11.69 (6th Edition)

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Chloe Likwong 2K
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Joined: Fri Sep 28, 2018 12:23 am

11.69 (6th Edition)

Postby Chloe Likwong 2K » Tue Jan 15, 2019 11:49 pm

Equation: CO(g) + H2O(g) ⇌ CO2(g) + H2(g)

Question: (d) If the concentration of H2O is decreased, what happens to the equilibrium constant for the reaction?

Answer: (d) The equilibrium constant for the reaction is unchanged because it is unaffected by any change in concentration.

Can someone please explain why doesn't the concentration decrease even if H2O is a gas in the equation? Thanks!

Jane Burgan 1C
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Joined: Fri Sep 28, 2018 12:15 am

Re: 11.69 (6th Edition)

Postby Jane Burgan 1C » Tue Jan 15, 2019 11:56 pm

Changing the concentration of reactants and products don't change K because K is a constant - the ratio of products to reactants. When the concentration of the reactant is changed, then the reaction shifts in the direction of the products until the original ratio of P to R is restored. When the concentration of the product is changed, then the reaction shifts in the direction of the reactants until the original P/R ratio is restored.

Elizabeth Gallmeister 1A
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Joined: Fri Sep 28, 2018 12:21 am

Re: 11.69 (6th Edition)

Postby Elizabeth Gallmeister 1A » Wed Jan 16, 2019 9:24 am

Changing the concentration of a product or reactant in the gas or aqueous phase only changes the Q value, as the reaction will no longer be at equilibrium. K does not change.

Melissa Bu 1B
Posts: 36
Joined: Fri Sep 28, 2018 12:19 am

Re: 11.69 (6th Edition)

Postby Melissa Bu 1B » Wed Jan 16, 2019 2:41 pm

The equilibrium constant K is a constant that describes the ratio of concentrations of products/concentrations of reactants at equilibrium (when the rate of the forward reaction is equal to the rate of the reverse reaction). Therefore, it is only "relevant" when the reaction is at equilibrium. Q is calculated in the same way as K, but it describes the ratio of concentrations of products/concentrations of reactants instantaneously (at any time) except when the reaction is at equilibrium.


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