K = kforward/kreverse
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K = kforward/kreverse
Just to make sure, there are times where kreverse is written as k', correct? So in this case the equilibrium constant K would be rate of forward reaction k over the rate of reverse reaction k'?
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Re: K = kforward/kreverse
Yes, and this relationship is important for the pre-equilibrium method in determining a rate law with a fast then slow step.
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Re: K = kforward/kreverse
That is correct. However I would be careful though. Sometimes k' is used to refer to the pseudo-rate constant of a reaction, although I don't believe that has been the case in this class, so I wouldn't worry too much about confusing the two types of k' on the formula sheet or anything.
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Re: K = kforward/kreverse
Yes that's correct. k' can also be used in pseudo rate laws, so don't confuse the two because they use the same notation.
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Re: K = kforward/kreverse
That's correct! However, it can also be used as the pseudo-rate constant in a reaction, so I would be careful and check the units to make sure.
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Re: K = kforward/kreverse
Yes that is correct. If you increase the rate of the forward reaction, then the equilibrium constant K will increase. If you increase the rate of the reverse reaction, the equilibrium constant K will decrease.
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Re: K = kforward/kreverse
Hello! That is correct, as kreverse can be expressed as k', however, k' with the " ' " notation utilized could also be used as the pseudo-rate constant in a reaction. It is crucial that you are aware of all units before moving forward so that you do not mix them up.
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Re: K = kforward/kreverse
yes, the k' is both the notation for both reverse and psuedo rates. Therefore, it is important to remember in context what you are working on. Also, remember that k/k' is equal to the K of the equilibrium will help for some problems
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Re: K = kforward/kreverse
This is correct, but make sure you acknowledge the units because k could also be used as the pseudo-rate constant in a reaction.
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Re: K = kforward/kreverse
Yes indeed, K (the equilibrium constant) is equivalent to the forward reaction rate over the reverse reaction rate. If the fwd reaction rate is favored, K will be large and favor products.
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Re: K = kforward/kreverse
Yes! The ratio of the forward k and reverse k equals the equilibrium constant
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Re: K = kforward/kreverse
This is correct, although there may be different notation for the rate constant of the reverse reaction. Just take note of the notation, as the question should make it clear what k' means or what value is the value for the reverse reaction.
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Re: K = kforward/kreverse
Yes, K can be classified as kfoward over kreverse. The shorthand of writing this would be K = k/k'. Make sure to note how if one changes, it will affect K. For instance, if you increase the temperature to an endothermic reaction, the reaction would go towards the product site, which increases kforward which will increase K overall.
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