Lecture18221

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Lecture18221

  1. 1. Thermochemistry: Stoichiometry Lecture 18
  2. 2. Thermochemical equation is a balanced equation that includes the heat of reaction (ΔH rxn ).
  3. 3. ΔH rxn value refers to the amounts of substances and their states of matter in that specific equation.
  4. 4. The enthalpy change has a sign. If reaction is exothermic, ΔH rxn is negative : S 8(s) + O 2(g)  8 SO 2(g) ΔH rxn < 0 If reaction is endothermic, ΔH rxn is positive : N 2(g) + O 2(g)  2N O (g) ΔH rxn > 0
  5. 5. The enthalpy change has a sign. A forward reaction has the opposite sign of the reverse reaction: 2H 2 O (l)  2H 2 O (g) + O 2(g) ΔH rxn = 572 kJ 2H 2 O (g) + O 2(g)  2H 2 O (l) ΔH rxn = -572 kJ
  6. 6. The enthalpy change has a magnitude. It is proportional to the amount of substance reacting: 2 H 2 O (g) + 1 O 2(g)  2 H 2 O (l) ΔH rxn = -572 kJ 1 H 2 O (g) + ½ O 2(g)  1 H 2 O (l) ΔH rxn = -286 kJ
  7. 7. Fractional coefficients are common in thermochemistry: 2 H 2 O (g) + 1 O 2(g)  2 H 2 O (l) ΔH rxn = -572 kJ 1 H 2 O (g) + ½ O 2(g)  1 H 2 O (l) ΔH rxn = -286 kJ
  8. 8. A sample problem using the heat of reaction to find mass of substance.
  9. 9. A sample problem using the mass of substance to find the heat of reaction.
  10. 10. <ul><li>Problem: When 1 mol of NO (g) forms from its elements, 90.29 kJ of heat is absorbed. How much heat is involved when 1.50 g of NO decomposes to its elements? </li></ul><ul><li>Plan : forward balanced equation  reverse balanced equation ; grams of NO decomposing  mols of NO decomposing  heat (kJ). </li></ul><ul><li>Solution: N 2(g) + O 2(g)  2NO (g) </li></ul><ul><li>½N 2(g) + ½O 2(g)  NO (g) ΔH rxn = 90.29 kJ </li></ul><ul><li>NO (g)   ½N 2(g) + ½O 2(g) ΔH rxn = -90.29 kJ </li></ul><ul><li>Amount of NO = mass of NO / molar mass of NO </li></ul><ul><li>ΔH = - 90.29 kJ/mol x 1.50 g / (14+16) g/mol = -4.515 kJ </li></ul>
  11. 11. THE END

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