Solveeit Logo
Login

Question

Question: At \( {25^o}C \) the heat of formation of \( {H_2}{O_{(l)}} \) is \( - 285.9KJmol{e^{ - 1}} \) and t...

At 25oC{25^o}C the heat of formation of H2O(l){H_2}{O_{(l)}} is −285.9KJmole−1- 285.9KJmol{e^{ - 1}} and that for H2O(g){H_2}{O_{(g)}} is −242.8KJmole−1- 242.8KJmol{e^{ - 1}} . The heat of vaporization of water at the same temperature is
(A) 43.1KJmole−143.1KJmol{e^{ - 1}}
(B) 242.8KJmole−1242.8KJmol{e^{ - 1}}
(C) −43.1KJmole−1- 43.1KJmol{e^{ - 1}}
(D) −242.8KJmole−1- 242.8KJmol{e^{ - 1}}

Explanation

Solution

Hint : Since the particles of a fluid are in consistent movement and have a wide scope of motor energies, at any second some portion of them has enough energy to escape from the outside of the fluid to enter the gas or fume stage. The Warmth of Vaporization (likewise called the Enthalpy of Vaporization) is the warmth needed to prompt this stage change.

Complete Step By Step Answer:
First let us write down the values that are given to us,
Heat of formation of H2O(l){H_2}{O_{(l)}} = −285.9KJmole−1- 285.9KJmol{e^{ - 1}} ,
Heat of formation of H2O(g){H_2}{O_{(g)}} = −242.8KJmole−1- 242.8KJmol{e^{ - 1}} .
We were given the change from liquid form to gaseous form in the question, that is:
H2O(l)→H2O(g){H_2}{O_{(l)}} \to {H_2}{O_{(g)}} .
We know that enthalpy of vaporization is the measure of heat needed to change over one mole of a liquid into its gaseous state at its boiling point.
Hence,
Heat of Vaporization = Heat of Formation of H2O(g){H_2}{O_{(g)}} - Heat of Formation of H2O(l){H_2}{O_{(l)}} .
⇒ΔHvop=ΔHf(H2O(g))−(ΔHf(H2O(l))\Rightarrow \Delta {H_{vop}} = \Delta {H_f}({H_2}{O_{(g)}}) - (\Delta {H_f}({H_2}{O_{(l)}})
As we were given the values in the question
ΔHvop=−242.8+285.9\Delta {H_{vop}} = - 242.8 + 285.9
⇒ΔHvop=43.1KJmole−1\Rightarrow \Delta {H_{vop}} = 43.1KJmol{e^{ - 1}}
Hence, we got the heat of vaporization as 43.1KJmole−143.1KJmol{e^{ - 1}}
Thus, the correct option to the above solution is option A.

Additional Information:
Enthalpy can be spoken to as the standard enthalpy, ΔHo\Delta {H^o} This is the enthalpy of a substance at standard state. The standard state is characterized as the pure substance held steady at 1 bar of pressure. Phase transitions, for example, ice to liquid water, require or assimilate a specific measure of standard enthalpy:
Standard Enthalpy of Vaporization (ΔHovap)(\Delta {H^o}_{vap}) is the energy that must be provided as heat at constant pressure per mole of atoms vaporized (liquid to gas).
Standard Enthalpy of Fusion (ΔHofus)(\Delta {H^o}_{fus}) is the energy that must be provided as heat at constant pressure per mole of molecules melted (solid to liquid).
Standard Enthalpy of Sublimation (ΔHosub)(\Delta {H^o}_{sub}) is the energy that must be provided as heat at constant pressure per mole of particles changed over to vapor to solid.

Note :
The Kinetic energy of the atoms in the gas and the liquid are the equivalent since the vaporization cycle occurs at steady temperature. Nonetheless, the addition of warm energy is utilized to break the possible energies of the intermolecular powers in the fluid, to produce atoms in the gas that are liberated from expected energy (for an ideal gas). Along these lines, while Heat of vapour is greater than Heat of liquid, the motor energies of the atoms are equivalent.