Question

18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ $mol^{-1}$. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?

Answer 1

The "amount of heat" absorbed or liberated during the chemical reaction is called enthalpy.

"Vaporization" is the conversion of liquid into vapour in the presence of heat.

18 g of ${ H }_{ 2 }O$ = 1 mole of ${ H }_{ 2 }O$

2 moles of ${ H }_{ 2 }O$ = ?

$1\quad mole\quad of\quad { H }_{ 2 }O\quad Enthalpy\quad of\quad vaporization,\\\\ \Delta { H }_{ vap\quad  }=\quad 40.79\quad kJ\quad mo{ l }^{ -1 }$

$2\quad mole\quad of\quad { H }_{ 2 }O\quad =\quad 2\quad \times \quad 40.79\quad =\quad 81.58\quad kJ$

$Standard\quad enthalpy\quad of\quad vaporisation\quad at\quad 100°C\quad and\quad 1\quad bar\quad of\quad pressure\quad is\quad\\\\ \Delta { { H }_{ vap }^{ o } }\quad =\quad +\quad 40.79\quad kJ\quad mo{ l }^{ -1 }$

Answer 2

a) The enthalpy change for vapourising two moles of water under the same conditions is 81.58 kJ

b) Standard enthalpy of vapourisation for water is 40.79 kJ/mol

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.

Given : The enthalpy of vaporization of 18.0 g of water at 100°C and 1 bar pressure = 40.79 kJ

n = number of moles =$\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{18.0g}{18g/mole}=1mole$

Thus 1 mole of water requires heat = 40.79 kJ

2 moles of water requires heat = $\frac{40.79}{1}\times 2=81.58kJ$

Standard enthalpy of a process or ∆H⁰ is the enthalpy change of the process when involved substances are in their standard states.

Hence, for water ∆H vaporization, will be equal to = 40.79 kJ/mol

Learn more about Latent heat of vaporization

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