Relacione el punto de ebullición del agua y la acetona (propanona), con la facilidad de a evaporación de estas sustancias (volat
ilidad), considerando las fuerzas intermoleculares.
1 answer:
Answer:
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Explanation:
1. Water
The O-H bond in water is highly polar.
It has about one-third ionic character.
The partially-positive H atoms and partially negative O atoms strongly attract each other by hydrogen bonds.
It takes a large amount of energy to separate the water molecules from each other.
Thus, the boiling point of water is relatively high (100 °C).
2. Acetone
The C=O bond in acetone is much less polar.
It is less than 20 % ionic.
The dipole-dipole attractions in acetone are much weaker than the hydrogen bonds in water.
It takes much less energy to separate the acetone molecules from each other.
Thus, even though the molar mass of acetone is more than three times that of water, the boiling point of acetone is only 56 °C.
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Answer:
At equilibrium, the forward and backward reaction rates are equal.
The forward reaction rate would decrease if is removed from the mixture. The reason is that collisions between
molecules and
molecules would become less frequent.
The reaction would not be at equilibrium for a while after was taken out of the mixture.
Explanation:
<h3>Equilibrium</h3>Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.
Whenever the forward reaction adds one mole of to the system, the backward reaction would have broken down the same amount of
. So is the case for
and
.
Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.
<h3>Collision Theory</h3>In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.
Assume that and
molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield
, only a fraction of the reactions will be fruitful.
Assume that molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.
Because fewer molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between
and
molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.
The backward reaction rate is likely going to stay the same right after was taken out of the mixture without changing the temperature or pressure.
The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.
Therefore, this reaction would not be at equilibrium immediately after the change.
As more and more gets converted to
and
, the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.