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# Lecture18221

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a supplemental resource for students

a supplemental resource for students

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### Transcript

• 1. Thermochemistry: Stoichiometry Lecture 18
• 2. Thermochemical equation is a balanced equation that includes the heat of reaction (ΔH rxn ).
• 3. ΔH rxn value refers to the amounts of substances and their states of matter in that specific equation.
• 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. 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. 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. 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. A sample problem using the heat of reaction to find mass of substance.
• 9. A sample problem using the mass of substance to find the heat of reaction.
• 10.
• 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?
• Plan : forward balanced equation  reverse balanced equation ; grams of NO decomposing  mols of NO decomposing  heat (kJ).
• Solution: N 2(g) + O 2(g)  2NO (g)
• ½N 2(g) + ½O 2(g)  NO (g) ΔH rxn = 90.29 kJ
• NO (g)   ½N 2(g) + ½O 2(g) ΔH rxn = -90.29 kJ
• Amount of NO = mass of NO / molar mass of NO
• ΔH = - 90.29 kJ/mol x 1.50 g / (14+16) g/mol = -4.515 kJ
• 11. THE END