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We can conclude that the evaporation of water when no vapor is present initially is a spontaneous process with ΔG<0, and the evaporation continues until the water vapor has reached the equilibrium vapor pressure, at which point ΔG = 0. Equivalently, the condensation of water vapor when the pressure is 1 atm is spontaneous with ∆G<0 until enough water vapor has condensed to reach the equilibrium vapor pressure, at which point again ∆G = 0. Thermodynamics tells us exactly what the conditions are for phase equilibrium!

Observation 2: thermodynamic description of reaction equilibrium

Now that we have developed a thermodynamic understanding of phase equilibrium, it is even more useful to examine the thermodynamic description of reaction equilibrium. We want to understand why the reactants and products come to equilibrium at the specific concentrations and pressures that are observed.

Recall that ΔG = ΔH – TΔS<0 for a spontaneous process, and ΔG = ΔH – TΔS = 0 at equilibrium. From these relations, we would predict that most (but not all) exothermic processes with ΔH<0 are spontaneous, because all such processes increase the entropy of the surroundings when they occur. Similarly, we would predict that most (but not all) processes with ΔS>0 are spontaneous.

We try applying these conclusions to synthesis of ammonia

N 2 (g) + 3 H 2 (g) → 2 NH 3 (g)

at 298 K, for which we find that ΔS˚ = 198 J/mol·K. Note that ΔS˚<0 because the reaction reduces the total number of gas molecules during the ammonia synthesis, thus reducing W , the number of ways of arranging the atoms in these molecules. ΔS˚<0 suggests that N 2 and H 2 should not react at all to produce NH 3 . On the other hand, ΔH˚ = -92.2 kJ/mol from which we can calculate that ΔG˚ = -33.0 kJ/mol at 298 K. This suggests that N 2 and H 2 should react completely to form NH 3 . But in reality, we learned in the Concept Development Study of reaction equilibrium that this reaction comes to equilibrium, with N 2 , H 2 , and NH 3 all present. We determined the equilibrium constant

K p = P N H 3 2 P N 2 · P H 2 3 = 6 × 10 5

Why does the reaction come to equilibrium without fully consuming all of the reactants?

The answer, just as before, lies in a more careful interpretation and application of the values given: ΔS˚, ΔH˚, and ΔG˚ are the values for this reaction at standard conditions, which means that all of the gases in the reactants and products are taken to be at 1 atm pressure. Thus, the fact that ΔG˚<0 for Reaction (6) at standard conditions means that, if all three gases are present at 1 atm pressure, the reaction will spontaneously proceed to increase the amount of NH 3 .

From this analysis, we can say that, since ΔG˚<0 for Reaction (6), reaction equilibrium results in the production of more product and less reactant than at standard pressures. This is consistent with our previous measurement that K p = 6×10 5 , a number much, much larger than 1. As a result, we can say that Reaction (6) is “thermodynamically favorable,” meaning that thermodynamics accurately predicts that the equilibrium favors products over reactants.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
Idrissa Reply
im all ears I need to learn
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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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
types of nano material
abeetha Reply
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Ramkumar Reply
what is nano technology
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what is system testing?
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
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
anybody can imagine what will be happen after 100 years from now in nano tech world
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
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
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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
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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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