Question

If the standard enthalpy of sublimation \( (\Delta_{sub}H^\circ) \) of solid \( CO_2 \), naphthalene, Li and Na are 25.2, 73.0, 162, 108 kJ mol\(^{-1} \) respectively, the order of sublimation of these substances with temperature is

• A. Na>Li>Naphthalene>Solid \( CO_2 \)
• B. Solid \( CO_2 \)>Na>Naphthalene>Li
• C. Solid \( CO_2 \)>Naphthalene>Na>Li
• D. Li>Na>Naphthalene>Solid \( CO_2 \)

Hint:

The ease of sublimation of a substance is inversely related to its enthalpy of sublimation. A lower enthalpy of sublimation means less energy is required for the phase transition from solid to gas, so the substance will sublime more readily at a given temperature. Arrange the substances in the order of increasing enthalpy of sublimation, and then reverse this order to get the order of sublimation with temperature.

Answer 1

The Correct Option is C

The enthalpy of sublimation \( (\Delta_{sub}H^\circ) \) is the energy required to transform one mole of a substance from the solid phase directly to the gaseous phase at a constant temperature and pressure (usually standard conditions).
A substance with a lower enthalpy of sublimation will require less energy to vaporize, meaning it will sublime more readily at a given temperature compared to a substance with a higher enthalpy of sublimation.
Therefore, the order of sublimation with temperature will be inversely related to the enthalpy of sublimation.
The substance with the lowest \( \Delta_{sub}H^\circ \) will sublime most readily at a given temperature, followed by the substance with the next lowest \( \Delta_{sub}H^\circ \), and so on.
Given the standard enthalpies of sublimation: - Solid \( CO_2 \): \( \Delta_{sub}H^\circ = 25.
2 \) kJ mol\(^{-1} \) - Naphthalene: \( \Delta_{sub}H^\circ = 73.
0 \) kJ mol\(^{-1} \) - Li: \( \Delta_{sub}H^\circ = 162 \) kJ mol\(^{-1} \) - Na: \( \Delta_{sub}H^\circ = 108 \) kJ mol\(^{-1} \) Arranging these values in increasing order: \( 25.
2<73.
0<108<162 \) kJ mol\(^{-1} \) This corresponds to the order: Solid \( CO_2 \)<Naphthalene<Na<Li (in terms of enthalpy of sublimation) Since the ease of sublimation is inversely proportional to the enthalpy of sublimation, the order of sublimation with temperature (i.
e.
, the ease with which these substances sublime) will be the reverse of this order: Solid \( CO_2 \)>Naphthalene>Na>Li Therefore, the correct order of sublimation of these substances with temperature is Solid \( CO_2 \)>Naphthalene>Na>Li.