- #1
maverick280857
- 1,789
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Hi again
Here's another problem:
When [tex]N_{2}O_{4}[/tex] is allowed to dissociate to form [tex]NO_{2}[/tex] at 15 degrees C at an equilibrium pressure of 1 atm, [tex]K_{p} = 0.41[/tex] for the reaction [tex]N_{2}O_{4}-->2NO_{2}[/tex].
(a) If [tex]N_{2}[/tex] is added to the system at constant volume will the equilbrium shift?
(b) If the system is allowed to expand an [tex]N_{2}[/tex] is added at constant pressure of 1 atm, what will be the equilbrium degree of dissociation when the partial pressure of [tex]N_{2}[/tex] is 0.6 atm.
My answer for part (a) (which I understand) is: the equilbrium will not shift either way if an inert gas is added at constant volume to the system because at constant volume, the partial pressures of the reactant product system would remain constant (the addition of inert gas in this way would result in an increase in the number of moles of gas and therefore in the total pressure).
For part(b), I figured that the addition of inert gas at constant pressure will reduce the partial pressures of the N2O4-NO2 system so the equilbrium shifts in the forward direction in this case, increasing the degree of dissociation.
[tex]K_{p} = 0.41atm = \frac{p_{NO_{2}}^2}{p_{N_{2}O_{4}}}[/tex]--(1)
Now if [tex]\alpha[/tex] is the degree of dissociation under the new conditions,
let initial number of moles of [tex]N_{2}O_{4}[/tex] = 1
hence, eq number of moles of [tex]N_{2}O_{4} = 1 - \alpha[/tex]
Since the system contained some moles (say a) of [tex]NO_{2}[/tex] previously, eq number of moles of [tex]NO_{2} = a + \alpha[/tex].
[tex]p_{NO_{2}}V = (a+\alpha)RT[/tex] -(2)
[tex]p_{N_{2}O_{4}}V = (1-\alpha)RT[/tex] --(3)
Also, [tex]1 atm = p_{NO_{2}} + p_{N_{2}O_{4}} + p_{N_{2}}[/tex]
so
[tex]0.4 atm = p_{NO_{2}} + p_{N_{2}O_{4}}[/tex] - (4)
In principle it is possible to solve for [tex]p_{N_{2}O_{4}}[/tex] using equations (1) and (4). Next we can use equation (2) to solve for alpha But we do not know volume so we will have to divide (2) by (3) first. Is this a correct approach?
I'd be happy if someone could tell me if this approach is correct and/or offer a better approach (a shorter one will be appreciated too :-)).
Thanks and cheers
VIvek
Here's another problem:
When [tex]N_{2}O_{4}[/tex] is allowed to dissociate to form [tex]NO_{2}[/tex] at 15 degrees C at an equilibrium pressure of 1 atm, [tex]K_{p} = 0.41[/tex] for the reaction [tex]N_{2}O_{4}-->2NO_{2}[/tex].
(a) If [tex]N_{2}[/tex] is added to the system at constant volume will the equilbrium shift?
(b) If the system is allowed to expand an [tex]N_{2}[/tex] is added at constant pressure of 1 atm, what will be the equilbrium degree of dissociation when the partial pressure of [tex]N_{2}[/tex] is 0.6 atm.
My answer for part (a) (which I understand) is: the equilbrium will not shift either way if an inert gas is added at constant volume to the system because at constant volume, the partial pressures of the reactant product system would remain constant (the addition of inert gas in this way would result in an increase in the number of moles of gas and therefore in the total pressure).
For part(b), I figured that the addition of inert gas at constant pressure will reduce the partial pressures of the N2O4-NO2 system so the equilbrium shifts in the forward direction in this case, increasing the degree of dissociation.
[tex]K_{p} = 0.41atm = \frac{p_{NO_{2}}^2}{p_{N_{2}O_{4}}}[/tex]--(1)
Now if [tex]\alpha[/tex] is the degree of dissociation under the new conditions,
let initial number of moles of [tex]N_{2}O_{4}[/tex] = 1
hence, eq number of moles of [tex]N_{2}O_{4} = 1 - \alpha[/tex]
Since the system contained some moles (say a) of [tex]NO_{2}[/tex] previously, eq number of moles of [tex]NO_{2} = a + \alpha[/tex].
[tex]p_{NO_{2}}V = (a+\alpha)RT[/tex] -(2)
[tex]p_{N_{2}O_{4}}V = (1-\alpha)RT[/tex] --(3)
Also, [tex]1 atm = p_{NO_{2}} + p_{N_{2}O_{4}} + p_{N_{2}}[/tex]
so
[tex]0.4 atm = p_{NO_{2}} + p_{N_{2}O_{4}}[/tex] - (4)
In principle it is possible to solve for [tex]p_{N_{2}O_{4}}[/tex] using equations (1) and (4). Next we can use equation (2) to solve for alpha But we do not know volume so we will have to divide (2) by (3) first. Is this a correct approach?
I'd be happy if someone could tell me if this approach is correct and/or offer a better approach (a shorter one will be appreciated too :-)).
Thanks and cheers
VIvek