If the enthalpy of combustion of diamond and graphite are `-395.4 kJ mol^(-1)` and `-393.6 kJ mol^(-1)`, what is enthalpy change for the `C("graphite") rarr C("diamond")`?

If the enthalpy of combustion of diamond and graphite are `-395.4 kJ m

1.	If the enthalpy of combustion of diamond and graphite are `-395.4 kJ mol^(-1)` and `-393.6 kJ mol^(-1)`, what is enthalpy change for the `C("graphite") rarr C("diamond")`?
Answer» `C("diamond") +O_(2)(g) rarr CO_(2)(g) DeltaH =- 395.4 kJ mol^(-1)………(i)` `C("graphite") +O_(2)(g) rarr CO_(2)(g) DeltaH =- 393.6 kJ mol^(-1) …..(ii)` `C("graphite") rarr C ("diamond")` Subtracting equation (i) from equation (ii), we get `C("graphite") rarr C("diamond")` `DeltaH = -393.6 kJ -(-395.4 kJ)` `DeltaH = +1.8 kJ mol^(-1)`.

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If the enthalpy of combustion of diamond and graphite are `-395.4 kJ mol^(-1)` and `-393.6 kJ mol^(-1)`, what is enthalpy change for the `C("graphite") rarr C("diamond")`?

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