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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

`-1.5KJ`

B

`+1.5KJ`

C

`+3.0KJ`

D

`-3.0KJ`

Text Solution

Verified by Experts

The correct Answer is:
B
Equaction (ii) - Equaction (i)
`C_(C) rarr C_(D): , Delta H = 1.5 kJ`
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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 :

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For the following reaction, {:(C("diamond")+O_(2)to CO_(2)(g),, DeltaH=-97.6kcal),(C("graphite")+O_(2)to CO_(2)(g),,DeltaH=-94.3kcal):} The heat change for the conversion of 1 g of C (diamond) to C(graphite) is:

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