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Enthalpy of Formation - AP Chemistry

  1. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given below to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: C2H4(g)+H2(g)→C2H6(g)\mathrm{C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)} Data (kJ/mol): ΔHf∘[C2H4(g)]=+52\Delta H_f^\circ[\mathrm{C_2H_4(g)}]=+52, ΔHf∘[H2(g)]=0\Delta H_f^\circ[\mathrm{H_2(g)}]=0, ΔHf∘[C2H6(g)]=−85\Delta H_f^\circ[\mathrm{C_2H_6(g)}]=-85.
    Answer: -137 kJ/mol This question tests the skill of calculating standard enthalpy of reaction using standard enthalpies of formation. Using the formula ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), we calculate: ΔH°rxn = [(-85)] - [(+52) + (0)] = -85 - 52 = -137 kJ/mol. This hydrogenation reaction is exothermic, as energy is released when the double bond in ethene is converted to a single bond in ethane. A common misconception is reversing the subtraction order, calculating reactants minus products instead, which would give +137 kJ/mol (answer A). To ensure correct calculation, always remember that products come first in the formula, then subtract reactants.
  2. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) Formation data (kJ/mol): ΔHf∘[N2(g)]=0\Delta H_f^\circ[\text{N}_2(g)] = 0, ΔHf∘[H2(g)]=0\Delta H_f^\circ[\text{H}_2(g)] = 0, ΔHf∘[NH3(g)]=−46\Delta H_f^\circ[\text{NH}_3(g)] = -46.
    Answer: -92 kJ/mol 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 ΔHfH_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.
  3. Given the following ΔHf∘\Delta H_f^\circ values at 298 K: ΔHf∘[H2O(l)]=−286 kJ mol−1\Delta H_f^\circ[\mathrm{H_2O(l)}]=-286\ \mathrm{kJ\,mol^{-1}}, ΔHf∘[H2O(g)]=−242 kJ mol−1\Delta H_f^\circ[\mathrm{H_2O(g)}]=-242\ \mathrm{kJ\,mol^{-1}}. What is ΔHrxn∘\Delta H_{\mathrm{rxn}}^\circ for H2O(l)→H2O(g)\mathrm{H_2O(l)\rightarrow H_2O(g)}?
    Answer: +44 kJ/mol 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.
  4. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. CH4(g)+2 O2(g)→CO2(g)+2 H2O(l)\mathrm{CH_4(g) + 2\,O_2(g) \rightarrow CO_2(g) + 2\,H_2O(l)} Given: ΔHf∘[CH4(g)]=−75 kJ/mol\Delta H_f^\circ[\mathrm{CH_4(g)}] = -75\ \text{kJ/mol}, ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\mathrm{O_2(g)}] = 0\ \text{kJ/mol}, ΔHf∘[CO2(g)]=−394 kJ/mol\Delta H_f^\circ[\mathrm{CO_2(g)}] = -394\ \text{kJ/mol}, ΔHf∘[H2O(l)]=−286 kJ/mol\Delta H_f^\circ[\mathrm{H_2O(l)}] = -286\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?​
    Answer: −891 kJ-891\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. The enthalpy of reaction is calculated using ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), where each value is multiplied by its stoichiometric coefficient. For products: [1×(-394)] + [2×(-286)] = -966 kJ; for reactants: [1×(-75)] + [2×(0)] = -75 kJ. Therefore, ΔH°rxn = -966 - (-75) = -891 kJ. A common error is forgetting to multiply by stoichiometric coefficients (like using only one mole of H2O instead of two), which would give -604 kJ (choice B). To avoid mistakes, always write out the full calculation showing all coefficients and double-check that you've included all species in the balanced equation.
  5. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given below to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: 2Na(s)+Cl2(g)→2NaCl(s)2\text{Na}(s) + \text{Cl}_2(g) \rightarrow 2\text{NaCl}(s) ΔHf∘[Na(s)]=0 kJ/mol\Delta H_f^\circ[\text{Na}(s)] = 0\ \text{kJ/mol} ΔHf∘[Cl2(g)]=0 kJ/mol\Delta H_f^\circ[\text{Cl}_2(g)] = 0\ \text{kJ/mol} ΔHf∘[NaCl(s)]=−411 kJ/mol\Delta H_f^\circ[\text{NaCl}(s)] = -411\ \text{kJ/mol}
    Answer: -822 kJ/mol This question tests the skill of calculating the standard enthalpy of reaction using standard enthalpies of formation. To find ΔH_rxn°, we use the formula ΔH_rxn° = Σ n ΔHfH_f°(products) - Σ m ΔHfH_f°(reactants), where n and m are stoichiometric coefficients. For products, 2 NaCl contributes 2 × -411 = -822 kJ/mol. For reactants, 2 Na contributes 0 kJ/mol and Cl2 contributes 0 kJ/mol, summing to 0 kJ/mol, so ΔH_rxn° = -822 - 0 = -822 kJ/mol. A tempting distractor is -411 kJ/mol (choice A), which arises from the misconception of forgetting to multiply the ΔHfH_f° of NaCl by its coefficient of 2. A transferable strategy is to always multiply each ΔHfH_f° by its coefficient, sum products and reactants separately, and ensure the subtraction is products minus reactants while accounting for signs.
  6. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. HCl(g)+NaOH(s)→NaCl(s)+H2O(l)\mathrm{HCl(g) + NaOH(s) \rightarrow NaCl(s) + H_2O(l)} Given: ΔHf∘[HCl(g)]=−92 kJ/mol\Delta H_f^\circ[\mathrm{HCl(g)}] = -92\ \text{kJ/mol}, ΔHf∘[NaOH(s)]=−426 kJ/mol\Delta H_f^\circ[\mathrm{NaOH(s)}] = -426\ \text{kJ/mol}, ΔHf∘[NaCl(s)]=−411 kJ/mol\Delta H_f^\circ[\mathrm{NaCl(s)}] = -411\ \text{kJ/mol}, ΔHf∘[H2O(l)]=−286 kJ/mol\Delta H_f^\circ[\mathrm{H_2O(l)}] = -286\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?
    Answer: −179 kJ-179\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. Using ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), for products: (-411) + (-286) = -697 kJ, and for reactants: (-92) + (-426) = -518 kJ. Therefore, ΔH°rxn = -697 - (-518) = -179 kJ. A common error is adding all the enthalpies of formation without considering the products minus reactants relationship, which might give -1103 kJ (choice C) by adding all four values. Remember to group products and reactants separately before subtracting, and be especially careful with signs when subtracting negative numbers.
  7. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given below to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: CO(g)+12O2(g)→CO2(g)\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) ΔHf∘[CO(g)]=−111 kJ/mol\Delta H_f^\circ[\text{CO}(g)] = -111\ \text{kJ/mol} ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\text{O}_2(g)] = 0\ \text{kJ/mol} ΔHf∘[CO2(g)]=−394 kJ/mol\Delta H_f^\circ[\text{CO}_2(g)] = -394\ \text{kJ/mol}
    Answer: -283 kJ/mol This question tests the skill of calculating the standard enthalpy of reaction using standard enthalpies of formation. To find ΔH_rxn°, we use the formula ΔH_rxn° = Σ n ΔHfH_f°(products) - Σ m ΔHfH_f°(reactants), where n and m are stoichiometric coefficients. For products, CO2 contributes -394 kJ/mol. For reactants, CO contributes -111 kJ/mol and 1/2 O2 contributes 0 kJ/mol, summing to -111 kJ/mol, so ΔH_rxn° = -394 - (-111) = -283 kJ/mol. A tempting distractor is +283 kJ/mol (choice C), which results from the misconception of switching the subtraction order between products and reactants. A transferable strategy is to always multiply each ΔHfH_f° by its coefficient, sum products and reactants separately, and ensure the subtraction is products minus reactants while accounting for signs.
  8. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. 2 SO2(g)+O2(g)→2 SO3(g)\mathrm{2\,SO_2(g) + O_2(g) \rightarrow 2\,SO_3(g)} Given: ΔHf∘[SO2(g)]=−297 kJ/mol\Delta H_f^\circ[\mathrm{SO_2(g)}] = -297\ \text{kJ/mol}, ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\mathrm{O_2(g)}] = 0\ \text{kJ/mol}, ΔHf∘[SO3(g)]=−396 kJ/mol\Delta H_f^\circ[\mathrm{SO_3(g)}] = -396\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?
    Answer: −198 kJ-198\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. Using ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), for products: 2×(-396) = -792 kJ, and for reactants: 2×(-297) + 1×(0) = -594 kJ. Therefore, ΔH°rxn = -792 - (-594) = -198 kJ. A common mistake is forgetting to account for the stoichiometric coefficient of 2 for both SO2 and SO3, which might lead to calculating with just one mole of each and getting -99 kJ (choice C). To ensure accuracy, always write out each term with its coefficient explicitly shown, and remember that O2 in its elemental form has ΔH°f = 0 kJ/mol.
  9. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. CO(g)+12 O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}\,O_2(g) \rightarrow CO_2(g)} Given: ΔHf∘[CO(g)]=−111 kJ/mol\Delta H_f^\circ[\mathrm{CO(g)}] = -111\ \text{kJ/mol}, ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\mathrm{O_2(g)}] = 0\ \text{kJ/mol}, ΔHf∘[CO2(g)]=−394 kJ/mol\Delta H_f^\circ[\mathrm{CO_2(g)}] = -394\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?
    Answer: −283 kJ-283\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. Applying ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), for products: 1×(-394) = -394 kJ, and for reactants: 1×(-111) + 0.5×(0) = -111 kJ. Therefore, ΔH°rxn = -394 - (-111) = -283 kJ. A common mistake is using the enthalpy of formation of CO2 directly as the answer (-394 kJ, choice D), forgetting that you must subtract the enthalpies of formation of the reactants. Remember that even though O2 has ΔH°f = 0, you still need to account for CO's enthalpy of formation in the calculation.
  10. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: 2SO2(g)+O2(g)→2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g) Formation data (kJ/mol): ΔHf∘[SO2(g)]=−297\Delta H_f^\circ[\text{SO}_2(g)] = -297, ΔHf∘[SO3(g)]=−396\Delta H_f^\circ[\text{SO}_3(g)] = -396, ΔHf∘[O2(g)]=0\Delta H_f^\circ[\text{O}_2(g)] = 0.
    Answer: -198 kJ/mol This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. ΔH_rxn° is the difference between summed, coefficient-multiplied ΔHfH_f° of products and reactants. Products are 2 SO3 at -396 kJ/mol each, totaling -792 kJ/mol; reactants are 2 SO2 at -297 kJ/mol (-594 kJ/mol) plus O2 at 0. Subtracting yields -792 - (-594) = -198 kJ/mol, exothermic as expected for oxidation. A tempting distractor is -99 kJ/mol, from the misconception of not multiplying coefficients for both SO2 and SO3. When calculating enthalpies, always scale each species by its coefficient and consider the reaction's balanced equation for accurate results.
  11. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given below to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: 2SO2(g)+O2(g)→2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g) ΔHf∘[SO2(g)]=−297 kJ/mol\Delta H_f^\circ[\text{SO}_2(g)] = -297\ \text{kJ/mol} ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\text{O}_2(g)] = 0\ \text{kJ/mol} ΔHf∘[SO3(g)]=−396 kJ/mol\Delta H_f^\circ[\text{SO}_3(g)] = -396\ \text{kJ/mol}
    Answer: -198 kJ/mol This question tests the skill of calculating the standard enthalpy of reaction using standard enthalpies of formation. To find ΔH_rxn°, we use the formula ΔH_rxn° = Σ n ΔHfH_f°(products) - Σ m ΔHfH_f°(reactants), where n and m are stoichiometric coefficients. For products, 2 SO3 contributes 2 × -396 = -792 kJ/mol. For reactants, 2 SO2 contributes 2 × -297 = -594 kJ/mol and O2 contributes 0 kJ/mol, summing to -594 kJ/mol, so ΔH_rxn° = -792 - (-594) = -198 kJ/mol. A tempting distractor is -99 kJ/mol (choice B), which arises from the misconception of not multiplying by coefficients, using single values instead. A transferable strategy is to always multiply each ΔHfH_f° by its coefficient, sum products and reactants separately, and ensure the subtraction is products minus reactants while accounting for signs.
  12. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given to calculate ΔHrxn∘\Delta H_{rxn}^\circ for the reaction below. C(s,graphite)+H2O(g)→CO(g)+H2(g)\text{C}(s,\text{graphite}) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + \text{H}_2(g) Given: ΔHf∘[C(s,graphite)]=0 kJ/mol\Delta H_f^\circ[\text{C}(s,\text{graphite})] = 0\ \text{kJ/mol}, ΔHf∘[H2O(g)]=−242 kJ/mol\Delta H_f^\circ[\text{H}_2\text{O}(g)] = -242\ \text{kJ/mol}, ΔHf∘[CO(g)]=−111 kJ/mol\Delta H_f^\circ[\text{CO}(g)] = -111\ \text{kJ/mol}, ΔHf∘[H2(g)]=0 kJ/mol\Delta H_f^\circ[\text{H}_2(g)] = 0\ \text{kJ/mol}.
    Answer: +131 kJ/mol This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. Using ΔHrxn∘=∑(ΔHf∘ products)−∑(ΔHf∘ reactants)\Delta H^\circ_{\text{rxn}} = \sum (\Delta H^\circ_f \text{ products}) - \sum (\Delta H^\circ_f \text{ reactants}) correctly handles this water-gas shift reaction. For products: CO(g) has ΔHf∘=−111 kJ/mol\Delta H^\circ_f = -111 \, \text{kJ/mol} and H2_2(g) has ΔHf∘=0 kJ/mol\Delta H^\circ_f = 0 \, \text{kJ/mol}, giving a total of -111 kJ/mol. For reactants: C(graphite) has ΔHf∘=0 kJ/mol\Delta H^\circ_f = 0 \, \text{kJ/mol} and H2_2O(g) has ΔHf∘=−242 kJ/mol\Delta H^\circ_f = -242 \, \text{kJ/mol}, giving a total of -242 kJ/mol. Therefore, ΔHrxn∘=−111−(−242)=+131 kJ/mol\Delta H^\circ_{\text{rxn}} = -111 - (-242) = +131 \, \text{kJ/mol}. A common mistake (choice C) is getting the sign wrong, reporting -131 kJ/mol. The positive value indicates this endothermic reaction requires energy input.
  13. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. N2(g)+3 H2(g)→2 NH3(g)\mathrm{N_2(g) + 3\,H_2(g) \rightarrow 2\,NH_3(g)} Given: ΔHf∘[N2(g)]=0 kJ/mol\Delta H_f^\circ[\mathrm{N_2(g)}] = 0\ \text{kJ/mol}, ΔHf∘[H2(g)]=0 kJ/mol\Delta H_f^\circ[\mathrm{H_2(g)}] = 0\ \text{kJ/mol}, ΔHf∘[NH3(g)]=−46 kJ/mol\Delta H_f^\circ[\mathrm{NH_3(g)}] = -46\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?
    Answer: −92 kJ-92\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. Using ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), we calculate for products: 2×(-46) = -92 kJ, and for reactants: 1×(0) + 3×(0) = 0 kJ. Therefore, ΔH°rxn = -92 - 0 = -92 kJ. A common misconception is using the enthalpy of formation value without considering stoichiometric coefficients, which would give -46 kJ (choice C) by using only one mole of NH3 instead of two. Remember that elements in their standard states (N2 and H2) always have ΔH°f = 0 kJ/mol, and always multiply enthalpies of formation by the coefficients from the balanced equation.
  14. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. 2 Al(s)+32 O2(g)→Al2O3(s)\mathrm{2\,Al(s) + \tfrac{3}{2}\,O_2(g) \rightarrow Al_2O_3(s)} Given: ΔHf∘[Al(s)]=0 kJ/mol\Delta H_f^\circ[\mathrm{Al(s)}] = 0\ \text{kJ/mol}, ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\mathrm{O_2(g)}] = 0\ \text{kJ/mol}, ΔHf∘[Al2O3(s)]=−1676 kJ/mol\Delta H_f^\circ[\mathrm{Al_2O_3(s)}] = -1676\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?
    Answer: −1676 kJ-1676\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. Using ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), for products: 1×(-1676) = -1676 kJ, and for reactants: 2×(0) + 1.5×(0) = 0 kJ. Therefore, ΔH°rxn = -1676 - 0 = -1676 kJ. A common error is dividing the result by 2 thinking the equation shows formation of 2 moles, giving -838 kJ (choice C), but the equation as written forms exactly 1 mole of Al2O3. When elements in their standard states are reactants (Al(s) and O2(g) here), the calculation simplifies to just the enthalpy of formation of the product.
  15. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given to calculate ΔHrxn∘\Delta H_{rxn}^\circ for the reaction below. 2H2O(l)→2H2(g)+O2(g)2\text{H}_2\text{O}(l) \rightarrow 2\text{H}_2(g) + \text{O}_2(g) Given: ΔHf∘[H2O(l)]=−286 kJ/mol\Delta H_f^\circ[\text{H}_2\text{O}(l)] = -286\ \text{kJ/mol}, ΔHf∘[H2(g)]=0 kJ/mol\Delta H_f^\circ[\text{H}_2(g)] = 0\ \text{kJ/mol}, ΔHf∘[O2(g)]=0 kJ/mol\Delta H_f^\circ[\text{O}_2(g)] = 0\ \text{kJ/mol}.
    Answer: +572 kJ/mol This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. This is the reverse of water formation, so we apply ΔHrxn∘=Σ(ΔHf∘ products)−Σ(ΔHf∘ reactants)\Delta H^\circ_{\text{rxn}} = \Sigma (\Delta H^\circ_f \text{ products}) - \Sigma (\Delta H^\circ_f \text{ reactants}). For products: 2H2(g)2\text{H}_2(g) and O2(g)\text{O}_2(g) are elements in standard states with ΔHf∘=0\Delta H^\circ_f = 0 kJ/mol each, giving a total of 0 kJ/mol. For reactants: 2H2O(l)2\text{H}_2\text{O}(l) has ΔHf∘=2(−286)=−572\Delta H^\circ_f = 2(-286) = -572 kJ/mol. Therefore, ΔHrxn∘=0−(−572)=+572\Delta H^\circ_{\text{rxn}} = 0 - (-572) = +572 kJ/mol. A common mistake (choice C) is using the ΔHf∘\Delta H^\circ_f for only 1 mole of water, giving +286 kJ/mol. The positive value confirms that decomposing water requires energy input, as expected for breaking stable bonds.
  16. Use standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction below. CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)} Given: ΔHf∘[CaCO3(s)]=−1207 kJ/mol\Delta H_f^\circ[\mathrm{CaCO_3(s)}] = -1207\ \text{kJ/mol}, ΔHf∘[CaO(s)]=−635 kJ/mol\Delta H_f^\circ[\mathrm{CaO(s)}] = -635\ \text{kJ/mol}, ΔHf∘[CO2(g)]=−394 kJ/mol\Delta H_f^\circ[\mathrm{CO_2(g)}] = -394\ \text{kJ/mol}. What is ΔHrxn∘\Delta H_{\text{rxn}}^\circ?
    Answer: +178 kJ+178\ \text{kJ} This question tests the skill of calculating enthalpy of reaction using standard enthalpies of formation. Applying ΔH°rxn = Σ(ΔH°f products) - Σ(ΔH°f reactants), we get for products: (-635) + (-394) = -1029 kJ, and for reactants: -1207 kJ. Therefore, ΔH°rxn = -1029 - (-1207) = +178 kJ. A common error is reversing the subtraction, calculating reactants minus products instead of products minus reactants, which would give -178 kJ (choice B). The positive value indicates this decomposition reaction is endothermic, which makes sense as energy is required to break apart the stable carbonate. Always remember the correct order: products minus reactants, and be careful with signs when subtracting negative numbers.
  17. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: H2(g)+Cl2(g)→2HCl(g)\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) Formation data (kJ/mol): ΔHf∘[H2(g)]=0\Delta H_f^\circ[\text{H}_2(g)] = 0, ΔHf∘[Cl2(g)]=0\Delta H_f^\circ[\text{Cl}_2(g)] = 0, ΔHf∘[HCl(g)]=−92\Delta H_f^\circ[\text{HCl}(g)] = -92.
    Answer: -184 kJ/mol This question tests the skill of calculating the standard enthalpy change of a reaction using standard enthalpies of formation. Using Hess's law, ΔHrxn∘\Delta H_{\text{rxn}}^\circ = Σ\Sigma (coefficients × ΔHf∘\Delta H_f^\circ products) - Σ\Sigma (coefficients × ΔHf∘\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.
  18. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given to calculate ΔHrxn∘\Delta H_{rxn}^\circ for the reaction below. CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) Given: ΔHf∘[CaCO3(s)]=−1207 kJ/mol\Delta H_f^\circ[\text{CaCO}_3(s)] = -1207\ \text{kJ/mol}, ΔHf∘[CaO(s)]=−635 kJ/mol\Delta H_f^\circ[\text{CaO}(s)] = -635\ \text{kJ/mol}, ΔHf∘[CO2(g)]=−394 kJ/mol\Delta H_f^\circ[\text{CO}_2(g)] = -394\ \text{kJ/mol}.
    Answer: +178 kJ/mol This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. Using ΔHrxn∘=Σ(ΔHf∘ products)−Σ(ΔHf∘ reactants)\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 ΔHf∘=−635\Delta H^\circ_f = -635 kJ/mol and CO2_2(g) has ΔHf∘=−394\Delta H^\circ_f = -394 kJ/mol, giving a total of -1029 kJ/mol. For reactants: CaCO3_3(s) has ΔHf∘=−1207\Delta H^\circ_f = -1207 kJ/mol. Therefore, ΔHrxn∘=−1029−(−1207)=+178\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.
  19. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given to calculate ΔHrxn∘\Delta H_{rxn}^\circ for the reaction below. N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) Given: ΔHf∘[N2(g)]=0 kJ/mol\Delta H_f^\circ[\text{N}_2(g)] = 0\ \text{kJ/mol}, ΔHf∘[H2(g)]=0 kJ/mol\Delta H_f^\circ[\text{H}_2(g)] = 0\ \text{kJ/mol}, ΔHf∘[NH3(g)]=−46 kJ/mol\Delta H_f^\circ[\text{NH}_3(g)] = -46\ \text{kJ/mol}.
    Answer: -92 kJ/mol This question tests your ability to calculate the standard enthalpy of reaction using standard enthalpies of formation. Using ΔHrxn∘=Σ(ΔHf∘ products)−Σ(ΔHf∘ reactants)\Delta H_{rxn}^\circ = \Sigma(\Delta H_f^\circ \text{ products}) - \Sigma(\Delta H_f^\circ \text{ reactants}), we need to multiply each ΔHf∘\Delta H_f^\circ by its stoichiometric coefficient. For products: 2NH3_3(g) has ΔHf∘=2(−46)=−92\Delta H_f^\circ = 2(-46) = -92 kJ/mol. For reactants: N2_2(g) and 3H2_2(g) both have ΔHf∘=0\Delta H_f^\circ = 0 kJ/mol (elements in their standard states), giving a total of 0 kJ/mol. Therefore, ΔHrxn∘=−92−0=−92\Delta H_{rxn}^\circ = -92 - 0 = -92 kJ/mol. A common mistake (choice A) is using the ΔHf∘\Delta H_f^\circ of NH3_3 without accounting for the coefficient 2, giving only -46 kJ/mol. Always multiply enthalpies of formation by their stoichiometric coefficients before summing.
  20. Use the standard enthalpies of formation, ΔHf∘\Delta H_f^\circ, given below to calculate ΔHrxn∘\Delta H_{\text{rxn}}^\circ for the reaction at 298 K: CO(g)+H2O(g)→CO2(g)+H2(g)\mathrm{CO(g) + H_2O(g) \rightarrow CO_2(g) + H_2(g)} Data (kJ/mol): ΔHf∘[CO(g)]=−111\Delta H_f^\circ[\mathrm{CO(g)}]=-111, ΔHf∘[H2O(g)]=−242\Delta H_f^\circ[\mathrm{H_2O(g)}]=-242, ΔHf∘[CO2(g)]=−394\Delta H_f^\circ[\mathrm{CO_2(g)}]=-394, ΔHf∘[H2(g)]=0\Delta H_f^\circ[\mathrm{H_2(g)}]=0.
    Answer: -41 kJ/mol This question tests the skill of calculating standard enthalpy of reaction using standard enthalpies of formation. Using the standard formula: ΔH°rxn = [(-394) + (0)] - [(-111) + (-242)] = -394 - (-353) = -394 + 353 = -41 kJ/mol. This water-gas shift reaction is slightly exothermic and is important in industrial hydrogen production. A common misconception is reversing the sign due to calculation error, which would give +41 kJ/mol (answer A). To avoid sign errors, write out each step clearly and double-check that you're subtracting the sum of reactant enthalpies from the sum of product enthalpies.

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