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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
By (ii)-(i)
`C_G 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 :

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

For the following reaction, C_("diamond")+O_(2)toCO_(2)(g), DeltaH=-94.3" "kcal C_("graphite")+O_(2)toCO_(2)(g), DeltaH=-97.6" "kcal the heat require to change 1 g of C_("diamond")toC_("graphite") is:

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:

If C(s)+O_(2)(g)rarrCO_(2)(g),DeltaH=X and CO(g)+1//2O_(2)(g)rarrCO_(2)(g),DeltaH=Y , then the heat of formation of CO is

C_("graphite")+O_(2)(g)rarrCO_(2)(g) deltaH=-94.05Kcalmol^(-1) C_("diamond")+O_(2)(g),DeltaH=-94.50Kcalmol^(-1) Therefore,

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