1 / 11

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

felix
Download Presentation

Answers Chapter 7

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

  11. Page 208, #11

More Related