Equilibrium of Reaction of Gases

In summary: Better, but you are still using itex tax for an expression that is put on its own line. This is perhaps not a serious error, but itex generates smaller formulas (\frac{(\frac x 2)^{3}}{[\frac{4-x} 2]^{3}[\frac {4-\frac{4x}{3}} 2]^{4}} = 1 ), as it is designed to fit in text.
  • #1
Qube
Gold Member
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1

Homework Statement



Nitrous oxide reacts with ozone to form dinitrogen pentoxide. There are four moles of each reactant and the volume of the container in which this reaction takes place is 2 L. T is constant. The equilibrium constant for this reaction is 1. How many moles of product are present when the reaction achieves equilibrium?

Homework Equations

Balanced equation: [itex]3N_{2}[/itex][itex]O[/itex] + [itex]4O_{3}[/itex] <--> [itex]3N_{2}[/itex][itex]O_{5}[/itex]

The Attempt at a Solution



Wrong method:

[itex]K_{c}[/itex][itex]=[/itex][itex]\frac{[N2O5]^{3}}{[N2O]^{3}[O3]^{4}}[/itex]

All the components are gaseous, so they can go into the equation. First, however, we have to convert moles to concentrations. So we divide moles by volume (2 L). Also the equilibrium constant is equal to 1. Then we can plug in everything to the equation:

[itex]1[/itex][itex]=[/itex][itex]\frac{[N2O5]^{3}}{[2]^{3}[2]^{4}}[/itex]

From here we can solve for the concentration of dinitrogen pentoxide. [itex]1[/itex] [itex]=[/itex][itex]\frac{[N2O5]^{3}}{128}[/itex]

We get the concentration of dinitrogen pentoxide as the third root of 128, and we must multiply this value by 2 liters to get the number of moles of dinitrogen pentoxide since by dimensional analysis the liters cancel out (moles/liter * liter = moles).

New and hopefully correct method!

Let's surmise that x moles of product (dinitrogen pentoxide) were made.

From the balanced equation we can see that the production of x moles of product consumes x moles of nitrous oxide and (4/3)x moles of ozone.

Now, we have the equilibrium, rather than the initial, number of moles.

Equilibrium number of moles:

Nitrous oxide: 4 - x moles
Ozone: 4 - (4/3)x moles
Dinitrogen pentoxide: x moles (as I surmised).

From here we can plug stuff into our equilibrium constant equation! Volume is constant so we'll just plug in the number of moles.

[itex]\frac{x^{3}}{(4-x)^{3}(4-4x/3)^{4}}[/itex] [itex]=1[/itex] Questions:

1) I first thought nitrous oxide referred to nitrogen monoxide (NO). I was wrong. Is nitrous oxide the scientific name or just a common name? I'm suspecting it's just a common name because I can't think of a rule for "-ous" compounds outside acids (e.g. sulfuorous acid).

2) I'm still learning equilibrium here and I'm wondering if my work is correct. I'm pretty sure it's correct after consulting a textbook (which actually refers to N2O as dinitrogen monoxide). Is my answer correct? More importantly, is my process and logic correct?

3) Are the initial moles of reactant equal to the equilibrium moles of reactant? Is this relevant? I'm suspecting what I did above is wrong because I didn't use the equilibrium moles but instead the initial moles (I just re-read the textbook).
 
Last edited:
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  • #2
Qube said:
[itex]1[/itex][itex]=[/itex][itex]\frac{[N2O5]^{3}}{[2]^{3}[2]^{4}}[/itex]

That would be correct if the concentrations of reactants after the reaction were 2. They are not - some of the reactants were consumed.

Note: you are doing strange things to Latex. No need to treat each part separately, combine everything together and use one set of tex tags:

[tеx]1=\frac{[N_2O_5]^{3}}{[2]^{3}[2]^{4}}[/tеx]

to get

[tex]1=\frac{[N_2O_5]^{3}}{[2]^{3}[2]^{4}}[/tex]

itex tags are for inline formulas (like [itex]e=\sum_0^\infty\frac 1 {n!}[/itex]).

You can always right-click a formula and select "show math as tex commands" to see how it was entered. Or you can quote the message to see the [itex]\LaTeX[/itex] code.
 
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  • #3
Gotcha! I fixed it in the problem above while preserving the incorrect work so other people can hopefully see what I did wrong and avoid the mistake of assuming the initial concentrations to be the equilibrium concentrations!
 
  • #4
Qube said:
From here we can plug stuff into our equilibrium constant equation! Volume is constant so we'll just plug in the number of moles.

[itex]\frac{x^{3}}{(4-x)^{3}(4-4x/3)^{4}}[/itex] [itex]=1[/itex]

This looks much better, but you can't ignore the volume. It won't cancel out.

(BTW, your latex is still slightly messed up - why two separate expressions, each in its own tags, instead of a single expression? Compare our LaTeX tutorial.)
 
  • #5
Borek said:
This looks much better, but you can't ignore the volume. It won't cancel out.

(BTW, your latex is still slightly messed up - why two separate expressions, each in its own tags, instead of a single expression? Compare our LaTeX tutorial.)

You're right on both points; I'll make sure to look at more LaTeX and learn it. I really appreciate the power of LaTex, and I've been lazy by using two separate LaTex expressions instead of combining them.

Also, yes, volume won't cancel out here; volume is cubed on top and ... raised to the seven power in the denominator. If the degrees of the numerator and denominator were the same, however, then volume would cancel out, I think, correct?

[itex]\frac{(x/2)^{3}}{[(4-x)/2]^{3}[(4-4x/3)/2]^{4}} = 1 [/itex]

(Only used one pair of latex tags above! I'm getting it! :) )
 
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  • #6
Qube said:
If the degrees of the numerator and denominator were the same, however, then volume would cancel out, I think, correct?

Yes.

[itex]\frac{(x/2)^{3}}{[(4-x)/2]^{3}[(4-4x/3)/2]^{4}} = 1 [/itex]

(Only used one pair of latex tags above! I'm getting it! :) )

Better, but you are still using itex tax for an expression that is put on its own line. This is perhaps not a serious error, but itex generates smaller formulas ([itex]\frac{(\frac x 2)^{3}}{[\frac{4-x} 2]^{3}[\frac {4-\frac{4x}{3}} 2]^{4}} = 1 [/itex]), as it is designed to fit in text. tex tags on the other hand generate images that are larger and easier to read:

[tex]\frac{(\frac x 2)^{3}}{[\frac{4-x} 2]^{3}[\frac {4-\frac{4x}{3}} 2]^{4}} = 1 [/tex]
 

Related to Equilibrium of Reaction of Gases

1. What is the equilibrium of a reaction of gases?

The equilibrium of a reaction of gases is the state at which the forward and reverse reactions occur at equal rates, resulting in no overall change in the concentrations of reactants and products.

2. How is equilibrium achieved in a reaction of gases?

Equilibrium is achieved in a reaction of gases when the concentrations of reactants and products reach a constant value and the rates of the forward and reverse reactions are equal. This can be achieved by altering temperature, pressure, or concentration of reactants and products.

3. What is Le Chatelier's principle and how does it apply to the equilibrium of a reaction of gases?

Le Chatelier's principle states that when a system at equilibrium is subjected to a stress, it will respond by shifting in a direction that minimizes the effect of that stress. In the case of a reaction of gases, this means that if the concentration of reactants or products is changed, the equilibrium will shift in the direction that reduces the change in concentration.

4. Can the equilibrium of a reaction of gases be affected by changes in temperature?

Yes, the equilibrium of a reaction of gases can be affected by changes in temperature. Increasing the temperature will shift the equilibrium in the endothermic direction, while decreasing the temperature will shift it in the exothermic direction.

5. How does pressure affect the equilibrium of a reaction of gases?

According to Le Chatelier's principle, an increase in pressure will cause the equilibrium to shift in the direction that reduces the number of gas molecules. This means that if there are more moles of gas on the reactant side, the equilibrium will shift towards the product side to balance the pressure.

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