Reverse Reactions

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Evonne Hsu 1J
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Reverse Reactions

Postby Evonne Hsu 1J » Wed Jan 06, 2021 6:41 am

Why do we have to take the reciprocal of the Kc/Kp when using the reverse reactions?

Taylor K 2D
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Re: Reverse Reactions

Postby Taylor K 2D » Wed Jan 06, 2021 8:19 am

I believe so!

Kelly Singh
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Re: Reverse Reactions

Postby Kelly Singh » Wed Jan 06, 2021 9:22 am

Evonne Hsu 1J wrote:Why do we have to take the reciprocal of the Kc/Kp when using the reverse reactions?


we have to take the reciprocal because it is always products/reactants, and when the reaction is reversed the original products become reactants and the original reactants become products. You can think about it like the reaction going from right to left instead of left to right.

OwenSumter_2F
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Re: Reverse Reactions

Postby OwenSumter_2F » Wed Jan 06, 2021 9:28 am

If it makes more sense, try working out the K value of a reaction one way, and then try working out the K value when it is reversed. You will find that the two K values are reciprocals of each other.

Anna Yang 1A
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Re: Reverse Reactions

Postby Anna Yang 1A » Wed Jan 06, 2021 11:42 am

Since Kc is [products]/[reactants], the value will change depending on which are the products and which are the reactants for the reaction. The reverse reaction will have these values swapped, so therefore, the reciprocal must be used.

Annika Tamaki 1E
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Re: Reverse Reactions

Postby Annika Tamaki 1E » Wed Jan 06, 2021 12:06 pm

If you reverse the reaction, the original reactants are now the product and the original products are now the reactants. If you were to plug those numbers into the Kc equation, you would get the reciprocal of the original reaction Kc. Hope this helps!

Emmeline Phu 1G
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Re: Reverse Reactions

Postby Emmeline Phu 1G » Wed Jan 06, 2021 12:27 pm

Hi! For reverse reactions, we take the reciprocal because to calculate K we always use the equilibrium concentration of reactants as the denominator and the equilibrium concentration of the products as the numerator. Thus, if the reaction was in reverse the reactants would become the products and the products would become the reactants. Therefore, when calculating K for the reverse reaction given that we already know K for the forward reaction, we use 1/K. Hope this helps! :)

Nadya Higgins 3F
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Re: Reverse Reactions

Postby Nadya Higgins 3F » Wed Jan 06, 2021 12:31 pm

Kc or Kp is essentially products over reactants. ([P]/[R]). If you want to find the reverse reaction, you will have to flip that ratio as well, making it reactants over products ([R]/[P]). This is why when finding the equilibrium constant for a reverse reaction, you will need to find the reciprocal of K from the forward reaction.

Eileen Quach Dis 2A
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Re: Reverse Reactions

Postby Eileen Quach Dis 2A » Wed Jan 06, 2021 1:36 pm

Since K is the concentration of products over the concentration of the reactants, and the reverse reaction is the same as the forward reaction but the previous "reactants" are now on the products side and vice versa, K for the reverse reaction would just be the reciprocal of K for the forward reaction.

Inderpal Singh 2L
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Re: Reverse Reactions

Postby Inderpal Singh 2L » Wed Jan 06, 2021 1:40 pm

You are essentially "flipping" the reactants and products of the forward reaction, and when you would "plug" this into the equation, you would end up with the reciprocal of the forward reaction. This applies in the other direction as well.

Samiha Molla 3G
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Re: Reverse Reactions

Postby Samiha Molla 3G » Wed Jan 06, 2021 3:04 pm

We have to do this because the reactants and products are essentially switching places. So, what goes on top and what goes on the bottom of the equilibrium constant equation also switch places, which is the same as taking the reciprocal!

AmyHo2K
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Re: Reverse Reactions

Postby AmyHo2K » Wed Jan 06, 2021 10:14 pm

You are switching the reactants and products of the forward reaction, and when you put this into the K equation, you would end up with the reciprocal of the first recation.

Uyen Trinh 3C
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Re: Reverse Reactions

Postby Uyen Trinh 3C » Wed Jan 06, 2021 11:12 pm

Professor Lavelle does a really good of explaining why you do this in his first lecture (I think it's 3/4 of the way through the lecture). Because the equilibrium constant is the ratio of the concentration of product to that of reactant, when you "flip" the reaction, you are "flipping the ratio". Thus, you now have the reciprocal of the first reaction, which is essentially the inverse of your first K. The inverse is understood as 1/K.


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