TBR Gen Chem Example 3.9 Equilibrium

[eshara]

Hey all - having trouble with this question:

At 92.2 C, the Kp for the following reaction is 0.2000 atm-1. If you were to place exactly 0.200 atm of N2O4 into a 1.00 L vessel, what would the partial pressure of NO2 be once equilibrium was established?
2NO2 --------> N2O4

A) 0.025 atm NO2
B) 0.200 atm NO2
C) 0.350 atm NO2
D) 0.400 atm NO2

I understand how to set up the equilibrium table, but like the solution says - using intuition would save time. How can you determine the change in NO2 partial pressure just from the Kp value given and the amount of product? Is there a general concept like if Kp is less than 1, the reaction favors the formation of reactants? Also how did BerkReviewTeach find out the N2O4 amounts for each choice? Thanks!!

BR General Chemistry Sample Question 3.9




Perhaps a distribution chart can help here. Let's consider the four answer choices.

· · · · · · · · NO2 · · · · N2O4
start · · · · · 0 · · · · 0.200
choice A · 0.025 · · · · 0.1875
choice B · 0.200 · · · · 0.1000
choice C · 0.350 · · · · 0.0250
choice D · 0.400 · · · · 0

Choice A can't be equilibrium, because there are more products than reactants. Given that K = 0.20, there must be more NO2 than N2O4.

Choice B can't be equilibrium, because plugging into the K expression yields 0.1/(0.2)exp2 = 0.1/0.04 = 10/4 > 0.200. Given that K = 0.20, that can't be right.

Choice D can't be equilibrium, because there must be at least some product.

Only choice C remains standing, thus it must be the best answer.

As the answer explanation states (BR page 174 of General Chemistry I), this question is designed to demonstrate that you don't really need to employ a great deal of math to solve numerical-based questions in a multiple-choice format.

BTW, if it makes you feel any better, the questions you sk are the usual suspects that most of our students ask about. Getting those questions correct is far less important to your preparation than figuring out a logical way to solve the questions quickly. You are doing quite well from what it seems. Three chapters down and seven to go.

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[BeatMCAT]

yeah i had question regarding this as well..i was confused while reading this the other day.

 

[Futurecalipt]

Well the Kp basically tells you a ratio of product over reactant at equilibrium. you could do the long way and find out for yourself how much NO2 will be created based on the Kp and doing an ICE table, but yea you could use intuition. Since the answers give you possible figures, you really don't need to look into solving anything in this case. Kp is .2, so go through the answers and find the one that shows N2O4 at 1/5th of NO2. (Kp= product/reactants). Or wait, are the answers only for NO2? Or are both included like in your highlighted box? That makes things different.

 

[BerkReviewTeach]

Hey all - having trouble with this question:

At 92.2 C, the Kp for the following reaction is 0.2000 atm-1. If you were to place exactly 0.200 atm of N2O4 into a 1.00 L vessel, what would the partial pressure of NO2 be once equilibrium was established?
2NO2 --------> N2O4

A) 0.025 atm NO2
B) 0.200 atm NO2
C) 0.350 atm NO2
D) 0.400 atm NO2

I understand how to set up the equilibrium table, but like the solution says - using intuition would save time. How can you determine the change in NO2 partial pressure just from the Kp value given and the amount of product? Is there a general concept like if Kp is less than 1, the reaction favors the formation of reactants? Also how did BerkReviewTeach find out the N2O4 amounts for each choice? Thanks!!

The numbers in that box I made come from the ice box method. You start with 0.200 atm N2O4. For every 1 N2O4 you lose, you gain two NO2s.

• In choice A, the formation of 0.025 atm NO2 would have consummed 0.0125 atm N2O4, leaving you with 0.1875 atm N2O4.

• In choice B, the formation of 0.200 atm NO2 would have consummed 0.100 atm N2O4, leaving you with 0.100 atm N2O4.

• In choice C, the formation of 0.350 atm NO2 would have consummed 0.175 atm N2O4, leaving you with 0.025 atm N2O4.

• In choice D, the formation of 0.400 atm NO2 would have consummed 0.200 atm N2O4, leaving you no N2O4.

So the choices can be viewed as:
A) 0.025 atm NO2 and 0.1875 atm N2O4
B) 0.200 atm NO2 and 0.100 atm N2O4
C) 0.350 atm NO2 and 0.025 atm N2O4
D) 0.400 atm NO2 and 0 atm N2O4

 

[eshara]

Thanks so much BerkReviewTeach. This method makes sense. I was looking over this problem again though and hopefully you or someone else can answer this as well. The solution says "First you must use the equilibrium expression to estimate the magnitude of x. In this case, Keq is less than 1.0 and the reaction starts with all products. The value of x is going to be significant (more than half shifts over). [...] the value of NO2 is greater than 0.200 atm...but less than 0.400."

Am I explaining this correctly? If Keq < 1, the reaction favors the formation of reactants, so we should have more NO2 at equilibrium. Assuming that all the 0.200 atm of N2O4 shifted, we would get 0.400 atm (2x the amount of N2O4 b/c of mole ratios given in the reaction), but since we want to know the partial pressure at equilibrium, we would assume that some N2O4 would not have shifted and so the value should be < 0.400 atm. As for why the amount is > 0.200 atm -- the reaction is not 1:1 molar ratio for reaction: products, so we would assume that we need more NO2 than N2O4 for equilibrium since it's 2:1. If I'm right, this problem seems a lot simpler now - I just had to look at the coefficients of rxts and products and see if Keq was >/< than 1.

Are equilibrium problems on the MCAT likely to require ICE tables in which we need to consider x (and therefore know the conditions in the table on the previous page)? I initially did this problem using the ICE table method and I noticed that in this problem we start with all products and Kp is not less than 10^-3, so as per the table we don't need to consider the x-term, but Kp is not greater than 10^3, so we're not supposed to ignore it either - I am probably overthinking this or overlooking something, but what's the indication that x is significant? Thanks in advance!

 

[gunj122]

i love this question lol

 

[dreamtobeMD]

Hey guys!

I'm also having trouble with this question. I still don't understand how we know that more than half shifts over? I understand the rest (or at least I think I do, lol) I just don't understand how they just know this is the case. Sorry if my question is dumb but I took gen chem like 5 yrs ago so I'm having a tough time with this

Thanks!

 

[Cruzerthebruzer]

Hey guys!

I'm also having trouble with this question. I still don't understand how we know that more than half shifts over? I understand the rest (or at least I think I do, lol) I just don't understand how they just know this is the case. Sorry if my question is dumb but I took gen chem like 5 yrs ago so I'm having a tough time with this

Thanks!

Kp = [products]/[reactants]

In this case [N2O4]/[NO2]^2 so if the Kp value is <1 the denominator is bigger than the numerator, and signifying that NO2 is favored.

If Kp value is >1 the numerator is bigger than the denominator, and signifying that the products, or N2O4 in this case is favored.

Hope that helps.

 

[dreamtobeMD]

Thanks Cruzerthebruzer,

I understand that there is more reactant formed than the product but I'm not understanding how we are supposed to know that more than half shifts over. If Keq<1 do we always just assume that more than half shifted over (same thing if keq>1)?

Thanks so much! I really appreciate it

 

[Cruzerthebruzer]

Yes, if Keq>1 then there is more concentration on the product side so >50% product, if Keq<1 there is less concentration on the reactant side, so >50% for the reactant side.

Hope this helps!

ctb

 

[dreamtobeMD]

Thanks!

 

[Golaso]

Hi guys,

alright the way I might do this question is first set up my equilibrium expression which will be
Kp= (0.2-x)/(2x)^2 ......now by looking at the answers you must understand what each mean, that is when the equilibrium is established NO2 partial pressure must equal to 2x.

Find your x value by dividing by 2


Now take that x value and plug it in the equlibrium equation and see if it's going to equal Kp value.....so let's take answer B, which is 0.200 atm, that answer must be equal to 2x (again you must understand why 2x= 0.200) .......therefore x=0.1 atm

now plug 0.1 atm in the equation above: (0.2-0.1)/(2*0.1)^2 and ask yourself will the answer be equal to the Kp value (0.2 atm)?


Alright to know whether x should be significant or not, you wana compare K to Q...if K is SIMILAR to Q then it's not significant
if K is DIFFERENT than Q then it's significant

In this example: k=0.200
Q=Product/Reactant=0.2/0 = infinity

Infinity is Different than 0.200 therefore x must be significant

I hate equilibrium questions lol

 

[deleted760180]

Hi @BerkReviewTeach,

Very glad to see you're still active on SDN. I'd still appreciate a point of clarification for this question (I understand the post is super old, but the example question is still there in my edition of TBR Gen Chem (2016) as Example 10.9 in the Equilibrium chapter).

It's not intuitive to me how we come to the understanding that more than half of the N2O4 is consumed.

I get that if ALL of the N2O4 is consumed, all of it would be converted to NO2 reactant (i.e., 0.4 atm NO2, which is D). But not all of it reacts since we're in equilibrium, so PNO2 must be less than 0.4 atm. I also understand that if half of it shifts over, we would arrive at the fact that PNO2 would be 0.2 atm, again according to the ICE box.

However: how do we know that more than half is consumed to arrive at an answer of 0.350 atm? What's the trend between amount consumed and PNO2 here? Is it as simple as saying that if less than half of the N2O4 was consumed, we'd have a Keq > 1?? Idk.

I know I must be overthinking it somehow, so I'd really appreciate a big dose of reality lol...

 

[BerkReviewTeach]

No problem with the age of this question. I remember this question from when I first started, and it has caused trouble for people year in and year out.

The important thing here is that Kp = 0.20, which means P/R < 1 at equilibrium, so at equilibrium, R > P.
Kp = (PN2O4)/(PNO2)^2

We start with all P, so to reach a state where R > P, over half of the original P at the start has to convert to R.

2 NO2 --> N2O4

0 .200 initial
+ 2x - x shift
2x .2-x at equilibrium

Let's plug in some random value for x, starting with 0.10 (half of the 0.20 we started with)

If x = 0.20, then PNO2 = 0.2 and PN2O4 = 0.1
That would give Q = 0.1/(0.2)^2 = 0.1/0.04 = 2.5
Q is too large (larger than K), so the reaction needs to shoot to reactants.
This means that x = 0.10 is too small, so x must be bigger (although it can be no bigger than 0.20)

We know 0.10 < x < 0.20, which means we know 0.20 < PNO2 < 0.40.
Only choice C is in that range.

This is kind of saying the same thing as the explanation in the book, so sorry if it's not helping with a new perspective. I hope it's different enough to help. Like I said at the start, this one has bothered people for as long as this book has been out. Crazy thought, but a doctor that will interview you for medical school may have done this same question years before you.

 

[deleted760180]

No problem with the age of this question. I remember this question from when I first started, and it has caused trouble for people year in and year out.

The important thing here is that Kp = 0.20, which means P/R < 1 at equilibrium, so at equilibrium, R > P.
Kp = (PN2O4)/(PNO2)^2

We start with all P, so to reach a state where R > P, over half of the original P at the start has to convert to R.

2 NO2 --> N2O4

0 .200 initial
+ 2x - x shift
2x .2-x at equilibrium

Let's plug in some random value for x, starting with 0.10 (half of the 0.20 we started with)

If x = 0.20, then PNO2 = 0.2 and PN2O4 = 0.1
That would give Q = 0.1/(0.2)^2 = 0.1/0.04 = 2.5
Q is too large (larger than K), so the reaction needs to shoot to reactants.
This means that x = 0.10 is too small, so x must be bigger (although it can be no bigger than 0.20)

We know 0.10 < x < 0.20, which means we know 0.20 < PNO2 < 0.40.
Only choice C is in that range.

This is kind of saying the same thing as the explanation in the book, so sorry if it's not helping with a new perspective. I hope it's different enough to help. Like I said at the start, this one has bothered people for as long as this book has been out. Crazy thought, but a doctor that will interview you for medical school may have done this same question years before you.

Thank you so much for the reply. It made more sense to me when realizing that x = 0.1 was just a value to test out. And by looking at the equilibrium expression, we see that larger and larger values for x mean smaller K's (exponential increase in the denominator and the numerator gets smaller too).

So we need an x that's not so large such that ALL of the N2O4 is shifted over, but not so small either (which was the case with x = 0.10).

I also had a question about 2 concepts in the book that I'd like to understand a bit more.

1. First, is K only unitless in instances where we have equal moles of reactant and product on both sides? Or just unitless in general? Because several questions' answer choices in this chapter do have units (e.g., atm^-1).
2. Secondly, with regards to K only changing with temperature, the chapter mentions that this is due to the fact that K is a measure of energy. I assume this is related to the expression of K in terms of the free energy we get from ∆G˚ = -RTln(Keq)

Thank you again . Your books are truly second to none for C/P. The poop analogy for equilibrium at the start was absolutely hilarious. Also, it's such a weird thing to think about attendings having also struggled with this! At the same time that's really cool. It's almost like a tradition hahaha...