does the strength of London forces out weigh permanent dipole forces?

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#1
hey guys, if two molecules have london and permanent forces inbetween the molecule, given that one molecule has a higher london force than the other and the other molecule has higher permanent dipole forces what one would have a greater effect
0
1 year ago
#2
london forces are weaker than permanent dipole forces
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1 year ago
#3
Permanent dipole-dipole interactions occur "over and above" London forces, so a molecule with a permanent dipole will have both London forces and permanent dipole interactions (it's not one instead of the other). So the one with the permanent dipole will have stronger intermolecular forces overall.
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1 year ago
#4
(Original post by ebeck7049)
london forces are weaker than permanent dipole forces
This is an oft repeated mantra, which is actually incorrect.

If you compare the boiling points of butane and propanone (for example)

Butane, Mr = 60, has a boiling point of 273K
propanone, Mr = 58, has a boiling point of 329K

The only intermolecular forces in butane are London dispersion, which have increased the b.p. from absolute zero to 273K, i.e. an increase of 273K
Propanone has both London dispersion forces AND dipole dipole interactions.

As demonstrated above, the London dispersion component can be expected to increase the boiling point by about 273 kelvin and therefore the dipole-dipole component has caused a FURTHER increase of only 56 Kelvin.

Therefore the dipole-dipole component is actually weaker than the London dispersion component.
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1 year ago
#5
(Original post by charco)
This is an oft repeated mantra, which is actually incorrect.

If you compare the boiling points of butane and propanone (for example)

Butane, Mr = 60, has a boiling point of 273K
propanone, Mr = 58, has a boiling point of 329K

The only intermolecular forces in butane are London dispersion, which have increased the b.p. from absolute zero to 273K, i.e. an increase of 273K
Propanone has both London dispersion forces AND dipole dipole interactions.

As demonstrated above, the London dispersion component can be expected to increase the boiling point by about 273 kelvin and therefore the dipole-dipole component has caused a FURTHER increase of only 56 Kelvin.

Therefore the dipole-dipole component is actually weaker than the London dispersion component.
It doesn't work like that. The strengths of intermolecular forces aren't measured by the difference from absolute zero and the effect is a lot more complex than just adding temperatures together.
1
1 year ago
#6
(Original post by GreenCub)
It doesn't work like that. The strengths of intermolecular forces aren't measured by the difference from absolute zero and the effect is a lot more complex than just adding temperatures together.
Without any attraction between particles a substance would be a gas at absolute zero.

The absolute temperature is directly proportional to the average energy of the particles present.

As the forces of attraction increase, more energy is needed to break them.

The boiling point is a good guide as to the strength of the intermolecular forces.
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1 year ago
#7
(Original post by charco)
Without any attraction between particles a substance would be a gas at absolute zero.

The absolute temperature is directly proportional to the average energy of the particles present.

As the forces of attraction increase, more energy is needed to break them.

The boiling point is a good guide as to the strength of the intermolecular forces.
Still, not all London forces are equal as the previous poster was implying.
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1 year ago
#8
London is the most important city is the UK, the force it project in Europe and the UK is unparalleled
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