Reversing Reactions
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Re: Reversing Reactions
The K for a reverse equation is the reciprocal of K, so 1/K. Hope this helps!
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Re: Reversing Reactions
It would just be the reciprocal 1/K because the original K is the forward reaction's products over reactants. The reverse would be reactants over products of the forward reaction.
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Re: Reversing Reactions
Another way you can think of this is that when you reverse a reaction, you are multiplying the stoichiometric coefficients of the equation by -1, so the new K is K^-1. This is the same as how if you double the stoichiometric coefficients, the new K is K^2.
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Re: Reversing Reactions
When you reverse a reaction, the proportions of K are flipped which is why you take the inverse of K (1/K).
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Re: Reversing Reactions
To calculate the K for a reverse reaction, you would take the 1/K of the forward reaction because the numerator and reactants are now switched.
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Re: Reversing Reactions
The K of the reverse reaction is the inverse of the forward reaction.
A + B <-> C
Forward reaction: K = [C] / [A][B]
Reverse Reaction: K = [A][B] / [C] = 1/K
A + B <-> C
Forward reaction: K = [C] / [A][B]
Reverse Reaction: K = [A][B] / [C] = 1/K
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Re: Reversing Reactions
When you reverse a reaction, you would take the reciprocal of K so it would be 1/k.
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Re: Reversing Reactions
Hi! When you reverse a reaction, K becomes the reciprocal, so you would calculate 1/K.
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Re: Reversing Reactions
It is 1/K. Also remember that K is going to be different than enthalpy, in enthalpy its a sign change.
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Re: Reversing Reactions
K is the reciprocal of its forward reaction value. It'll just be 1/K for the reverse reaction.
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Re: Reversing Reactions
When K represents the forward reaction, then 1/K represents the reverse reaction.
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Re: Reversing Reactions
When you reverse a reaction that you know K, you can assume that the equilibrium constant for that reaction will be 1/K if the temperature remains the same.
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Re: Reversing Reactions
When you reverse a reaction, the k value will simply be the inverse of the original forward reaction. 1/k or k^-1
Re: Reversing Reactions
usually they just simply inverse and from k it goes to 1/k and then we can solve for it from there.
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Re: Reversing Reactions
When you reverse a reaction, the equilibrium constant of the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction (Kreverse = 1/Kforward). Hope this helps!
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Re: Reversing Reactions
When a reaction is reversed the proportions of K are flipped so you take the inverse of K (1/K)
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Re: Reversing Reactions
Additionally, when multiplying a whole equation, its K is raised that power as well.
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Re: Reversing Reactions
When you reverse a reaction the new reaction coefficient is the reciprocal of the original reaction coefficient. This makes sense if you consider writing K with the new "products" and "reactants"
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Re: Reversing Reactions
Hi, when you reverse a reaction K would become the reciprocal of itself, so it would be 1/K.
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Re: Reversing Reactions
The reverse of a reaction reverses which group is the products and which are the reactants. So for R <--> P, instead of P/R, K would be R/P for the reverse reaction. That means K would now be 1/K of the forward reaction.
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Re: Reversing Reactions
For the forward reaction, K = [products] / [reactants]. The K for the reverse reaction equals the inverse of the equilibrium constant (K) for the forward reaction. K forward = 1 / K reverse.
Re: Reversing Reactions
Can someone clarify the following manipulations to the equilibrium constant Kc?
- Reverse reaction: 1/Kc
- Multiplying reaction by a coefficient: (Kc)^coefficient
- Summing 2 reactions with difference K constants: K(K)
Thanks in advance.
- Reverse reaction: 1/Kc
- Multiplying reaction by a coefficient: (Kc)^coefficient
- Summing 2 reactions with difference K constants: K(K)
Thanks in advance.
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Re: Reversing Reactions
K would be now become the inverse of what it was initially. Hence, it would be 1/K because the expression would be flipped once you reverse the reaction.
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Re: Reversing Reactions
A reverse reaction means you take the reciprocal of K, as calculating this K value would mean that the numerator and denominator are swapped. Therefore, K for a reverse reaction would just be 1/k.
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Re: Reversing Reactions
For the reverse reaction, the products are now reactants and the reactants are now products, meaning that k is "flipped," or 1/K.
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Re: Reversing Reactions
As many others have written, a forward reaction equilibrium constant K can be used to calculate that for the reverse by determining the reciprocal of K (K' = K^-1). In terms of how this can be rationalized (not memorized), the "reverse" equilibrium mixture would have products in place of reactants and vice versa compared to the "forwards" equilbrium. As such, the numerators and denominators are flipped.
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Re: Reversing Reactions
The K value will change to its inverse: 1/K. This makes sense mathematically since the numerators and denominators will be switched as the products and reactants are switched, repectively.
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Re: Reversing Reactions
When you reverse a reaction just as the products will move to the denominator so will K because K represents products/reactants thus reactants/products would indicate 1/K.
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Re: Reversing Reactions
When you reverse a reaction, your equilibrium constant turns into 1/K, K being the constant of the forward reaction.
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Re: Reversing Reactions
Reversing a reaction means that the reaction would switch the reactants with the products, and the products with the reactants. This means that K would go from being [products]/[reactants] to being [reactants]/[products]. To calculate this more efficiently, the new K of the reverse reaction would be 1/K from the initial reaction.
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Re: Reversing Reactions
Hello! When you reverse a reaction you are essentially doing the same thing but flipping the products and reactants. This in turn will lead you to find the inverse of your K in order to get the reverse reaction value. You would find this by using 1/K. Hope this helps.
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Re: Reversing Reactions
Prof. Lavelle stated this on Lecture 1, it's around the midpoint, maybe a little less than the midpoint of the lecture. He explained that the reverse of the reactions eq constant should just be the inverse of K so 1/K. He shows that it is pretty much flipping the structure of the equilibrium constant equation by changing place of the reactants and the products.
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Re: Reversing Reactions
When you reverse a reaction, the equilibrium constant would be the inverse of K (1/K).
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Re: Reversing Reactions
Hello! When you reverse a reaction, the equilibrium constant would be the inverse of K, so 1/K. You flip the products and reactants.
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Re: Reversing Reactions
If you are looking at the reverse reaction, take the reciprocal of K. This means the K for the reverse reaction is 1/K.
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Re: Reversing Reactions
For the reverse reaction, the K of that will be the reciprocal of the forward reaction (1/K)
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Re: Reversing Reactions
Hello!
When you reverse a reaction, the products of the old reaction become the new reactants (vice versa for the old reactants). Using the K formula, it would mean that the K value of the reverse reaction would be the reciprocal (1/K) of the old reaction. Hope this helps!
When you reverse a reaction, the products of the old reaction become the new reactants (vice versa for the old reactants). Using the K formula, it would mean that the K value of the reverse reaction would be the reciprocal (1/K) of the old reaction. Hope this helps!
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