TBR Chem Chapter 3 #34

[HOOTiamOWL]

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Please help anyone! This one is driving me crazy

Relevant info from Passage (all components are in Gas phase, and all at 500 degrees C):
Reaction X: 1 H2O + 1 Cl2O <--> 2HOCl, Kp = 8.61 x 10^-3 atm-1
Reaction Y: 1 HCl + 1 CO <--> 1 HCOCl, Kp = 3.26 x 10^-6 atm-1


#34)
Which Keq value does NOT match the reaction (all components are in Gas phase, and all at 500 degrees C)?
A) 2HOCl <--> H2O + 1 Cl2O where Kp= 8.61 x 10^-3
B) 0.5 H20 + 0.5 Cl20 <--> 1 HOCl where Kp= 9.28 x 10^-2
C) 0.5 HCOCl <--> 0.5 HCl + 0.5 CO Kp= 5.54 x10^2
D) 2HCl + 2CO <--> 2HCOCl Kp= 1.06 x 10^-11

Correct answer: A

I understand that the Kp for the reverse reaction (i.e. answer choice A) will be the reciprocal of the Kp for the forward reaction. In the explanation it also says: since answer choice b is the same reaction as reaction X except with the values cut in half, the Kp for answer choice B is the square root of the Kp for reaction X. Similarly, the Kp for answer choice D is just the square of the Kp for reaction Y.

How did they figure that out? Could someone please explain why halving/doubling the stoichiometric values of each reactant and product causes the Kp to be square rooted/squared?

Thank you!!!!

 

[MMa23]

How did they figure that out?
Yes, your reasoning is correct you will have to do 1/kp (the reciprocal) to get the correct answer. Remember Keq is just the concentrations of the products raised to the power of their coefficients. But when you take the backward reaction you do it the other way (the reciprocal).

Could someone please explain why halving/doubling the stoichiometric values of eachreactant and product causes the Kp to be square rooted/squared?
See above the reasoning to this question, when you double the coefficients i.e go from 1 to 2 you will take that concentration to the squared or if you half it you will take it to the 1/2 power which is the square root.

Good luck with your studying!

 

[HOOTiamOWL]

Thanks! That sounds so simple when you put it that way... and logically it totally makes sense. Its just that I was having difficulty working out the math to prove that to myself and I couldn't get it to work out mathematically the way that I was expecting it to. But I think this is good enough for me! Now I can move on, thanks again!

Although, if anyone is willing to try and explain the math out I would greatly appreciate that as well 😀

 

[MMa23]

What is it you need help with in the math?

 

[Jay2910]

I think HOOTiamOWL means, it would be nice to get full math procedures .. so that we can double check ourselves mathwise even though we understood the question conceptwise.

Yeah I agree . . .I think that about half the challenge in the equilibrium chapter is getting the math done on time and getting it done right.

I have a question too and it would be great to get some input:

My question pertains to #19:
If there is initally 1atm of NOCL in a flask what is the partial of NO gas once at equilibrium?
1) .0230
B).0460
C).0025
D).005

I get the fact that you have to do the ICE table . .. but I don't understand how we can take recipricols of numbers within a short amount of time.

Like here they have 1/(2.1*10^4)=4.9*10^-5 . .. how did they get that so fast?
Likewise for 34 Answer choice B) It states that the result is that the Keq as written should be square root of 8.61*10^-3. How do you know sqrt(8.61*10^-3)=9.28*10^-2?

 

[MMa23]

I am fairly certain that doing an ICE table (initial change equilibrium table) is not going to be on the MCAT. As for your questions though:

Like here they have 1/(2.1*10^4)=4.9*10^-5 . .. how did they get that so fast?
To do (1 / 2.1x10^4) you will have to change 1 to (1x10^0/2.1x10^4) so when you divide exponents, you divide 1/2.1 then 10^0/10^4, so 1/2.1 is roughly ~0.5 and when you divide exponents you minus the exponents (if you need more clarification on this google it) so it is 0-4 so it is 0.5x10^-4, traditionally you write the first number to be greater than 0 but less than 10 so it is 5x10^-1x10^-4 (add the exponents when you multiply) so 5x10^-5.

Likewise for 34 Answer choice B) It states that the result is that the Keq as written should be square root of 8.61*10^-3. How do you know sqrt(8.61*10^-3)=9.28*10^-2?
To square root that number you need to change the numbers so you can take the square root of both, for this particular instance it is VERY hard to do that without a calculator so I wouldn't sweat it.

 

[circulus vitios]

To square root that number you need to change the numbers so you can take the square root of both, for this particular instance it is VERY hard to do that without a calculator so I wouldn't sweat it.

I found this thread because the TBR ICE table problems are ridiculous and are pissing me off.

Anyway, there is a trick to finding square roots: modify the coefficient and the exponent such that you know (or know a reasonable approximation of) the coefficient and the exponent divides evenly by 2.

Example:
sqrt(8.61E-5)
I know a reasonable approximation of sqrt(86.1).
8.61E-5 = 86.1E-6.
6 divides evenly by 2, so that checks out.
Take the square root of 86.1, divide -6 by 2.
sqrt(86.1E-6) = ~9E-3. True value is 9.3E-3. Close enough.

Example:
sqrt(1.73E-17)
I know a reasonable approximation of sqrt(17.3).
1.73E-17 = 17.3E-18.
18 divides evenly by 2, so that checks out.
Take the square root of 17.3, divide -18 by 2.
sqrt(17.3E-18) = ~4E-9. True value is 4.2E-9. Close enough.

Example:
sqrt(360236)
I know a reasonable approximation of sqrt(36.0236).
360236 = 3.60236E5 = 36.0236E4
4 divides evenly by 2, so that checks out.
Take the square root of 36.0236 and divide 4 by 2.
sqrt(36.0236E4) = ~6E2 = ~600. True value is 600.2. Close enough.