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Answers Chapter 7

Answers Chapter 7. Page 208, Question #1. NO 2(g)  NO (g) + ½ O 2(g) Δ H = +57.0 kJ C (s) + 2H 2(g)  CH 4(g) Δ H = -74.7 kJ C 12 H 22 O 11(s) +12O 2 12CO 2 + 11H 2 O Δ H = -5156 kJ Hints: Multiply the second equation by 11, Multiply the third by equation by 12

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Answers Chapter 7

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  1. AnswersChapter 7

  2. Page 208, Question #1 • NO2(g)  NO(g) + ½ O2(g) ΔH = +57.0 kJ • C(s) + 2H2(g)  CH4(g)ΔH = -74.7 kJ • C12H22O11(s)+12O2 12CO2 + 11H2O ΔH = -5156 kJ Hints: Multiply the second equation by 11, Multiply the third by equation by 12 The answer is also kJ/mol of reactant. For kJ/mol of CO2 ÷ by 12, for kJ/mol of H2O ÷ 12 • H2SO4  H2O + SO2 + ½ O2 ΔH = +275.4 kJ • 2 H2S + 3O2 2H2O + 2 SO2ΔH = -1036.0 kJ Hint: The first and third reaction are multiplied by 2 If you want to calculate the ΔH per mole of product: Divide the whole equation by 2

  3. Page 208, #2 2CO2 • C6H12O6 + 6O2 6CO2 + 6H2O ΔH=-2803.1kJ • C2H5OH + 3O2  2CO2 + 3H2O ΔH=-1366.8kJ • 4CO2 + 6H2O 2C2H5OH+ 6O2ΔH=+2733.6kJ • C6H12O6 2C2H5OH+ 2CO2 ΔH= -69.5 kJ The overall heat of the reaction is 69.5 kJ of energy released, but that is for the formation of 2 moles of ethanol. For a single mole, divide by 2 to give: Answer: molar heat of reaction of ethanol is 34.8 kJ/mol

  4. Page 208, #3 • 2 Al + 3/2O2 Al2O3ΔH= -1675.7kJ • Fe2O3  2Fe + 3/2O2ΔH= +824.2kJ • Fe2O3 + 2Al  Al2O3 + 2Fe Target (note: no manipulation of the equations was needed before addition this time) Fe2O3 + 2Al  Al2O3 + 2 Fe ΔH=-851.5kJ

  5. Page 208, #4 • The equation given is for the formation of two moles of hydrogen atoms from normal molecular hydrogen: H2(g) 2H(g) ΔH=+434.6 kJ • The standard molar heat of formation would be found by dividing by 2, to give one mole of H atoms: ½ H2 H ΔH=+217.3 kJ/mol Answer: The molar heat of formation of atomic hydrogen is 217.3 kJ/mol.

  6. Page 208, #5 • a) 4 simple reactions • b) 3 intermediates • c) 4 activated complexes • d) 70 kJ • e) 150 kJ • f) +50 kJ • g) -40 kJ • h) endothermic • i) -10 kJ • j) exothermic 2 Simple Reaction1 Simple Reaction 2 Simple Reaction 3 Simple Reaction 4 3 1 4 1 3 2

  7. Page 209, #6a • Look up standard enthalpy of formation on page 418, then work out formation equation • ½N2+ O2 NO2ΔH=+33.2 • ½N2 + ½O2 NOΔH=+90.2 • NO + ½O2NO2 Target ½ O2 NO  ½ N2 + ½ O2 ΔH=–90.2 NO + ½ O2 NO2ΔH=–57.0

  8. Page 209, #6b • Look up standard enthalpy of formation on page 418, then work out formation equation • C + O2 CO2ΔH=-393.5 • C+ ½O2 NOΔH=-110.5 • CO + ½O2CO2 Target ½ O2 R: CO  C+ ½ O2 ΔH=+110.5 CO + ½ O2 CO2ΔH=–283

  9. Page 209, #7 • C2H2 + 5/2 O2 2 CO2 + H2O -1299.6 • C + O2  CO2 -393.5 • H2 + ½ O2  H2O -285.8 • 2 C + H2  C2H2 Target! R: 2 CO2 + H2O  C2H2 + 5/2 O2 +1299.6 2x: 2C + 2O2  2CO2 -787.0 5/2 same as 2 ½ 2 C + H2+ 2 ½ O2  C2H2+2 ½ O2 +226.8

  10. Page 208, #9 Rewrite the equations using ΔH notation • CO + H2 H2O + C -130 kJ • C + O2  CO2 -393.5 kJ • H2O  H2 + ½O2 +241.8 • 3C + 4H2  C3H8-104.7 • CO + ½ O2  CO2 -283 • C3H8 + 5O2 3 CO2 + 4 H2O Target Eliminate the unnecessary equations (equations 1 and 5) x3: 3C + 3O2  3CO2 -1180.5 Rx4: 4H2+ 2O2  4H2O -967.2 R: C3H8  3C + 4H2+104.7 Add: C3H8 + 5O2  3 CO2 + 4 H2O -2043.7

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