reversing reactions
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Re: reversing reactions
In a forward reaction, you do products over reactants for the equilibrium equation. For the reverse it is technically reactants over products, so it makes sense that your K would be 1/K for the reverse.
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Re: reversing reactions
If you reverse the reaction simply take the reciprocal of the the original K value so 1/K.
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Re: reversing reactions
The value of K in a reverse reaction is the inverse of the forward reaction. If the value of K in a forward reaction is K, then the equilibrium constant in the reverse reaction is 1/K.
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Re: reversing reactions
The K of the reverse reaction will be the inverse of the K of the forward reaction
Kreverse = 1/Kforward
A + B -> C
Kforward = [C]/[A][B]
Kreverse = [A][B]/[C]
Kreverse = 1/Kforward
A + B -> C
Kforward = [C]/[A][B]
Kreverse = [A][B]/[C]
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Re: reversing reactions
When you reverse the reaction, you have to invert K. Therefore, for a forward reaction, the equilibrium constant is equal to K. For the reverse reaction, the equilibrium constant is equal to 1/K.
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Re: reversing reactions
The equilibrium constant (K) for a reverse reaction is simply the inverse of the K for the forward reaction. If you think about it logically it makes sense -- a reaction that heavily favors the products would not have much reactants being formed, and vice versa.
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Re: reversing reactions
Hi! In the reverse reaction, it would be 1/K since the fraction is flipped. Hope this helps :)
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Re: reversing reactions
When you reverse a reaction the equilibrium constant becomes the reciprocal of K so it becomes 1/k.
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Re: reversing reactions
For the reverse reaction, k of forward reaction will become 1/k in the reverse reaction.
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Re: reversing reactions
For the reaction P <-> R (reversible reaction), [P]/[R] will give us our K value. However, if we reverse what we define as our P and R, we would simply be flipping the equation, so the old R is on top and the old P is on the bottom. This will give us the inverse of our original K value, or simply put, 1/K.
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Re: reversing reactions
For the reverse reaction, K will become its inverse or reciprocal, which is 1/K.
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Re: reversing reactions
K becomes inverted. The way to remember this is that when an equation is reversed, what was once the products (numerator) becomes the reactants (denominator) and vice versa so the K equation would flip.
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Re: reversing reactions
when you take a reverse reaction, you will inverse the K value (1/K) to get the new K value.
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Re: reversing reactions
The equilibrium constant for a reverse reaction is the reciprocal of the one for the forward reaction, so it would be 1/K.
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Re: reversing reactions
An easy way to calculate the K value for the reverse reaction is to find the reciprocal of the forward reaction, which is K = (1/Forward). I find this calculation quick and simple instead of flipping the formula to K = (R)/(P) and plugging in values from there.
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Re: reversing reactions
When you reverse the reaction, k is inverted so it would then be 1/k or k^-1
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Re: reversing reactions
Reversing the reaction flips all Reactants to products and vice versa, thus when u are calculating K the top and bottom are flipped which is why we put 1/k when we reverse the reaction.
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Re: reversing reactions
K becomes 1/K when you reverse a reaction. This shows that the concentrations are the same during. equilibrium. This. can also be written as K-1
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Re: reversing reactions
Reverse reaction move towards the reactants, right? Does K only refer to the forward reaction or can we specify K(reverse)
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Re: reversing reactions
When you reverse a reaction, the K value of it is the inverse of the forward reaction. So to find K for the reverse reaction solve for 1/K.
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Re: reversing reactions
The K of a reverse reaction is the inverse of the K for the forward reaction.
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Re: reversing reactions
Hi! The K of the reverse reaction will be the reciprocal of the K of the forward reaction (1/K).
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Re: reversing reactions
K would just be 1/K when you reverse a reaction. I think Dr.Lavelle had an example of this in his first lecture.
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Re: reversing reactions
I was confused on this as well, all the above replies really helped me.. Thank you everyone :)!
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Re: reversing reactions
The reverse reaction K value is just the inverse of K for the forward reaction. Therefore it would be 1/K
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Re: reversing reactions
It would be 1/K because in a forward reaction, K = [P]/[R]. In a reverse reaction, the product in the forward reaction would become the reactant in the reverse reaction, and the same goes with the reactant in the forward reaction. Therefore, the K for the reverse reaction equals to [R]/[P]. [R]/[P] is the 1/([P]/[R]). Therefore, it would be 1/K.
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Re: reversing reactions
K is products over reactants so if you use the reverse reaction, K would be the reactants over the products. That's why K (reverse) is 1/K (forward)
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Re: reversing reactions
When reversing a reaction, the value of K becomes the inverse of the forward reactions K (1/K).
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Re: reversing reactions
When K is reversed, it becomes the inverse of the forward reaction, therefore 1/K. Hope this helps!
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Re: reversing reactions
K would also be inversed. It will become 1/K because the equilibrium constant for the inversed reaction is adopting a reversed product-reactant relationship.
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Re: reversing reactions
When you reverse a reaction, K is just equal to the reciprocal of K instead. This is 1 divided by K.
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Re: reversing reactions
The K of the reverse reaction would be the inverse. 1/K or K-1.
EX. [A] + [B] <-> [C]
KFORWARD = [C] / [A][B]
KREVERSE =[A][B] / [C] , which is the reciprocal/inverse of the forward reaction.
EX. [A] + [B] <-> [C]
KFORWARD = [C] / [A][B]
KREVERSE =[A][B] / [C] , which is the reciprocal/inverse of the forward reaction.
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Re: reversing reactions
It is the inverse of the forward reaction, so you would take 1/K to find the reverse constant.
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Re: reversing reactions
just like how you reversed the reaction, you would have k inversed so it would be 1/K
Re: reversing reactions
One good video that expressed an example of reversing a reaction was on this link https://www.youtube.com/watch?v=aJ0KNQ5-KaI
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Re: reversing reactions
K for a reversed reaction is equal to the inverse of the forward reaction's k.
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Re: reversing reactions
When you reverse the reaction, the K now becomes 1/K which would essentially be the inverse of the original K.
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Re: reversing reactions
When the reaction of K is reversed, K becomes the inverse of the reaction, 1/K.
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Re: reversing reactions
The K for the reverse reaction is the inverse (1/K) of the forward reaction, and vice versa :)
Re: reversing reactions
When you reverse the reaction, K simply becomes the inverse of itself (1/K).
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Re: reversing reactions
When you reverse a reaction, you should take the inverse of K, which means computing 1/K.
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Re: reversing reactions
For a reverse reaction, you should take the inverse of the original K in order to find the equilibrium constant (AKA: new K value)
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Re: reversing reactions
When you reverse reactions, you are going from products to reactants. Because of that, you would get K = 1/K because K is products over reactants and you would want the reverse, reactants over products because that is the new products/reactants.
Use inverse K (1/k, or K^-1) for a reverse reaction. Think of doing the problem backwards, flip the K.
Use inverse K (1/k, or K^-1) for a reverse reaction. Think of doing the problem backwards, flip the K.
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Re: reversing reactions
When you have a reaction A+B-->C and have its K value and then reverse the reaction to C-->A+B, the K value of this reverse reaction will be the inverse of the first K value.
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Re: reversing reactions
When you reverse a reaction you take the inverse of K, 1/K, because you are essentially just flipping the products and reactants of the reaction. So in the reverse reaction the products are the reactants from the original reaction and the reactants are the products.
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