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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)

best_answer

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 $\Rightarrow$ Fusion, Vaporisition, Sublimation

$\Delta H\Rightarrow$ Enthalpy

$\Delta E\Rightarrow$ Internal Energy

$T\Rightarrow$ Transitional 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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