. A common mistake (choice B) is getting the sign wrong by subtracting in the wrong order, giving -178 kJ/mol. The positive value makes sense because decomposition reactions typically require energy input.