# Two equilibrium problems

1. May 1, 2005

### poodlefarm

1) I just want to check if this one is correct

reaction => CCl4(g) <==> C(s) = 2Cl2(g)

Kp= .76

find the initial partial pressure of CCl4 that will produce an equilibrium total pressure of 1.20 atm

p.total = p.CCl4 + p.Cl2

CCl4 <=> 2Cl2
i. press x 2x
change 1.20 - x 1.20 - 2x

kp = .76 = (p.Cl2^2)/(p.CCL4) = (1.2-2x)^2/(1.2-x)
x= .8557 or x= .1543
.8557 will not work so take x = .1543

p.CCl4 = 1.20- .1543 = 1.05 atm

the rub is that if I use x to find p.Cl2 I get .892 atm
1.05 + .892 =/= 1.2 ?

Q2.

Ok
constant temp =25 C constant volume

reaction; Nh4HS (g)<==> NH3(g) + H2S(g)

step 1; some NH4HS decomposed in an evacuated container to give a total pressure at equililbrium of .659 atm

step 2; extra NH3 is added. re-established equilibrium gives a partial pressure for NH3 of .750 atm

find Kp

I'm stuck on this one.
.

2. May 1, 2005

### Gokul43201

Staff Emeritus
Q1 ) CCl4(g) <==> C(s) + 2Cl2(g)

Let the dissociation constant of CCl4 be x. If you start with 1 mole of CCl4, then in equilibrium, how many moles of CCl4, Cl2 will you have ? Remember, CCl4 is being consumed and Cl2 is being produced. Now use the equation for Kp to arrive at a quadratic in x which you can solve to find x.

Next, if the container had some volume V, and there are n moles of CCl4 to start, what will the pressure be at some temperature T ? Write a similar equation for the equilibrium number of moles and call this pressure Po. From these two, you can eliminate V/T and find the value of Po.

3. May 1, 2005

### Gokul43201

Staff Emeritus
let the initial number of moles of NH4HS be n, and let the dissociation constant be x. Now proceed from there ...

4. May 1, 2005

### poodlefarm

Q1)

I'm a little lost.

assuming 1 mole of CCl4

initial ; CCl4 = 1 Cl2 = 0
concentration CCl4 = -x cl2 = +2x
equilibrium CCl4 1-x Cl2 2x

Kp = .76 = 4x^2/(1-x)

x = .35

not sure if I'm doing this right.

but here is where I really get lost.

I think you are refering to PV =nRT

P= 1.2 atm

so 1.2 = (nRT)/V

and Po = (.65RT)/V // .65= 1-x

I still don't see it

5. May 2, 2005

### Gokul43201

Staff Emeritus
So far, so good !

Umm. Let's try again.

Initially, you have n moles at a pressure Po. Finally, you have n(1 - x + 2x) = n(1+x) moles at a pressure of 1.2 atm, where the value of x is known from above. Make sure you understand where the n(1+x) comes from.

Initial : Po = nRT/V
Final :1.2 = n(1.35)RT/V

From this, find Po (as well as the final partial pressures to make sure they add to give 1.2 atm)

6. May 2, 2005

### GCT

The second one's a #\$%^&. If the initial pressure was known, you could calculate Kp simply by using the percent dissociation, no need for the second aspect of the information given. It'll need to be worked out mathematically, if a solution actually exist to the exact form of this problem. I'm able to deduce several meaningful equations so far, and the best one has been

$$P_{total,eq2}=.75atm + P_{0}$$ where $$P_{0}$$ is the initial pressure at the very beginning of the situation. If $$.75atm$$ were valid as the net change in pressure, than one could calculate the Kp by finding the moles of each compound in undergoing the reaction.

Nevertheless I'll wait for both of you to get to this part before proceeding.

7. May 2, 2005

### GCT

I was hoping for a more creative way to solve the second question, yet the only sure way to solve it seems to involve a messy equation at the end.

$$P_{total,eq1}=.659=P_{0}+x,~x=P_{NH3}=P_{H2S}$$

$$Kp= \frac{[x][x]}{[P_{0}-x]}=\frac{[x][x]}{[.659atm-2x]}$$

also

$$Kp= \frac{[.750atm][x- \Delta y]}{.659atm-2x+ \Delta y]}=$$

set the two equal to each other and solve for y in terms of x, plug back in the second Kp...set this Kp equal to the first Kp and solve for x.

perhaps a better way will involve the incorporating the temperature