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Enduring Understanding 5.C.2: Reaction Energies

  • During chemical reactions, chemical bonds are broken and/or formed.
  • The energy required or released in a chemical reaction can be estimated by adding up:
  • The energy required to break the bonds in reactants.
  • The energy released by the newly formed bonds in the products
  • If the energy released by the newly formed bonds is more than that required to break bonds in the reactants, the reaction will be exothermic. If the converse, the reaction will be endothermic.
  • Potential energy:
  • In an exothermic reaction, the products are at a lower potential energy than the reactants
  • In and endothermic reaction, the products are at a higher potential energy than the reactants.
  • Since energy is conserved in a closed system, in an endothermic reaction, the products are at a lower potential energy; they therefore must have more kinetic energy.
  • This means products in an exothermic reaction are at a higher temperature.
  • Over time, they will exchange heat (thermal energy) with their surroundings and come to thermal equilibrium.
  • Hess's Law states that the heat of any chemical reaction is equal to the sum of the heats of reaction of any combination of reactions that, when added up, are equivalent to the overall reaction.
  • Because of Hess's Law, tables of standard enthalpies of formation can be used to calculate the enthalpy of reactions.
  • Reversing a reaction reverses the enthalpy change, e.g. a reaction that has a ΔH of - 200 kJ/mol in the forward direction has a ΔH of +200 kJ/mol in the reverse direction.
  • Example: Given the following reactions:
    • C(s) + O2(s) → CO2(g) ΔH = -393.5 kJ/mol
    2H2(g) + O2(g) → 2H2O(g) ΔH = -483.6 kJ/mol
  • Calculate the enthalpy of the following reaction:
    • C(s) + 2H2O(g) → CO2(g) + 2H2(g)
  • To determine the enthalpy from the given reactions, simply reverse the second reaction:
    • C(s) + O2(s) → CO2(g) ΔH = -393.5 kJ/mol
    2H2O(g) →2H2(g) + O2(g) ΔH = +483.6 kJ/mol
  • Add the two reactions, and the enthalpies:
    • C(s) + O2(s) + 2H2O(g) → 2H2(g) + O2(g) + CO2(g) ΔH = -393.5 kJ/mol + 483.6 kJ/mol
  • Removing oxygen from both sides:
    • C(s) + 2H2O(g) →CO2(g) + 2H2(g) ΔH = + 90.1 kJ/mol
  • Sample Problem: Calculate the enthalpy of the reaction:
    • H2(g) + Cl2(g) → 2HCl(g)
  • Given the following bond dissociation energies:
  • H2, 436 kJ/mol; Cl2, 242 kJ/mol; HCl, 431 kJ/mol.
  • The bond dissociation reactions in question are:
  • H2(g) → 2H(g), ΔH = +436 kJ/mol
  • Cl2 (g) → 2Cl(g), ΔH = +242 kJ/mol
  • HCl (g) → H(g) + Cl(g), ΔH = +431 kJ/mol
  • Invert the last reaction and multiply by 2 (enthalpy becomes negative, and doubles)
    • 2H(g) + 2Cl(g) → 2 HCl (g), ΔH = -862 kJ/mol
  • Now add this to the two other reactions:
    • 2H(g) + 2Cl(g) → 2 HCl (g), ΔH = -862 kJ/mol
    H2(g) → 2H(g), ΔH = +436 kJ/mol
    Cl2 (g)→ 2Cl(g), ΔH = +242 kJ/mol
    H2(g) + Cl2 (g) + 2H(g) + 2Cl(g) → 2 HCl (g) + 2H(g) + 2Cl(g) ΔH = -184 kJ/mol
  • The enthalpy of the reaction is ΔH = -184 kJ/mol



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Bond lengths and dissociation energies



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