If the enthalpy of combustion of diamond and graphite are – 395.4 mo

1.	If the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ mol-1. What is the enthalpy change for the C (graphite) → C (diamond)?
Answer» C(diamond) + O₂(g) → CO₂(g) ΔH = -395.4 kJmol^-1 …(1) C(graphite) + O₂(g) → CO₂(g) ΔH = -393.6 kJmol^-1 … (2) C(graphite) → C(diamond ) substracting (1) from (2), we get C(graphite) → C(diamond) ΔH = -393.6 kJ – (-395.4 kJ), ΔH = +1.8 kJ mol^-1

If the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ mol-1. What is the enthalpy change for the C (graphite) → C (diamond)?

Answer»

C(diamond) + O₂(g) → CO₂(g) ΔH = -395.4 kJmol^-1 …(1)

C(graphite) + O₂(g) → CO₂(g) ΔH = -393.6 kJmol^-1 … (2)

C(graphite) → C(diamond ) substracting (1) from (2), we get

C(graphite) → C(diamond) ΔH = -393.6 kJ – (-395.4 kJ), ΔH = +1.8 kJ mol^-1

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If the enthalpy of combustion of diamond and graphite are – 395.4 mol-1 and -393.6kJ mol-1. What is the enthalpy change for the C (graphite) → C (diamond)?

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