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Offline raffster2

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equilibrium question..
« on: December 07, 2008, 01:06:05 AM »
"How will the equilibrium shift if Helium gas is added to the following reaction: 4NH3 + 3O2 <-> 6H2O + 2N2?"

Wouldn't the reaction shift towards the reactants since the pressure is increased and there are more mols of gas on the right? The answer book says no shift would be made.

Offline Astrokel

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Re: equilibrium question..
« Reply #1 on: December 07, 2008, 01:31:07 AM »
Quote
since the pressure is increased

Is the partial pressure affected? The question should be more specified at constant volume or constant pressure.
« Last Edit: December 07, 2008, 01:43:18 AM by Astrokel »
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Offline raffster2

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Re: equilibrium question..
« Reply #2 on: December 07, 2008, 02:49:49 AM »
well the exact wording of the question is "How will the equilibrium of the following reaction shift if 4 moles of Helium gas is added? 4NH3 + 3O2 <-> 6H2O + 2N2?"

The answer is "The equilibrium will not change.  To change a gaseous equilibrium, the partial pressures of at least one gas need to be changed.  Since He is unreactive with all of the gases present, the total pressure will change but the partial pressures will not."

The thing that confused me was that in another part of the problem, it asks what happens when reaction's volume is increased, which led to the equilibrium shifting to the right.

thanks for the help

Offline Astrokel

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Re: equilibrium question..
« Reply #3 on: December 07, 2008, 03:08:01 AM »
Quote
The thing that confused me was that in another part of the problem, it asks what happens when reaction's volume is increased, which led to the equilibrium shifting to the right.
This is what i was talking about on constant pressure situation. Assuming constant pressure, What will happened to the concentration of the species in equilibrium when reaction's volume is increased?

Also, note the difference in constant pressure and constant volume situation, where the latter introducing helium gas will not affect equilibrium system while the formal does.
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Offline Hunt

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Re: equilibrium question..
« Reply #4 on: December 08, 2008, 11:07:26 AM »
This is an interesting question. I was working the other day on a similar problem where some random chemical reaction is considered. It took me some time to prove that indeed an addition of some inert gas changes the total pressure but keeps the partial pressures constant. This however works only for ideal gases. For real gases equilibrium does not remain unaffected. Unfortunately in general chemistry textbooks the proof is omitted and the reader is left wondering why equilibrium does not shift to the left.

Offline Astrokel

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Re: equilibrium question..
« Reply #5 on: December 08, 2008, 12:07:37 PM »
Hey Hunt, how does your prove sound? I know that as long as volume is constant, adding an inert gas doesn't affect partial pressure because it doesn't affect the concentration of the species in equilibrium. Correct me if i am wrong! :D
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Offline Hunt

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Re: equilibrium question..
« Reply #6 on: December 08, 2008, 12:45:47 PM »
Hi Astrokel,

You are correct.

Usually in textbooks it is written that increasing the pressure of a chemical system shifts the equilibrium to the less number of particles. This is in accordance with le chatelier's principle. What many writers do not stress on is your point , the fact that it is the volume that is decreasing i.e. the pressure increases by compression. But what if one changes the pressure of the system not mechanically ? Say by the addition of an inert gas. Obviously for ideal gases there's no change in equilibrium position as you explained, ( P = M RT , M is a const then P must be a constant too ). With real gases , however, I had some trouble in proving that a different situation arises. A real gas in this case occupies a volume less than V, which changes when an inert gas like helium is added to the system. If you are interested I can share the proof with you when I have some time .

Offline Astrokel

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Re: equilibrium question..
« Reply #7 on: December 08, 2008, 01:01:34 PM »
Sure, I'm interested to hear from your proof soon!
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Offline Borek

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Re: equilibrium question..
« Reply #8 on: December 08, 2008, 03:52:57 PM »
This is my version of the proof that partial pressures wont change after inert gas is added, posted here over three years ago:

http://www.chemicalforums.com/index.php?topic=4442.0

Where do the problem arise? When we assume that pressure is additive. For ideal gas (or any gas that is sufficiently close to ideal behavior)

p=nRT/V

and I have used this identity to calculate the pressure after inert gas was added. But if we are far from the ideality, this equation doesn't hold, so whole proof can be thrown away.
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Offline Astrokel

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Re: equilibrium question..
« Reply #9 on: December 08, 2008, 04:37:03 PM »
Borek, the proof is awesome!(snack) A random question: Which inert gas behaves most ideally under standard conditions(not high temperature or low pressure)? I think it has to be He or Ne?
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Offline Borek

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Re: equilibrium question..
« Reply #10 on: December 08, 2008, 05:51:04 PM »
That's what I would expect.
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Offline Hunt

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Re: equilibrium question..
« Reply #11 on: December 09, 2008, 07:10:26 AM »
That is a very interesting method Borek . Infact I used a very similar method for ideal gases but for a more general chemical reaction. The problem with real gases is that you dont obtain easy analytical expressions for fugacity coefficients. So I used equilibrium thermodynamics to relate the fugacity to a specific equation of state, not a great thing but still better than PV=nRT. Unfortunately I couldnt find a more general approach. I'll post this soon using Latex.

Offline Hunt

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Re: equilibrium question..
« Reply #12 on: December 25, 2008, 06:28:58 PM »
For ideal gases it is best to prove this using the equation of state pV = nRT.
This expression gives a linear relation between the concentration and pressure : p = M RT which can then be used to show that neither equilibrium nor partial pressures change. A further advantage is the reduction of Kc to Kp using the same equation , then one uses partial pressures to quantify equilibria. Finally the entire process must be isothermal ( ideal gases undergo free expansion at const tempreature ) otherwise there would be shift in equilibrium.

An alternative method is the use of dalton's law alone. It can be shown using this method that the partial pressures do not change when the equilibrium does not shift after the addition of an inert gas.

Consider a system in which a certain gas-phase chemical reaction rests at a dynamic equilibrium state. If the gases obey dalton's law of partial pressures, then it can be shown that the addition of an inert gas does not change the partial pressures of the components as long as the total volume is constant.

For the ith component, let * denote its initial state. Dalton's law at thermal equilibrium :



Suppose the system is perturbed by an addition of an inert gas, then since no chemical reaction occurs the mole number of each gas remains a constant. Let ( small ) delta p represent the pressure of the inert gas in the closed system, and ( small ) delta n the amount of the gas added. Again according to dalton's law :




The absence of * signifies the final equilibrium state reached by the gases. Pi is the difference in total pressure between the initial and final states.

The difference between the partial pressures for the ith component is eq (1) :



Applying dalton's law for the inert gas :



The type of an ideal gas is irrelevant. Therefore for an amount delta n of a gas present initially in the mixture ( with constant composition ) must have a partial pressure delta p :



Combining the last two equations leads to :



Therefore the change in total pressure of the system is due to the addition of the inert gas. The partial pressure of each gas has not changed as can be verified by substituting the value of small delta p into eq (1) which would vanish.

For real gases :

The study of real gas-phase systems must include a quantitative analysis of fugacities. In general fugacities are thermodynamic pressures related to real pressures by the relation :



Phi is called the fugacity coefficient and it is a function of T , p , the type of gas, and the nature of its surrounding medium. If the gas is dominated by attractive forces in the domain ( p , T ) of application , then phi < 1 . Similarily if the gas is dominated by repulsive forces then phi > 1. Ofcourse for ideal behaviour the total net forces are zero and phi is unity.

Let "id" represent the ideal state of a gas. Using the equation dG = Vdp - SdT , at constant temperature :




mu is the chemical potential. From the definition of fugacity :



Replacing this formula in the previous equation :



Substituting the compressibility/compression factor and knowing that:



Integrating :



To solve this integral Z must be known. In general Z can be expressed as a power series in terms of pressure. As long as the pressure is not significantly high the series can be truncated at the 2nd term. It follows then :



Since the virial coefficient is independent of pressure then the precise expression for the fugacity coefficient becomes :



This equation shows that the fugacity coefficient is a function of pressure and the 2nd virial coefficient which is a function of temperature and the potential energy of the medium. In general the thermodynamic equilibrium constant for real gases can be written in the form :



Kp is the equilibrium constant in terms of real pressures. v is the stoichiometric number. Based on a specific equation of state beta is determined in terms of certain parameters.

Adding a small amount of an inert gas changes the temperature and pressure of the system as well as its potential energy. However if one assumes the gas interacts weakly with the species present, then the 2nd virial coefficient becomes a function of temperature alone. Furthermore, assuming that the thermodynamic equilibrium constant changes slightly with temperature ( i.e. the change in enthalpy for adding an inert gas is approximately zero ) will make the calculations easier.

Since the total volume is constant , the partial pressures can be calculated if the change in temperature is known from an equation of state f(P,V,T) = 0. However such an equation must include volume corrections for all the species involved.



The 2nd virial coefficient can then be easily calculated.
For example for the Van der waal eq :



Therefore the total volume, initial partial pressures, and initial temperature must be known as well as the final temperature of the system. The new fugacities can then be calculated. How equilibrium shifts is easily determined.

The great failure of this method is in neglecting the interactions between the molecules of one gas and another. This is because the fugacity coefficients were determined for each gas as if it were not interacting with another gas. Correcting this requires a completely different approach and a modified equation of state to account for the attractive forces among the gases. I haven't given this much thought for now, but I suppose some results from statistical thermodynamics can be used to take into account the potential energy difference.

Offline Astrokel

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Re: equilibrium question..
« Reply #13 on: January 14, 2009, 09:28:54 AM »
A cookie for you! :P Apparently, it is not what i could comprehend at my level, i will definitely revisit this page again when i get into university, thanks heaps!
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

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