For the process, CO2(s)→CO2(g):

  • Option 1)

    Both \delta H and \delta S are positive

  • Option 2)

    \delta H is negative and \delta S is positive

  • Option 3)

    \delta H is positive and \delta S is negative

  • Option 4)

    Both \delta H and \delta S are negative

 

Answers (1)

As learnt in

Enthalpy of Sublimation -

Amount of enthalpy change to sublimise 1 mole solid into 1 mole vapour at a temperature below its melting point

- wherein

H_{2}O_{(s)}\rightarrow H_{2}O_{(g)}

\Delta H_{sublimation}= 46.6\, kj/mol

 

 and

Entropy for phase transition at constant pressure -

\Delta S= \frac{\Delta H_{Transition}}{T}

- wherein

Transition \RightarrowFusion, Vaporisition, Sublimation

\Delta H\RightarrowEnthalpy

\Delta E\RightarrowInternal Energy

T\RightarrowTransitional temperature

 

Both \Delta S and \Delta H are positive


Option 1)

Both \delta H and \delta S are positive

This option is correct

Option 2)

\delta H is negative and \delta S is positive

This option is incorrect

Option 3)

\delta H is positive and \delta S is negative

This option is incorrect

Option 4)

Both \delta H and \delta S are negative

This option is incorrect

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