The molar enthalpy of vapourisation of acetone is less than that of water. Why?

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Solution

Molar enthalpy of vapourisation of a liquid is the amount of heat required to convert 1 mole of the liquid to gas/vapour.

Acetone does not consist of any hydrogen bond like water that is why the intermolecular attractive forces in acetone will be less that that of water molecules which makes acetone boil faster. Dipole-dipole interactions are there in acetone, which are relatively weaker.
So, due to this lesser energy will be required to convert 1 mol of liquid to vapour for acetone than water.

Hence, the molar enthalpy of vapourisation of acetone is less than that of water.

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The molar enthalpy of vaporisation of acetone is less than that of water. Why?

One mole acetone requires less heat to vapourise than 1 mole of water. Which of the two liquids has higher enthalpy of vapourisation?

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); {Δ H}_{295 K} = - 285.8 k J .

The molar enthalpy of vapourisation of water at 1 atm and

25^{o} C

is 44 kJ. The standard enthalpy of formation of 1 mole of water vapour at

25^{o} C

is:

Assuming that water vapour is an ideal gas, the internal energy change $(Δ U)$ when 1 mol of water is vapourised at 1 bar pressure and 100 $^{0}$ C will be:

[Given molar enthalpy of vapourisation of water at 1bar and 373 K=41 kJ mol $^{- 1}$ ]

The enthalpy of vapourisation of water is 186.5 KJ/mol,then what is it's entropy of vapourisation

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