Chemical Equilibrium Problem #2

[maverick280857]

Hi again

Here's another problem:

When $$N_{2}O_{4}$$ is allowed to dissociate to form $$NO_{2}$$ at 15 degrees C at an equilibrium pressure of 1 atm, $$K_{p} = 0.41$$ for the reaction $$N_{2}O_{4}-->2NO_{2}$$.

(a) If $$N_{2}$$ is added to the system at constant volume will the equilbrium shift?
(b) If the system is allowed to expand an $$N_{2}$$ is added at constant pressure of 1 atm, what will be the equilbrium degree of dissociation when the partial pressure of $$N_{2}$$ 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.

$$K_{p} = 0.41atm = \frac{p_{NO_{2}}^2}{p_{N_{2}O_{4}}}$$--(1)

Now if $$\alpha$$ is the degree of dissociation under the new conditions,

let initial number of moles of $$N_{2}O_{4}$$ = 1
hence, eq number of moles of $$N_{2}O_{4} = 1 - \alpha$$

Since the system contained some moles (say a) of $$NO_{2}$$ previously, eq number of moles of $$NO_{2} = a + \alpha$$.

$$p_{NO_{2}}V = (a+\alpha)RT$$ -(2)
$$p_{N_{2}O_{4}}V = (1-\alpha)RT$$ --(3)

Also, $$1 atm = p_{NO_{2}} + p_{N_{2}O_{4}} + p_{N_{2}}$$
so
$$0.4 atm = p_{NO_{2}} + p_{N_{2}O_{4}}$$ - (4)

In principle it is possible to solve for $$p_{N_{2}O_{4}}$$ 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

 

[Gokul43201]

Since the system contained some moles (say a) of $$NO_{2}$$ previously, eq number of moles of $$NO_{2} = a + \alpha$$.

That should be $a + 2\alpha$

Each mole of N2O4 that dissociates, makes 2 moles of NO2.

Also, it looks like a = 0, in this problem, but that doesn't matter. If some non-zero 'a' is given, then you can not assume 1 mole of N2O4. You must consider some 'b' moles, so that after equilibrium you have $b(1- \alpha)$ moles of N2O4.

 

[maverick280857]

Yes you're right (I made a typo).

Here's what I eventually did:

I computed the partial pressures as 0.249 and 0.150 atm for NO2 and N2O4 respectively. Now as the partial pressures are directly proportional to the number of moles,

$$\frac{0.249}{0.150} = \frac{2\alpha}{1-\alpha}$$

from which we get $$\alpha = 0.45$$ which is greater than the original degree of dissociation (=0.30). Is this okay?

Here's another one:

Solid $$NH_{4}HS$$ is introduced in a cold vessel and heated upto 500K. At equilibrium, the pressure of the mixture is 1 atm. Equal weights of $$NH_{3}$$(g) and $$H_{2}S$$(g) along with 0.5 mol of an inert gas is introduced in a vessel of capacity 8.2 litres at an initial pressure of 4 atm. Find the maximum weight of solid formed.

My solution:

From the data, it is clear that at initial equilbrium, $$p_{NH_3} = p_{H_{2}S} = 0.5 atm$$ so that

$$K_{p} = 0.25 atm^2$$

If a moles of hydrogen sulfide are taken, then no of moles of ammonia initially present = 2a and no of moles of the inert gas = 0.5. The initial total pressure (due to these three gases) equals 4 atm, so using the ideal gas equation, we can find the value of a, which turns out to be 0.1 mol.

Now, the objective is to find x, the number of moles of solid formed. Then, (2a-x) = (0.2-x) is the number of moles of NH3 left and (a-x) = (0.1-x) is the number of moles of hydrogen sulfide left (or more precisely, these are the moles at equilibrium).

Now at equilbrium, the total moles of gas = (0.2-x + 0.1 -x + 0.5) = (0.8-2x). Hence, the partial pressures are,

$$p_{NH_{3}} = \frac{0.2-x}{0.8-2x}P_{eq}$$
$$p_{H_{2}S} = \frac{0.1-x}{0.8-2x}P_{eq}$$

I can use the same equilbrium constant as the temperature is constant but I don't know how to proceed further. Please help.

Thanks and cheers
Vivek

 

[Gokul43201]

I get 0.45 too...but I'm not sure how you got the 0.15, 0.25 numbers.

What I did :

$$(1 - \alpha + 2\alpha) (RT/V) = 0.4 => RT/V = 0.4/(1+\alpha)~~~~~(1)$$

$$(RT/V)\frac{4 \alpha ^2}{1-\alpha} = 0.41 ~~~~~(2)$$

Plugging in for RT/V from (1) :

$$\frac{4 \alpha ^2}{1-\alpha^2} = 0.41/0.4 = (about)~1 => 4\alpha ^2 = 1 - \alpha^2$$

$$=> \alpha ^2 = 1/5 => \alpha = 0.45$$

 

[Gokul43201]

For the second problem, all you have to do (the final step) is :

$$K_p = 0.25 = (RT/V)^2 (0.2-x)(0.1-x)$$

Since T, V are known this gives you a quadratic in x, which you can solve.

 

[maverick280857]

Thanks Gokul... I just didn't think what I could do after the last step! (Searching for the nearest wall to bang my head against it...)

Cheers
Vivek
PS-Please summarize the first problem mentioned in the earlier post.