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At 5xx10^(4) bar pressure density of dia...

At `5xx10^(4)` bar pressure density of diamond and graphite are `3 g//c c` and `2g//c c` respectively, at certain temperature `'T'`.Find the value of `DeltaU-DeltaH` for the conversion of 1 mole of graphite to 1 mole of diamond at temperature `'T' :`

A

`100kJ // mol`

B

`50kJ//mol`

C

`-100kJ//mol`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`C("graphite") rarr C("diamond")`
`DeltaH=DeltaU+P_(2).DeltaV`
`V_(m)("diamond")=(12)/(3)mL`
`v_(m)("graphite")=(12)/(2)mL`
`DeltaH-DeltaU=(500xx10^(3)N//m^(2))((12)/(3)-(12)/(2))xx10^(-6)`
`=100kJ//mol^(-1)`
`DeltaU-DeltaH=+100kJ//mol`
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Knowledge Check

  • At 5 xx 10^(5) bar pressure, density of diamond and graphite are 3g/cc and 2g/cc respectively, at a temperature T. What will be the value of triangleU-triangleH for the conversion of 1 mole grpahite to 1 mole diamond at diamond at temperature T?

    A
    100 kJ/mole
    B
    50 kJ/mole
    C
    `-100" kJ/mole"`
    D
    300 kJ/mole

  • Densities of diamond and graphite are 3.5 and 2.3 g mL^(-1) , respectively. The increase of pressure on the equilibrium C_("diamond") hArr C_("graphite")

    A
    Favours backward reaction
    B
    Fovours forward reaction
    C
    Have no effect
    D
    Increases the reaction rate

  • The enthalpies of combustion of C_(("graphite")) and C_(("diamond")) are -393.5 and -395.4kJ//mol respectively. The enthalpy of conversion of C_(("graphite")) to C_(("diamond")) in kJ/mol is:

    A
    `-1.9`
    B
    `-788.9`
    C
    `1.9`
    D
    `788.9`

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