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Introduction

In our study of phase equilibrium, we have examined only pure materials. However, we will eventually want to study chemical reactions, which will mean understanding solutions with many components that might react with one another. Before considering reactions then, we need to consider what happens to phase equilibrium when there is more than one component present. How does mixing things together change equilibrium? This is actually a quite general question that we will address in many contexts including chemical reactions. For now, we’ll consider the phase equilibrium first, since that is where we have discovered equilibrium and that is where we have developed an understanding based on the concept of dynamic equilibrium.

There are many types of solutions to consider. We can mix together gases with gases, liquids with liquids, gases with liquids, solids with liquids, and so on. Each of these present different challenges, but we will find that there are similarities amongst them as well. In particular, we’ll find that dynamic equilibrium can be applied in each of these cases to understand the phase equilibrium that exists amongst the different components of the solutions.

Foundation

We will assume some understanding of solutions. Recall that the solvent is the major component in a solution, and is typically but not always a liquid. Far more often than not in Chemistry we use water as our solvent. A solute is a minor component of a solution, and in a single solution there may be more than one solute. The solute, in its pure form, can be a solid, another liquid, or even a gas. Once in solution, the solute usually has very different properties and is generally no longer recognizable from its pure form. Think about dissolving sugar or salt in water. The solution formed shows no evidence of the original crystalline solid solute.

Solutions are defined in large part by the concentration of the solute in the solvent. There are many ways to define and measure the concentration. The most common is the “molarity” of the solution, meaning the number of moles of solute per liter of solution. The units of molarity are “molar” with a capital M: a solution with 1.0 mole of solute in 1.0 L of solution is a 1.0 M solution. In this study we will also discuss the mole fraction of the solute. This is simply the number of moles of solute divided by the total number of moles of particles of all types in the solution.

We will lean very heavily on the concept of dynamic equilibrium. The idea will show up in all of our explanations that two competing processes at equilibrium must have the same rate. We will examine several types of processes and the factors that determine their rates.

Observation 1: lowering of the vapor pressure in solution

To begin studying solutions, our first task is to observe what impact, if any, the presence of a solute has on the properties of the solvent. We will begin with a simple two-component solution, with a solvent and a single solute. The type of solute will matter to us, as we will observe different behaviors for different solutes, particularly whether the solute is, in its pure form, a solid, a liquid, or a gas. To start, we will consider solutions formed by dissolving a solid solute into a liquid solvent. This choice is easiest to start with because the solid solute will be assumed to be non-volatile. That is, it does not readily evaporate and therefore has zero vapor pressure. Solids do have a vapor pressure, but for most solids, the vapor pressures are sufficiently small that we can ignore them. As a first guess, then, we might assume that the solution formed from a volatile solvent and a non-volatile solute would have the same vapor pressure as the solvent alone, since the solute seems to contribute nothing to the equilibrium vapor pressure.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Concept development studies in chemistry 2013. OpenStax CNX. Oct 07, 2013 Download for free at http://legacy.cnx.org/content/col11579/1.1
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