Problem: Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine ΔH  for C(diamond) ⟶ C(graphite)(1) C(diamond) + O2(g) ⟶ CO2(g)          ΔH = −395.4 kJ(2) 2CO2(g) ⟶ 2CO(g) + O2(g)              Δ H = 566.0 kJ(3) C(graphite) + O2(g) ⟶ CO2(g)          Δ H = −393.5 kJ(4) 2CO(g) ⟶ C(graphite) + CO2(g)      ΔH = −172.5 kJ

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Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine ΔH  for 

C(diamond) ⟶ C(graphite)
(1) C(diamond) + O2(g) ⟶ CO2(g)          ΔH = −395.4 kJ
(2) 2CO2(g) ⟶ 2CO(g) + O2(g)              Δ H = 566.0 kJ
(3) C(graphite) + O2(g) ⟶ CO2(g)          Δ H = −393.5 kJ
(4) 2CO(g) ⟶ C(graphite) + CO2(g)      ΔH = −172.5 kJ

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.