• Funniest Story on the Job Contest Starts Now!

    Contest starts now and ends September 27th. Winner will receive a special user banner and $10 Amazon Gift card!

    JOIN NOW
  • Site Updates Coming Next Week

    Site updates are coming next week on Monday and Friday. Click the button below to learn more!

    LEARN MORE

need help with chem equilibrium

ASDIC

The 9th Flotilla
15+ Year Member
Apr 7, 2003
961
9
261
New Jersey
  1. Medical Student
here was a question that totally threw me off...it appeared on one of my practice tests:

Given (in the passage): A(g) + B (g) <==> C (g) + 2D (g)

One mole of A and one mole of B are placed in a 0.4 L container. After equilibrium has been established, 0.2 mole of C is present in the container. What is the equilibrium constant for the reaction?
a) 0.0625
b) 0.125
c) 0.5
d) 2

The answer is (a) 0.0625

Does anyone know why?
 
About the Ads

TripleDegree

Joker Doctor
15+ Year Member
Nov 6, 2003
687
0
251
Here's thinking out loud....


If 0.2 mole of C was produced, then from the stoichiometric ratios, 0.2 moles of A and 0.2 moles of B must have been consumed. Also 0.4 moles of D must have been produced.

Therefore, with the assumption that since volume is the common denominator, and it can be left out of the equation:

[A] = 0.8
= 0.8
[C]= 0.2
[D] =0.4

K = 0.2*[0.4]^2/0.8*0.8 = 0.05

*shrug*
 

researchprof

Senior Member
7+ Year Member
15+ Year Member
Feb 27, 2004
110
0
0
Here is my reasoning:

K= [C][D]^2 / [A]

Find the molarity of all constituent gases. Volume is constant at 0.4L so you can always divide the mole values by that factor.

[A] = (1-0.2)mol /0.4L = 2M (since 0.2 mol is consumed in the formation of C meaning A and B have 0.8 mol remaining).

= [A]

[C] = 0.2mol/0.4L = 0.5M
[D] = 2*[C] = 1M

Now plug all these values into the equation provided and K= 0.125.

But as I said before, I don't know where the 0.0625 comes from. Maybe someone can take this up from here. Or maybe I am wrong, who knows?
 

InfraMan

Member
7+ Year Member
15+ Year Member
Dec 26, 2003
56
1
0
I just started studying this equilibrium stuff, so I may be way off here, BUT

I thought the equations everybody is using only apply to liquid and solid SOLUTIONS.

That is, for gases, you should use the formula

Kp = Kc (RT)^DeltaN (from memory, might be slightly different)

It's barely mentioned in the EK Chem book.
 

AlternateSome1

Burnt Out
15+ Year Member
Aug 2, 2002
988
30
296
NY
  1. Attending Physician
Originally posted by researchprof
Here is my reasoning:

K= [C][D]^2 / [A]

Find the molarity of all constituent gases. Volume is constant at 0.4L so you can always divide the mole values by that factor.

[A] = (1-0.2)mol /0.4L = 2M (since 0.2 mol is consumed in the formation of C meaning A and B have 0.8 mol remaining).

= [A]

[C] = 0.2mol/0.4L = 0.5M
[D] = 2*[C] = 1M

Now plug all these values into the equation provided and K= 0.125.

But as I said before, I don't know where the 0.0625 comes from. Maybe someone can take this up from here. Or maybe I am wrong, who knows?


:laugh:
Wow, yeah. This looks right. I assumed the values where in M and not moles from the get go. This makes the values you are working with more even as well, which is a good hint that it's the right strategy.

~AS1~
 
About the Ads

Smitty3L

Senior Member
7+ Year Member
15+ Year Member
Aug 31, 2002
230
10
0
40
Oklahoma
  1. Medical Student
This is out of notes I have takenn (maybe from Kaplans MCAT review, maybe from some textbook)

"Kc indicates the equilibrium constant is in terms of concentration (molar units). When reactants and products are all gases we can use partial pressures:

aA + bB <---> cC + dD

Kp = [(PC)^c * (PD)^d] /[(PA)^a * (PB)^b]"


If ONLY .2 moles of C was formed (none of D) you can get exactly .0625!

The partial pressures of A, B and C added together equals the total pressure (Dalton's law). To simplify the math we can use the 4:1 ratio between A,B and C. Whatever the partial pressure of A is it equals the partial pressure of B and is 4 times larger than C.

Kp = 1/4*4 = .0625
 

researchprof

Senior Member
7+ Year Member
15+ Year Member
Feb 27, 2004
110
0
0
Originally posted by premyo2002
if C=0.2mol than D=0.4mol because the ratio is 1:2,
this means that both A and B, combined are losing 0.6mol (0.3 each: ratio 1:1)

then, If we assume P=1 atm, and use moles = partial pressure we could do this

A=0.7
B=0.7
C=0.2
D=0.4

[.2][.4]^2 / [.7] [.7] ~ 0.0625

Excellent. I knew we were missing something. I am sure you are going to destroy the sumuvabitch called the MCAT.
 

CanIMakeIt

It will get better once you are...still waiting :)
15+ Year Member
Oct 8, 2003
1,501
30
296
  1. Attending Physician
Originally posted by jarrod_dale
Sorry, but the answer above doesn't make sense. We are dealing with concentration, yes, we start off with 1mol of A and B, but that is in a container of only 0.4 L. You have to divied 1/0.4 to get initial concentrations. You mole amounts would give us an equation of (.5*1^2)/(1.75*1.75) = 0.1633 - Not the right answer.

That being said, I don't understand how to determine the concentrations on the reactants side after the reaction has reached equilibrium. Do we just use the concentrations initially in the equilibrium equation?

Can anyone tell me.

My vote is for 0.125.

The best way is to create a reaction table in moles and then convert to concentration from "FINAL" line before putting in the K formula

A + B -----------> C + 2D
Initial 1 1 0 0
diff -0.2 -0.2 +0.2 +0.4
----------------------------------------------------------------------------
Final 0.8 0.8 0.2 0.4
----------------------------------------------------------------------------

Now these numbers are in moles, so convert them to M (moles/L) by dividing by 0.4 and put the number in the K equation
and I am still getting 0.125 ..... may be there is a typo in the key.



As for smitty's "If ONLY .2 moles of C was formed (none of D) you can get exactly .0625!" .... IMO it is not correct...... if C is produced then D will also be produced right...

Later
 
About the Ads

JDAD

1K Member
7+ Year Member
15+ Year Member
Oct 7, 2003
1,380
1
0
Thanks for the help. I agree with your way, and the numbers created are easy to work with.

So, to find the amount of reactants at equilibrium, you take the inital #mols and subtract the products.(stoichiometrically)sp?

Does it not matter that .4mole of D was produced. If you used that ratio, then A and B would be .6 and not .8

This would give us an answer of .222.

How do you know which ratio to use?
 

researchprof

Senior Member
7+ Year Member
15+ Year Member
Feb 27, 2004
110
0
0
Originally posted by researchprof
Excellent. I knew we were missing something. I am sure you are going to destroy the sumuvabitch called the MCAT.

Premyo, I disagree with your solution. I initially looked at it cursorily without paying attention to your method. Besides, succumbing to doing it your way I get 0.0653 which is different from 0.0625, so that is not a mathematical approximation as you indicated.

So let's continue waiting for the messiah who will get 0.0625 in a way we all agree :cool:
 

CanIMakeIt

It will get better once you are...still waiting :)
15+ Year Member
Oct 8, 2003
1,501
30
296
  1. Attending Physician
use the ratios...... for every 1 A & 1B, we get 1 C and 2 D..... in other words..... A:C = 1:1 and A: D = 1:2 similary B:C = 1:1 and B: D = 1:2

so if you want to use amt. of D as your reference, then to produce 2D we have to spend only half the amount for A and B thus your answer will still be 0.8 and not 0.6 .......

Hope that helped......otherwise post again I will try to explain it some other way

Later...
 

fun8stuff

*hiding from patients*
15+ Year Member
Apr 10, 2003
3,075
51
306
  1. Resident [Any Field]
Originally posted by premyo2002
if C=0.2mol than D=0.4mol because the ratio is 1:2,
this means that both A and B, combined are losing 0.6mol (0.3 each: ratio 1:1)

then, If we assume P=1 atm, and use moles = partial pressure we could do this

A=0.7
B=0.7
C=0.2
D=0.4

[.2][.4]^2 / [.7] [.7] ~ 0.0625

This method is wrong for many reasons. The answer is 0.125 as has been pointed out above days ago. I was surprised that there is still discussion on this. The key is wrong.
 

fun8stuff

*hiding from patients*
15+ Year Member
Apr 10, 2003
3,075
51
306
  1. Resident [Any Field]
Originally posted by CanIMakeIt
The best way is to create a reaction table in moles and then convert to concentration from "FINAL" line before putting in the K formula

A + B -----------> C + 2D
Initial 1 1 0 0
diff -0.2 -0.2 +0.2 +0.4
----------------------------------------------------------------------------
Final 0.8 0.8 0.2 0.4
----------------------------------------------------------------------------

Now these numbers are in moles, so convert them to M (moles/L) by dividing by 0.4 and put the number in the K equation
and I am still getting 0.125 ..... may be there is a typo in the key.



As for smitty's "If ONLY .2 moles of C was formed (none of D) you can get exactly .0625!" .... IMO it is not correct...... if C is produced then D will also be produced right...

Later

This is the correct explanation. No more time should be spent on this problem. :)
 

Smitty3L

Senior Member
7+ Year Member
15+ Year Member
Aug 31, 2002
230
10
0
40
Oklahoma
  1. Medical Student
My quote should be from the following website:

http://wine1.sb.fsu.edu/chm1046/chm1046.htm

Under lecture notes and then the "Equlibrium Constant" link on the left. Here are some other websites with reference Kp:

http://www.chemguide.co.uk/physical/equilibria/kp.html
http://members.aol.com/profchm/kc_kp.html
http://bilbo.chm.uri.edu/CHM112/lectures/KpKc.htm

It's same thing mentioned by InfraMan earlier

Originally posted by InfraMan
I just started studying this equilibrium stuff, so I may be way off here, BUT

I thought the equations everybody is using only apply to liquid and solid SOLUTIONS.

That is, for gases, you should use the formula

Kp = Kc (RT)^DeltaN (from memory, might be slightly different)

It's barely mentioned in the EK Chem book.

Using the Kc equation you get 0.125 no matter if any of D was formed or not because the molar concentration of D is 1 (.4mol/.4L). So there can be no doubt that 0.125 is the correct answer for Kc.


Now about the Kp equation. It seems as though I was wrong :oops:
The partial pressure of A IS equal to the partial pressure of B and they are both 4x as large as the partial pressure of C. But using

.2/.8*.8 = .3125

is not the same as

1/4*4 = .0625

And not only that, I did not divide by the total number of moles which gives

(.2/1.8)/
(.8/1.8)(.8/1.8) = .5625

And not only that the mole fraction (which is what .2/1.8 represents) needs to be multiplied by the total pressure. (partial pressure = XPtotal) I was assuming that it would cancel out, but that was also wrong, because it is multiplied twice in the denominator and only once in the numerator.

If we assume that .4mol of D is formed the Kp formula still doesn't work, because the total pressure is still not canceled, because the partial pressure of D would be squared, leaving the total pressure^3 in the numerator and total pressure^2 in the denominator.
 

premyo2002

Full Member
7+ Year Member
15+ Year Member
Feb 17, 2003
679
1
0
\\
I was looking for a rational reason that 0.0625 was the correct answer. I originally got 0.125, but instead of thinking that since I got 0.125, the answer MUST be 0.125 and the people that wrote the test are wrong; I looked for a way to justify the answer of 0.0625 and the closest explaination that I was able to come too with my feable mind i posted, hoping that it made since.... I?m sorry i challenged your intellectual superiority, best of luck on the mcat.


Smitty, I was talking about your signature, the clam eyes.. I heard it somewhere before, but couldn't remember from where.
 

AlternateSome1

Burnt Out
15+ Year Member
Aug 2, 2002
988
30
296
NY
  1. Attending Physician
Not having D produced when C is doesn't make sense. Imagine that A, B, C & D are sets of blocks.

R = red, B = blue, G = green

Now, A = RBG
B = RBB
C = GB
D = RB


If you took A + B and used them to make one C, you would have enough material left over to make 2 D.

RBG + RBB = GB + 2 RB

There is no way around this. You can't form C from A and B without leaving 2 D unless you create some other molecule, such as RBBR. As that is not an option in this problem, you can't assume it.

This argument also contests any thought to using a total of .6 moles of the reactants. The two moles of D came out of 1 mole of each of the A and B. C and 2 D are formed for each mole of A and B, The number of moles on each side of the equation do not have to equal each other. If that were true, then consider this problem...

1A -> 2B

You start with .5 moles of A
If you convereted all of the reactant to the product, you would end up with 1 mole of B. If you need to lose as many moles of reactant as you form of product, then you are going to end up with -.5 moles of A, which makes no sense.

~AS1~
 

elias514

Senior Member
10+ Year Member
15+ Year Member
Oct 20, 2003
529
2
201
This question is inherently flawed, because it doesn't specify which type of equilibrium constant you're supposed to find--the concentration equilibrium constant or the partial pressure equilibrium constant. Nonetheless, we must assume that the question asks for the concentration equilibrium constant, simply because no information is given regarding the temperature of the container.

Since .2 moles of C form, as per the stoichiometric relationships in the equation, .4 moles of D must form and .8 moles of both A and B must remain at equilibrium. The equilibrium constant expression for this equation takes the form

(molarity of C) x (molarity of D)^2 / (molarity of A) x (molarity of B)

Since the total volume of the container is .4 L, the following values can be substituted into the equation:

(.5)(1)^2/ (2)(2) =.125

Thus, the correct answer is .125, not .0625.

BTW, if you were to assume standard temperature on this problem (i.e., 273 K) and determine the partial pressure equilibrium constant, you would arrive at the value 2.79, which is not an option.

Don't pay too much attention to this problem, because it's too vague.
 
This thread is more than 17 years old.

Your message may be considered spam for the following reasons:

  1. Your new thread title is very short, and likely is unhelpful.
  2. Your reply is very short and likely does not add anything to the thread.
  3. Your reply is very long and likely does not add anything to the thread.
  4. It is very likely that it does not need any further discussion and thus bumping it serves no purpose.
  5. Your message is mostly quotes or spoilers.
  6. Your reply has occurred very quickly after a previous reply and likely does not add anything to the thread.
  7. This thread is locked.