Explaining Q<K and Q>K  [ENDORSED]

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NatBrown1I
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Joined: Fri Sep 28, 2018 12:17 am

Explaining Q<K and Q>K

Postby NatBrown1I » Thu Jan 10, 2019 2:12 pm

Why is it true that if Q is smaller than K there are more reactants than products and therefore the forward reaction is favoured. If there are more reactants than products, wouldn't that mean that the reverse reaction is favoured?

(Like in example 5I.11b in the textbook)

Artin Allahverdian 2H
Posts: 76
Joined: Fri Sep 28, 2018 12:26 am

Re: Explaining Q<K and Q>K  [ENDORSED]

Postby Artin Allahverdian 2H » Thu Jan 10, 2019 2:19 pm

First, remember that Q=[products]/[reactants] at an instantaneous given time not at equilibrium. So when Q<K, it indicates that there is a higher concentration of reactants in the denominator than for the ratio for K, thus resulting in a smaller number for Q. Since we want the reaction to proceed to equilibrium, we want to create more product so that the ratio of products to reactants increases from Q to K, so the forward reaction is favored in order to generate more product in order to reach the desired K value.

Layal Suboh 1I
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Joined: Fri Sep 28, 2018 12:23 am

Re: Explaining Q<K and Q>K

Postby Layal Suboh 1I » Thu Jan 10, 2019 2:27 pm

If Q<K, more reactants are present in the reaction (Dr. Lavelle said that there may be some other source adding reactant) than products being removed, so in order to decrease the amount of reactant to reach equilibrium, the forward reaction is favored to produce more product.

If Q>K, there is a higher concentration of products than reactants, so the reverse reaction is favored so that products decompose to reactant in order to reach equilibrium.

Melody P 2B
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Joined: Fri Sep 29, 2017 7:06 am

Re: Explaining Q<K and Q>K

Postby Melody P 2B » Thu Jan 10, 2019 4:11 pm

So, when Q<K it just means the reaction isn't at equilibrium, but before it right? Because there are more reactants than products, so in order to reach K the reaction will proceed forward.

And when Q>K the reaction isn't at equilibrium, but after it? Because there are more products in the product/reactant ratio, so in order to reach K the reaction will favor the reverse rxn (reactants)?

pamcoronel1H
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Joined: Fri Sep 28, 2018 12:25 am

Re: Explaining Q<K and Q>K

Postby pamcoronel1H » Thu Jan 10, 2019 5:11 pm

If Q<K, that means that K is bigger and there is a higher value on the numerator of the k equation, i.e. there's more products at equilibrium than what we currently have. So, the reaction will need to proceed forward in order to get the amount of product that is favorable in equilibrium.

If Q>K, that means that K is snaller and so the denominator needs to be bigger, i.e. there's currently not enough reactants. So, the reaction will favor a reverse reaction in order to get more reactants and be in equilibrium.

LedaKnowles2E
Posts: 62
Joined: Fri Sep 28, 2018 12:27 am

Re: Explaining Q<K and Q>K

Postby LedaKnowles2E » Thu Jan 10, 2019 5:15 pm

If Q is less than K, it means that there are more reactants and less products than at equilibrium, which means more of the reactants have to react and become products in order to reach equilibrium.

Nicole Elhosni 2I
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Joined: Fri Sep 28, 2018 12:28 am

Re: Explaining Q<K and Q>K

Postby Nicole Elhosni 2I » Thu Jan 10, 2019 11:01 pm

If Q < K and the desire is to get the ratio to equal K as the reaction is in the process of reaching equilibrium, then we have to think about how we can increase Q. Both are calculated by dividing the concentration of products raised to their stoichiometric coefficients by the concentration of reactants raised to their stoichiometric coefficients. So in order to increase Q, the concentration of products must increase and therefore the forward reaction will be favored at that moment given those concentrations.

Ruiting Jia 4D
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Joined: Fri Sep 28, 2018 12:27 am

Re: Explaining Q<K and Q>K

Postby Ruiting Jia 4D » Sat Jan 12, 2019 8:54 pm

You can think of it like a pendulum. When K means the pendulum is in the middle (at equilibrium), Q lets you know how far it is from equilibrium based on its relation to K. If Q swings left, such as the arrow "<", you can visualize it as being driven to products because it is before reaching equilibrium (the middle). The reverse happens when Q swings right and Q > K.


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