The vapour pressures of water, acetone and ethanol at 293 K are 2.34 kPa, 12.36 kPa and 5.85 kPa respectively. Which of the following has the lowest boiling point?
Acetone
Vapour Pressure and Boiling Point Relationship
The boiling point of a liquid is the temperature at which its vapour pressure equals the surrounding atmospheric pressure. A liquid with a higher vapour pressure at a given temperature is more volatile, meaning it requires less energy to turn into a gas. Consequently, substances with higher vapour pressures at a specific temperature will generally have lower boiling points compared to substances with lower vapour pressures at the same temperature.
Comparing Vapour Pressures at 293 K
We are given the vapour pressures of water, acetone, and ethanol at 293 K:
- Water: 2.34 kPa
- Acetone: 12.36 kPa
- Ethanol: 5.85 kPa
To find which substance has the lowest boiling point, we need to identify the substance with the highest vapour pressure at the given temperature (293 K).
Comparing the values:
- Acetone (12.36 kPa) has a higher vapour pressure than Ethanol (5.85 kPa).
- Ethanol (5.85 kPa) has a higher vapour pressure than Water (2.34 kPa).
Therefore, Acetone has the highest vapour pressure among the three substances at 293 K.
Determining the Lowest Boiling Point
Since Acetone has the highest vapour pressure at 293 K, it means that Acetone is the most volatile among the three liquids under these conditions. A higher volatility corresponds to weaker intermolecular forces, which in turn leads to a lower boiling point.
Based on the vapour pressure data, the order of increasing vapour pressure is: Water < Ethanol < Acetone.
This implies the order of decreasing boiling point is: Water > Ethanol > Acetone.
Thus, Acetone has the lowest boiling point among water, acetone, and ethanol.
Important Questions from Chemical Equilibrium
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The equilibrium dissociation energy of a diatomic molecule is 4.75 eV and its stretching frequency corresponds to 0.5 eV. The minimum energy required to dissociate the molecule in eV is
For the reaction N2(g) + 3H2(g) → 2NH3(g), the thermodynamic quantity which depends upon the pressure at which equilibrium is arrived at is (superscript 'o'represents standard state)
The equilibrium constant for the formation of the complex \(\rm Zn (NH_3)_4^{2+}\) at 25°C is close to:
[Given: \(\rm E^0_{Zn^{+2}/Zn}=-0.763V\) and \(\rm E^0_{Zn(NH_3)^{+2}_4/Zn,NH_3}=-1.03V\)]
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