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4.8 g of C (diamond) on complete combust...

`4.8 g` of `C` (diamond) on complete combustion evolves`1584 kJ` of heat. The standard heat of formation of gaseous carbon is `725 kJ//mol`. The energy required for the process
C(graphite) `rarr` C(gas)
C(diamond) `rarr` C(gas) are

A

`725,727`

B

`727,725`

C

`725,723`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
The expression for standard heat of formation of gaseous carbon is
`C("graphite")rarrC("gas")`
`DeltaH=725KJ//mol`
As graphite is thermodynamically more stable than diamond so the heat required to convert graphic to gaseous mond so the heat required to convert graphic to gaseous carbon should be more:
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Knowledge Check

  • The heat of formation of CO_(2) is -407 kJ/mol. The energy required for the process 3CO_(2)(g) rarr 3C(g) + 2O_(3)(g) is

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    Less than 1221 kJ
    B
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  • The standard heat of combustion of graphite carbon is -393.5 kJ mol^(-1) . The standard enthalpy of CO_(2) is

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    `+393.5 kJ mol^(-1)`
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    C
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    D
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  • For conversion C(graphite) rarr C(diamond) the DeltaS is

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