For the process $Dry\,ice\, \rightleftarrows C{O_2}(g)$
A) $\Delta H$ is positive and $\Delta S$ is negative
B) Both $\Delta H$ and $\Delta S$ are negative
C) Both $\Delta H$ and $\Delta S$ are positive
D) $\Delta H$ is negative and $\Delta S$ is positive

Hint: If any process is carried out reversibly, so that dqrev is the heat absorbed by the system in the process at constant temperature (T), then the entropy change (dS) is given by
dS = \[\dfrac{{d{q_{rev}}}}{T}\]

Complete Step by step answer: In the given reaction,
Heat is absorbed by the system to convert solid dry ice into $C{O_2}$ .
\[\;{\Delta _{rr}}H\] = positive
As heat is absorbed by the system Dry ice is more organized, that is solid state. From moving to a gaseous state, entropy of the system increases.
$\therefore $$\Delta S$ is positive
Enthalpy (H) or heat content of the system is defined as the sum of the internal energy and the P-V energy of the system. Under a given set of conditions.
H = E + PV
H is a state function. Its value depends on the initial and final state of the system. As a result, absolute value of enthalpy cannot be determined but ΔH can be experimentally calculated as $ΔH = {{H_f} – {H_i}}$
For a reaction, the change in enthalpy (ΔH) is given by,
ΔH = ΔE + PΔV (at constant P)
From first law of thermodynamics
ΔE = q + w
ΔE = q – PΔV
Substituting this value of ΔE in equation ΔH = ΔE + PΔV, we get,
ΔH = q – PΔV + PΔV
ΔH = qP.
From the above discussion we can conclude that both change in enthalpy and entropy should be positive.

Therefore, option “C” is correct

Note: If the reaction is carried out at constant pressure, it involves the change in volume against constant pressure. Work is done by the system in expanding against the atmospheric pressure
For finite changes, ΔS = \[\dfrac{{{q_{rev}}}}{T}\]