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Hess's Law - AP Chemistry

  1. 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}
    Answer: ΔH=−114 kJ\Delta H = -114\ \text{kJ} 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.
  2. 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?​​
    Answer: ΔH=−198 kJ\Delta H = -198\ \mathrm{kJ} 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.
  3. Use Hess's law to determine ΔH\Delta H for the target reaction: Target: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)\mathrm{Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)} Given: 1) 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} 2) CO(g)+12O2(g)→CO2(g)\mathrm{CO(g) + \tfrac{1}{2}O_2(g) \rightarrow CO_2(g)} ΔH=−283 kJ\Delta H = -283\ \text{kJ}
    Answer: ΔH=−25 kJ\Delta H = -25\ \text{kJ} This question tests the application of Hess's law for reduction reactions. For Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g), reverse the first to Fe₂O₃(s) → 2Fe(s) + ³⁄₂O₂(g) with ΔH = +824 kJ, then add three copies of reverse second 3CO₂(g) → 3CO(g) + ³⁄₂O₂(g) with ΔH = 3*(+283) = +849 kJ. Wait, second is CO + 1/2O2 → CO2, ΔH=-283, reverse CO2 → CO + 1/2O2, +283. Adding: Fe₂O₃ → 2Fe + 3/2 O2, +824; 3CO2 → 3CO + 3/2 O2, +849; total Fe₂O₃ + 3CO2 → 2Fe + 3CO + 3O2, no: to get target, actually better: target is Fe2O3 + 3CO → 2Fe + 3CO2, so reverse first +824, then add three times the reverse of second, but reverse second is CO2 → CO +1/2O2 +283, but that would put CO on right, not left. To have CO on left, use the second as is but reverse the direction for combination. Standard way: the ΔH for target is ΔHf products - reactants, but using given: reverse first to cancel Fe, then use reverse second to cancel O2 but with CO to CO2. Let's calculate: if I take reverse first: Fe2O3 → 2Fe +3/2 O2 +824. To cancel 3/2 O2, I need to add reactions that consume O2, but with CO. The second is CO +1/2O2 → CO2 -283, so three times: 3CO +3/2 O2 → 3CO2 -849. Add to reverse first: Fe2O3 + 3CO +3/2 O2 → 2Fe +3/2 O2 + 3CO2, cancel 3/2 O2, left Fe2O3 +3CO → 2Fe +3CO2, ΔH = +824 -849 = -25 kJ. Yes, matches B -25 kJ.
  4. Use Hess's law to determine ΔH\Delta H for the target reaction: Target: CH4(g)+2O2(g)→CO2(g)+2H2O(l)\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)} Given: 1) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} ΔH=−394 kJ\Delta H = -394\ \text{kJ} 2) 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} 3) C(s)+2H2(g)→CH4(g)\mathrm{C(s) + 2H_2(g) \rightarrow CH_4(g)} ΔH=−75 kJ\Delta H = -75\ \text{kJ}
    Answer: ΔH=−891 kJ\Delta H = -891\ \text{kJ} This question tests the application of Hess's law using formation enthalpies. For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), reverse the third reaction to CH₄(g) → C(s) + 2H₂(g) with ΔH = +75 kJ, then add the first reaction C(s) + O₂(g) → CO₂(g) with ΔH = -394 kJ and the second 2H₂(g) + O₂(g) → 2H₂O(l) with ΔH = -572 kJ. This cancels C(s) and 2H₂(g), leaving the target with 2O₂(g) and ΔH = +75 kJ - 394 kJ - 572 kJ = -891 kJ. The large negative value indicates a highly exothermic combustion. A tempting distractor is -741 kJ, from mistakenly halving the water formation enthalpy. When using Hess's law, verify all coefficients align with the target before summing ΔH.
  5. 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)} given: (1) N2(g)+O2(g)→2NO(g)\mathrm{N_2(g) + O_2(g) \rightarrow 2NO(g)} ΔH=+180 kJ mol−1\Delta H = +180\ \mathrm{kJ\,mol^{-1}}; (2) N2(g)+2O2(g)→2NO2(g)\mathrm{N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)} ΔH=+66 kJ mol−1\Delta H = +66\ \mathrm{kJ\,mol^{-1}}.
    Answer: ΔH=−114 kJ mol−1\Delta H = -114\ \mathrm{kJ\,mol^{-1}} This question tests the skill of applying Hess's law to determine enthalpy changes. To find ΔH for 2NO(g) + O₂(g) → 2NO₂(g), we need to manipulate the given equations to produce our target reaction. We have equation (1) N₂(g) + O₂(g) → 2NO(g) with ΔH = +180 kJ/mol and equation (2) N₂(g) + 2O₂(g) → 2NO₂(g) with ΔH = +66 kJ/mol. To eliminate N₂(g) and get our target equation, we reverse equation (1) to get 2NO(g) → N₂(g) + O₂(g) with ΔH = -180 kJ/mol, then add equation (2). This gives us 2NO(g) + N₂(g) + 2O₂(g) → N₂(g) + O₂(g) + 2NO₂(g), which simplifies to 2NO(g) + O₂(g) → 2NO₂(g) with ΔH = -180 + 66 = -114 kJ/mol. A common mistake is adding the equations without reversing, which would give +246 kJ/mol (option A), failing to recognize that we need to reverse equation (1) to cancel N₂(g). The key strategy is to identify which species need to be eliminated and determine whether equations need to be reversed, multiplied, or divided to achieve the target equation.
  6. 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}
    Answer: ΔH=−97 kJ\Delta H = -97\ \mathrm{kJ} 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.
  7. Use Hess's law to determine ΔH\Delta H for the target reaction: Target: H2S(g)+32O2(g)→SO2(g)+H2O(l)\mathrm{H_2S(g) + \tfrac{3}{2}O_2(g) \rightarrow SO_2(g) + H_2O(l)} Given: 1) H2(g)+12O2(g)→H2O(l)\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)} ΔH=−286 kJ\Delta H = -286\ \text{kJ} 2) H2(g)+S(s)→H2S(g)\mathrm{H_2(g) + S(s) \rightarrow H_2S(g)} ΔH=−21 kJ\Delta H = -21\ \text{kJ} 3) S(s)+O2(g)→SO2(g)\mathrm{S(s) + O_2(g) \rightarrow SO_2(g)} ΔH=−297 kJ\Delta H = -297\ \text{kJ}
    Answer: ΔH=−562 kJ\Delta H = -562\ \text{kJ} This question tests the application of Hess's law using formation reactions. For H₂S(g) + ³⁄₂O₂(g) → SO₂(g) + H₂O(l), reverse the second to H₂S(g) → H₂(g) + S(s) with ΔH = +21 kJ, add the first H₂(g) + ½O₂(g) → H₂O(l) with ΔH = -286 kJ, and the third S(s) + O₂(g) → SO₂(g) with ΔH = -297 kJ. This cancels H₂ and S, leaving ³⁄₂O₂ with ΔH = +21 - 286 - 297 = -562 kJ. The exothermic nature is expected. A tempting distractor is -604 kJ, from using incorrect O₂ coefficient. Verify gas balances in Hess's law for accurate results.
  8. Use Hess's law to determine ΔH\Delta H for the target reaction: Target: CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)} Given: 1) Ca(s)+12O2(g)→CaO(s)\mathrm{Ca(s) + \tfrac{1}{2}O_2(g) \rightarrow CaO(s)} ΔH=−635 kJ\Delta H = -635\ \mathrm{kJ} 2) 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\Delta H = -1207\ \mathrm{kJ} 3) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} ΔH=−394 kJ\Delta H = -394\ \mathrm{kJ} What is ΔH\Delta H for the target reaction?​​
    Answer: ΔH=+178 kJ\Delta H = +178\ \mathrm{kJ} 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 CaCO₃(s) → CaO(s) + CO₂(g), we need to manipulate the given equations. We keep equation (1) as is: Ca(s) + ½O₂(g) → CaO(s), ΔH = -635 kJ. We reverse equation (2): CaCO₃(s) → Ca(s) + C(s) + 3/2O₂(g), ΔH = +1207 kJ. We keep equation (3) as is: C(s) + O₂(g) → CO₂(g), ΔH = -394 kJ. Adding these three manipulated equations and canceling common terms gives us the target reaction, with ΔH = -635 + 1207 + (-394) = +178 kJ. A common misconception is to forget to reverse equation (2), which would lead to an incorrect negative value like -178 kJ (choice A). When solving Hess's law problems, carefully identify which equations need to be reversed to place reactants and products on the correct sides.
  9. 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\ \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}
    Answer: ΔH=+114 kJ\Delta H = +114\ \text{kJ} This question tests the application of Hess's law for reverse of previous. For 2NO₂(g) → 2NO(g) + O₂(g), this is the reverse of question 6, so ΔH should be +114 kJ, matching B. Reverse the second to 2NO₂(g) → N₂ + 2O₂ with ΔH = -66 kJ, add the first N₂ + O₂ → 2NO with ΔH = +180 kJ, cancels N₂ and O₂ (net O₂), ΔH = -66 +180 = +114 kJ. Yes. A tempting distractor is -114 kJ, from not reversing. Strategy: note reverses have opposite signs.
  10. 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}
    Answer: ΔH=−198 kJ\Delta H = -198\ \mathrm{kJ} 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) + O2O_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 O2O_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.
  11. 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}
    Answer: ΔH=−852 kJ\Delta H = -852\ \text{kJ} 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.
  12. 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}}.
    Answer: ΔH=−852 kJ mol−1\Delta H = -852\ \mathrm{kJ\,mol^{-1}} 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.
  13. 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}
    Answer: ΔH=−114 kJ\Delta H = -114\ \mathrm{kJ} 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) + O2O_2(g) → 2NO_2(g), reverse the first reaction to get 2NO(g) → N2N_2(g) + O2O_2(g) with ΔH = -180 kJ. Add the second reaction N2N_2(g) + 2O_2(g) → 2NO_2(g) with ΔH = +66 kJ. The overall enthalpy change is -180 kJ + 66 kJ = -114 kJ, with N2N_2 and O2O_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.
  14. 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}
    Answer: ΔH=−196 kJ\Delta H = -196\ \text{kJ} 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.
  15. 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?​​
    Answer: ΔH=−196 kJ\Delta H = -196\ \mathrm{kJ} 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.
  16. 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}
    Answer: ΔH=−45 kJ\Delta H = -45\ \text{kJ} 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.
  17. 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}
    Answer: ΔH=−19 kJ\Delta H = -19\ \text{kJ} 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.
  18. Use Hess's law to determine ΔH\Delta H for the target reaction: Target: 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\Delta H = -1207\ \text{kJ} 2) Ca(s)+12O2(g)→CaO(s)\mathrm{Ca(s) + \tfrac{1}{2}O_2(g) \rightarrow CaO(s)} ΔH=−635 kJ\Delta H = -635\ \text{kJ} 3) C(s)+O2(g)→CO2(g)\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} ΔH=−394 kJ\Delta H = -394\ \text{kJ}
    Answer: ΔH=+178 kJ\Delta H = +178\ \text{kJ} This question tests the application of Hess's law for decomposition reactions. For CaCO₃(s) → CaO(s) + CO₂(g), reverse the first reaction to CaCO₃(s) → Ca(s) + C(s) + ³⁄₂O₂(g) with ΔH = +1207 kJ, then add the second Ca(s) + ½O₂(g) → CaO(s) with ΔH = -635 kJ and the third C(s) + O₂(g) → CO₂(g) with ΔH = -394 kJ. This cancels Ca(s), C(s), and ³⁄₂O₂(g), yielding the target with ΔH = +1207 kJ - 635 kJ - 394 kJ = +178 kJ. The endothermic value matches limestone decomposition. A tempting distractor is -178 kJ, from not reversing the first ΔH sign. Always ensure the direction of each reaction aligns with the target in Hess's law calculations.
  19. 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}
    Answer: ΔH=−198 kJ\Delta H = -198\ \text{kJ} 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.
  20. 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}
    Answer: ΔH=−283 kJ\Delta H = -283\ \mathrm{kJ} 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) + ½O2O_2(g) → CO_2(g), reverse the second reaction to get CO(g) → C(s) + ½O2O_2(g) with ΔH = +111 kJ. Add the first reaction C(s) + O2O_2(g) → CO_2(g) with ΔH = -394 kJ. The overall enthalpy change is +111 kJ - 394 kJ = -283 kJ, with C and ½O2O_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.

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