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### Reverse Reactions

Posted: **Sun Dec 03, 2017 10:32 pm**

by **welcometochillis**

If the k value for a system in equilibrium is a number, will it's reverse reaction be the same number? I assume yes because the rate at which products and reactants "go" into another should be the same?

### Re: Reverse Reactions

Posted: **Sun Dec 03, 2017 10:36 pm**

by **Carlos Gonzales 1H**

The K value of the reverse reaction will be the reciprocal of the forward reaction, so the reverse reaction will be 1/K.

### Re: Reverse Reactions

Posted: **Sun Dec 03, 2017 10:37 pm**

by **Kathleen Vidanes 1E**

No,the K value for the reverse reaction would be the inverse of the forward reaction. Using the example given in class, if the K value for the forward reaction = 61.0, the K value for the reverse reaction will = 1/61.0 or 0.016

### Re: Reverse Reactions [ENDORSED]

Posted: **Sun Dec 03, 2017 10:37 pm**

by **CameronJohari1J**

The k value of the reverse reaction is inverse of k.

If k is the equilibrium value for A+B<-->C K=C/(A*B)

then C<-->A+B 1/K=(A*B)/C