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Chemistry 12 - Keq & equilibirum - QUESTION

  • Thread starter 1calculus1
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Consider the following equilibrium:
CO(g) + 2H2(g) <-> CH3OH(g)
Some CO, H2 and CH3OH were placed in a 1.0L container. When equilibrium
was established, the [CO] had increased. Which of the following is true?

A. Trial Keq > Keq so reaction shifted left to reach equilibrium.
B. Trial Keq < Keq so reaction shifted left to reach equilibrium.
C. Trial Keq > Keq so reaction shifted right to reach equilibrium.
D. Trial Keq < Keq so reaction shifted right to reach equilibrium.


So, my answer is letter "b". However, I had found out from the provincial answer that this is wrong. Now, I'm boggled because keq = [products]/[reactants] therefore, IF the [CO] has increased which is a reactant, the products would be smaller than the reactant and therefore resulting a smaller keq than the original keq.

So, am I wrong? Or is the provincial answer wrong?
(The answer from the provincial is letter "a")
 

Answers and Replies

  • #2
Borek
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Sorry, I am not familar with Keq - is it just a reaction quotient? Becasue K is usually reserved for equilibrium constants, which is - as name implies - constant, so Keq = Keq no matter what you do. On the other hand reaction quotient is identical to K (ie [products]/[reactants]) but it describes current situation, so it can take any value, depending on circumstances.
 
Last edited:
  • #3
1,901
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Consider the following equilibrium:
CO(g) + 2H2(g) <-> CH3OH(g)
Some CO, H2 and CH3OH were placed in a 1.0L container. When equilibrium
was established, the [CO] had increased. Which of the following is true?

A. Trial Keq > Keq so reaction shifted left to reach equilibrium.
B. Trial Keq < Keq so reaction shifted left to reach equilibrium.
C. Trial Keq > Keq so reaction shifted right to reach equilibrium.
D. Trial Keq < Keq so reaction shifted right to reach equilibrium.


So, my answer is letter "b". However, I had found out from the provincial answer that this is wrong. Now, I'm boggled because keq = [products]/[reactants] therefore, IF the [CO] has increased which is a reactant, the products would be smaller than the reactant and therefore resulting a smaller keq than the original keq.

So, am I wrong? Or is the provincial answer wrong?
(The answer from the provincial is letter "a")
It's obviously "a". If trial Keq > Keq it means the products' concentrations are too high with respect to the reagents' conc., so the reaction shifts to the left.
 

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