Posted by Ana on .
Liquid chloroform (CHCl3) is placed in a closed container and allowed to equilibrate with chloroform vapor. With the temperature held constant, additional chloroform is introduced into the system, and the vapor-liquid equilibrium is reestablished. Following this, the chloroform vapor pressure will:
a) be higher
b) be lower
c) be the same if both liquid and vapor phases are present
d) be lower if the heat of vaporization is positive
e) be higher if the heat of vaporization is positive
The correct answer is C but I don't quite get it why so? Could you please explain? Thank you
(1-11) Chemistry - Science (Dr. Bob222) - DrBob222, Saturday, May 10, 2014 at 1:42pm
Yes, the answer is C. Any liquid placed in a closed system will come to equilibrium with its vapor at a particular temperature. The vapor pressure then is say x. Adding more liquid will NOT make the liquid evaporate more which would, of course, increase the vapor. The vapor pressure is a function of the temperature. Those molecules with sufficient energy will be able to jump out of the liquid and that depends upon T. Increasing T will give them more energy and increase the vapor pressure but that's the only way to increase p. You may say "but increasing volume will give it more room to evaporate" and that is very true, however, more molecules will enter the vapor phase but when the vapor pressure reaches equilibrium it will be the SAME it was at the smaller volume
(1-11) Chemistry - Science (Dr. Bob222) - Ana, Saturday, May 10, 2014 at 6:27pm
Will the Le Chatelier's principle work here? Because pure liquid has no effect on the equilibrium position, it will not move towards products; thus, chloroform vapor will not form.
Chemistry - Science (Dr. Bob222) -
That's an interesting proposition but I don't think Le Chatelier's principle is quite the one to use here. However, if you look at it as an equilibrium (which it is), then
H2O(l) --> H2O(g)
So Kp = pH2O(g). When pH2O(g) has reached equilibrium with the liquid, then Kp is satisfied, and no more water will vaporize. You can see from the equation that adding more water has no effect on the equilibrium; you're right about liquid water (as long as a drop is there) having no effect on the pressure of the water vapor.