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(1) C(s) + O2(g) → CO2(9) AH = -393.51 kJ mol1 (2) S(s) + O2(g) → SO2(g) A:H = -296.83 kJ mol1 (3) C(s) + 2S(s)→ CS2(t) A:H = +89.70 kJ mol-1 %3D (4) CS2(t) + 302(g) → CO2(g) + 2SO2(g)

(1) C(s) + O2(g) → CO2(9) AH = -393.51 kJ mol1 (2) S(s) + O2(g) → SO2(g) A:H = -296.83 kJ mol1 (3) C(s) + 2S(s)→ CS2(t) A:H = +89.70 kJ mol-1 %3D (4) CS2(t) + 302(g) → CO2(g) + 2SO2(g)

Physical Chemistry
Physical Chemistry
2nd Edition
ISBN:9781133958437
Author:Ball, David W. (david Warren), BAER, Tomas
Publisher:Ball, David W. (david Warren), BAER, Tomas
Chapter2: The First Law Of Thermodynamics
Section: Chapter Questions
Problem 2.79E: The enthalpy of combustion of diamond is -395.4 kJ/mol. C s, dia O2 g CO2 g Determine the fH of C...
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(1) C(s) + Oz(g) –→ CO2(g) A:H = -393.51 kJ mol1 (2) S(s) + O2(g) → SO2(g) A:H = -296.83 kJ mol1 (3) C(s) + 2S(s) CS2(t) A:H = +89.70 kJ mol-1 (4) CS2(t) + 302(g) → CO2(g) + 2SO2(g)
Transcribed Image Text:(1) C(s) + Oz(g) –→ CO2(g) A:H = -393.51 kJ mol1 (2) S(s) + O2(g) → SO2(g) A:H = -296.83 kJ mol1 (3) C(s) + 2S(s) CS2(t) A:H = +89.70 kJ mol-1 (4) CS2(t) + 302(g) → CO2(g) + 2SO2(g)
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