3
$\begingroup$

The following problem was asked in JEE Mains 2020 (Sept 2, Shift 1),

An open beaker of water in equilibrium with water vapor is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules:

(A) leaves the solution increases
(B) leaves the solution decreases
(C) leaves the vapor increases
(D) leaves the vapor decreases

According to the NCERT for Class XII, Part I, pg. 46, para 3,

In a pure liquid the entire surface is occupied by the molecules of the liquid. If anon-volatile solute is added to a solvent to give a solution [Fig. 2.4.(b)], the vapor pressure of the solution is solely from the solvent alone. This vapor pressure of the solution at a given temperature is found to be lower than the vapor pressure of the pure solvent at the same temperature. In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced.

So, after reading the last lines of above quoted text, I think that the vapor pressure is reduced, because the rate at which solvent molecules leaves the solution decreases, due to a decrease in exposed surface area. So, acc. to me option (B) should be correct. But, it was incorrect, acc. to the key.

Is there anything else, I'm missing?


The answer given is,

(C) leaves the vapor increases

$\endgroup$
  • $\begingroup$ It seems to me it is less about principles and more about the formulation of the task. In some ways both B and C are correct. B is correct as the rate of evaporation decreases. C is correct, as the net rate of the phase exchange increases in favor of vapour condensation. But I consider this interpretation as tricky, supporting B. $\endgroup$ – Poutnik Sep 14 at 8:58
3
$\begingroup$

I'd like to specifically commment on this:

According to the NCERT for Class XII, Part I, [pg. 46, para 3][1],

In a pure liquid the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. 2.4.(b)], the vapor pressure of the solution is solely from the solvent alone. This vapor pressure of the solution at a given temperature is found to be lower than the vapor pressure of the pure solvent at the same temperature. In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecules escaping from the surface is correspondingly reduced, thus, the vapor pressure is also reduced. [emphasis mine]

NCERT's explanation for why vapor pressure is lowered due to the presence of a dissolved solute is incorrect, because vapor pressure is independent of the surface area accesible to the solvent (assuming it's non-zero). Consider these two examples:

  1. You have two sealed containers, both of which contain an open beaker of identical solvent. In container A, the beaker is low and wide. In container B, the beaker is tall and narrow. The surface area of solvent in container B is thus smaller. Yet, assuming the conditions in the respective containers are identical, the equilibrium vapor pressures will be the same.

  2. You have two sealed containers, both of which contain an open beaker of identical solvent. The beakers are identical, so the surface area of liquid is the same. The conditions in the containers are also identical. Glucose is dissolved into the solvent in container A, while sucrose is dissolved into the solvent in container B. The final concentrations are the same. Sucrose is larger than glucose. Consequently, the fraction of the surface area covered by solvent in container A is larger than in container B. Yet, ignoring non-ideality, the vapor pressures are the same; they depend only on the concentration of solute, not its nature.

The surface area only matters for kinetics, i.e., for how quickly the solvent can escape the container to reach equilibrium. Clearly, with a larger surface area, the equilibrium vapor pressure will be reached more rapidly.

Feel free to send an email to NCERT with a link to this comment.

Here's the thermodynamic explanation: The addition of solute lowers the chemical potential of the solvent, due to the entropy of mixing. [There can also be energetic effects, in either direction, but with a soluble solute the entropy term is dominant.]

At equilibrium, the chemical potential of the solvent in the liquid and vapor phases must be equal. Thus, since the chemical potential of the solvent in the liquid phase has lowered, that of the the solvent in the vapor phase must lower as well.

And since the chemical potential of a gas increases with its partial pressure, the vapor pressure of the solvent in the gas phase will decrease until its chemical potential reaches the new (lowered) chemical potential of the solvent in the liquid phase.

As far as the question itself goes, I'm uncertain, because the question is asking for a kinetic, rather than a thermodynamic, explanation. But I can offer two alternative pictures:

Let's represent the movement between the liquid and vapor phases of the solvent ("X"), as follows:

$$\ce{X_{(l)}<=>X_{(g)}}$$

Picture (I): Both "B" and "C" are correct.

Let's start with the vapor in equilibrium with pure solvent. At equilibrium, the rates of the forward and backward reactions are equal.

Now suppose we add solute to the liquid phase. This shifts the reaction to the left, which means that the rate of the forward reaction initially decreases (which is answer "B") and the rate of the backward reaction initially increases (which is answer "C"). Eventually the system reaches its new equilibrium point, at which point the rates of the forward and reverse reactions are again equal.

Picture (II): "B" is correct.

Again, let's start with the vapor in equilibrium with pure solvent. At equilibrium, the rates of the forward and backward reactions are equal.

Now suppose we add solute to the liquid phase. The rate at which gas enters the liquid phase depends only on the concentration of gas, so the backwards reaction will be unaffected by the presence of the solute in the liquid. However, the rate of the forward reaction will be reduced, since it is now more favorable to for the solvent to remain in the liquid state. Hence the answer is "B".

As I mention in the Comments:

In Picture (I), the change in relative chemical potentials would raise the energetic barrier in the forward direction and lower it in the reverse direction, which would in turn increase the % of collisions that are succesful in the rev. direction, thus increasing the rate of the rev. reaction even though the collision frequency of the vapor with the liquid woudn't be changed by the presence of the dissolved solute.

But: If there isn't any barrier in the rev. direction, such that 100% of vapor molecules that collide with the liquid phase become liquid, then Picture (I) would not apply.

I'm afraid I don't know enough about the microscopic kinetics to have certainty on this. Perhaps someone that has done simulation work on this could provide an answer. [But to get this, you would want to vote to reopen the question.]

| improve this answer | |
$\endgroup$
  • $\begingroup$ If I've understood you clearly, then you're saying that my answer seems reasonable, as decrease in exposed SA takes it longer to reach the equilibrium vapor pressure. But, the reduction in vapor pressure isn't a consequence of decrease in exposed SA. Am I right? $\endgroup$ – Rahul Verma Sep 11 at 8:08
  • $\begingroup$ NCERT's explanation is supported here and here. I can't think of a way to refute your explantion. What do you think is the reason for lowering of vappour pressure? $\endgroup$ – Robin Singh Sep 11 at 10:18
  • $\begingroup$ @RobinSingh See the explanation I added. BTW, this is not my explanation, this is the standard thermodynamic explanation. $\endgroup$ – theorist Sep 12 at 6:07
  • $\begingroup$ @RahulVerma Your second sentence is correct. You can read what I added to the end of my answer. $\endgroup$ – theorist Sep 12 at 6:27
  • $\begingroup$ This doesn't sound right: "the rate of the backward reaction initially increases". Following your own logic there is no reason why the rate of collision of the gas molecules with the surface of the solution should increase simply because something was added to the solution. $\endgroup$ – Buck Thorn Sep 12 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.