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C("diamond")+O(2)(g)rarrCO(2)(g),DeltaH=...

`C_("diamond")+O_(2)(g)rarrCO_(2)(g),DeltaH=-395 kJ` ......`(i)`
`C_("graphite")+O_(2)(g)rarrCO_(2)(g),DeltaH=-393.5KJ" "…(ii)`
The `DeltaH` , when diamond is formed from graphite, is

A

`-1.5 kJ`

B

`+1.5 kJ`

C

`+3.0 kJ`

D

`-3.0 kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
Aim: `C_(("graphite"))rarrC_(("diamond")), eq. (ii)-(i)` gives the result.
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The enthalpy change for the following reactions are: C_("diamond")+O_(2(g)) rarr CO_(2(g))" "DeltaH=-395.3 KJ "mol"^(-1) C_("graphite")+O_(2(g)) rarr CO_(2(g)) " "DeltaH=-393.4 KJ "mol"^(-1) The enthalpy change for the transition C_("diamond")rarr C_("graphite") will be :

Given : C("diamond")+O_(2)rarrCO_(2),DeltaH=-395 kJ C("graphite")+O_(2)rarrCO_(2),DeltaH=-393 kJ The enthalpy of formation of diamond from graphite is

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