Equilibrium Question

[pnoybballin]

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After the addition of 0.10 atm. of H2 (g) to an equilibrium mixture of H2 (g), I2 (s), and HI (g) with an initial total pressure of 1.60 atm., the new total pressure once equilibrium is established will be:

H2 (g) + I2 (s) <--> 2 HI (g)

A. 1.66 atm.
B. 1.70 atm.
C. 1.76 atm.
D. 1.82 atm.

The answer is C. Can someone please give me both a qualitative and a mathematical explanation for this question? Thanks a lot everyone!

 

[Dr Gerrard]

I have no clue.

What does the solution say? Maybe that might help.

 

[matth87]

I couldn't fall alseep last night because I was thinking about this stupid problem.

 

[capn jazz]

I know. I tried it on paper. I tried googling how to do equilibrium problems with pressure. WTF.

 

[PRATFO]

damn. Just when I thought I had equillibrium down

 

[matth87]

Only way I can get that answer would be if you add .10H2

then 1/3 of it will stay H2 = .033atm
and 2/3 of it will go to 2HI = .066atm but you would have to multiply this by 2 because you have to account for the (s) turning into gas.

.03 + .1333 = .16 total.

 

[Seraph 84]

Seems like they left out the equilibrium constant.

 

[G1SG2]

haha, now I feel a bit better since I'm not the only one who can't figure this out...I felt incredibly dumb last night but I guess this is a weird problem?

 

[wanderer]

Seems like they left out the equilibrium constant.

To get the answer on your own you definitely need Kp or Kc. However after thinking about this for a while I think I might have something of use:
The equilibrium reaction is H2 <---> 2 HI
For every .1 atm that H2 goes down HI's partial pressure must go up by .2 and the total pressure goes up by .1. Since the mixture is initially at equilibrium at 1.6 atm, adding .1 atm of H2 makes the total pressure 1.7. However because there is an excess of reactants the system must shift to the right and increase the total pressure, so we know the pressure must be greater than 1.7 leaving answers C and D for now.
Now according to Le Chatelier's principle, the system tries to minimize the "disturbance" and reduces H2, but it wont reduce it by more than .1 atm, because doing so would mean that the system would no longer be in equilibrium. So the upper maximum for total pressure is 1.8 atm, leaving C as the only viable answer.
The premise of my reasoning is that even after Le Chatelier's principle takes effect you still end up with more of the specific reagent that you added than was initially present. I hope what I'm saying makes sense.

 

[pnoybballin]

To get the answer on your own you definitely need Kp or Kc. However after thinking about this for a while I think I might have something of use:
The equilibrium reaction is H2 <---> 2 HI
For every .1 atm that H2 goes down HI's partial pressure must go up by .2 and the total pressure goes up by .1. Since the mixture is initially at equilibrium at 1.6 atm, adding .1 atm of H2 makes the total pressure 1.7. However because there is an excess of reactants the system must shift to the right and increase the total pressure, so we know the pressure must be greater than 1.7 leaving answers C and D for now.
Now according to Le Chatelier's principle, the system tries to minimize the "disturbance" and reduces H2, but it wont reduce it by more than .1 atm, because doing so would mean that the system would no longer be in equilibrium. So the upper maximum for total pressure is 1.8 atm, leaving C as the only viable answer.
The premise of my reasoning is that even after Le Chatelier's principle takes effect you still end up with more of the specific reagent that you added than was initially present. I hope what I'm saying makes sense.

Wanderer you hit it right on the dot. This question was from one my TBR Packets in class but I just couldn't remember how to solve this problem but after reading your solution I just remembered that this is the way to go about it. Thanks a lot!

 

[pnoybballin]

Yea I feel pretty stupid for not remembering how to solve this

 

[Dr Gerrard]

To get the answer on your own you definitely need Kp or Kc. However after thinking about this for a while I think I might have something of use:
The equilibrium reaction is H2 <---> 2 HI
For every .1 atm that H2 goes down HI's partial pressure must go up by .2 and the total pressure goes up by .1. Since the mixture is initially at equilibrium at 1.6 atm, adding .1 atm of H2 makes the total pressure 1.7. However because there is an excess of reactants the system must shift to the right and increase the total pressure, so we know the pressure must be greater than 1.7 leaving answers C and D for now.
Now according to Le Chatelier's principle, the system tries to minimize the "disturbance" and reduces H2, but it wont reduce it by more than .1 atm, because doing so would mean that the system would no longer be in equilibrium. So the upper maximum for total pressure is 1.8 atm, leaving C as the only viable answer.
The premise of my reasoning is that even after Le Chatelier's principle takes effect you still end up with more of the specific reagent that you added than was initially present. I hope what I'm saying makes sense.

Where do you get this from? Does this not completely depend on Kp?

If Kp is very small, then it would not follow that trend.

 

[JohnnyBravo]

Where do you get this from? Does this not completely depend on Kp?

If Kp is very small, then it would not follow that trend.

They have this in the equilibrium section. The kp= 1.2^2/.4=3.6

 

[wanderer]

Where do you get this from? Does this not completely depend on Kp?

If Kp is very small, then it would not follow that trend.

Then it that case it would be something like 1.70000001, which is the same as 1.70, so there can be a situation where my reasoning wouldn't work, but out of the choices 1.70 and 1.76, 1.76 is a lot more likely.

 

[thebillsfan]

wait...why does kp have to be small?

 

[wanderer]

wait...why does kp have to be small?

It doesn't have to be but if that is the case, reactants will dominate, leading to little conversion to HI.