Chemical Equilibrium with Multiple Kc

In summary, the student tried to solve the homework equation A<->2B+C but both methods yielded incorrect results. The student redid their calculations and found that by alternating between the results of A<->C and A<->2B, they found a result that was closer to the correct answer.
  • #1
Kariga
3
0

Homework Statement


Compound A can react to form B and C according to the following

A<->2B Kc=1
A<->C Kc=1

If 1 mole of A is solved into 1 litre of water, what will the concentration of A be after equilibrium has been reached?

Homework Equations


ICE-tables(?)
A<->2B+C
Kc=(^2[C])/[A]

The Attempt at a Solution


Doing this homework the usual way, doing the ICE tables, I ended up with the solution c(A)=0.39, which is not correct according to the answer.
This made me think that perhaps this can't be solved assuming that
A<->2B+C , Kc=1, C(A) at start=1, but instead there is some other method?

I would be grateful to have some tips on how I should approach this problem.
 
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  • #2
This system can be described by three equations - two Kc and one mass balance. Write them all, solve.
 
  • #3
Hi
I forgot to mention this task is something "off course" so I don't have any books to get information from. Also I don't know the english term for this reaction so it's hard to find answers on the internet.

That being said, thank you for your advice. I would ask a bit more about what you meant with two Kc and Mass Balance.
I suppose Mass Balance refers to conservation of mass? Would that mean that since [A] at start is 1 mole, then [A]+[2B]+[C] at end =1?
I also made an attempt to solve this doing
A<->C K=1 and solving for c(C)
and
A-<->2B K=1 and solving for c(B)
but neither gave results that would be possible.
I think the trouble is that once A reacts to form B it reacts again to form C, lowering c(A)
and then A reacts again to form B in smaller amounts and so on in an endless loop.

If there is a particular term used for this "multiple Kc" reaction, could you tell it? I'm sure if I knew where to look I'd be able to finish this.
 
  • #4
Kariga said:
I suppose Mass Balance refers to conservation of mass? Would that mean that since [A] at start is 1 mole, then [A]+[2B]+[C] at end =1?

Yes.

I also made an attempt to solve this doing
A<->C K=1 and solving for c(C)
and
A-<->2B K=1 and solving for c(B)
but neither gave results that would be possible.

That's the approach I tried and it yielded something that looks quite reasonable.

I think the trouble is that once A reacts to form B it reacts again to form C, lowering c(A) and then A reacts again to form B in smaller amounts and so on in an endless loop.

Doesn't matter. System is described by the equations that were listed and they are enough to take care of the problem. General approach to every equilibrium problem is to write all equations describing the equilibrium and all mass balances, and then to solve this system for all unknowns.
 
  • #5
Hi

Thank you for your help.

I redid my calculations and it seems that by iterating between the result of
A<->C
and
A<->2B
I got results with [A] starting off at 0,5 0,375 0.335 0,355... nearing the correct answer 0,35.

Still it amazes me what this makes this reaction different from plain A<->2B+C

Thank you, now I am able to perform this calculation from now on.
 
  • #6
Kariga said:
Still it amazes me what this makes this reaction different from plain A<->2B+C

Have you tried to correctly balance this reaction?
 

1. What is chemical equilibrium with multiple Kc?

Chemical equilibrium with multiple Kc refers to a situation where a chemical reaction involves multiple steps and therefore has multiple equilibrium constants (Kc values).

2. How is chemical equilibrium with multiple Kc different from single Kc equilibrium?

In single Kc equilibrium, there is only one equilibrium constant that represents the ratio of products to reactants at equilibrium. In multiple Kc equilibrium, there are multiple equilibrium constants as there are multiple steps involved in the reaction.

3. How do you calculate the overall equilibrium constant (Kc) for a reaction with multiple steps?

The overall equilibrium constant is calculated by multiplying the individual equilibrium constants for each step in the reaction. For example, if a reaction has three steps with equilibrium constants Kc1, Kc2, and Kc3, then the overall equilibrium constant (Kc) would be Kc1 x Kc2 x Kc3.

4. Can the equilibrium constant (Kc) change during a reaction with multiple steps?

Yes, the equilibrium constant (Kc) can change during a reaction with multiple steps. This is because the concentrations of the reactants and products can change as the reaction progresses, leading to changes in the equilibrium constant values.

5. How does the equilibrium constant (Kc) affect the direction of a reaction with multiple steps?

The equilibrium constant (Kc) can be used to determine the direction of a reaction with multiple steps. If the overall equilibrium constant (Kc) is greater than 1, the reaction will favor the products at equilibrium. If the overall equilibrium constant (Kc) is less than 1, the reaction will favor the reactants at equilibrium.

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