If yes what does this mean ? What would the graph look like ?

56 minutes ago, HbWhi5F said:

What would the graph look like ?

Two points on a straight line at +1 and -1.

Can you post a scan of the source of your equation?

@sethoflagos

What the text is saying is that the equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward reaction. So, it should not be:

K_c = 1/K_c

but:

K_c^reverse = 1/K_c

In the text, I can't see whether they distinguished between the equilibrium constants of the forward and reverse reactions, but it would be wrong if they did not.

Edited September 13Sep 13 by KJW

2 hours ago, HbWhi5F said:

@sethoflagos

It's quite difficult to be sure of the subscripts, but I think there is an error in Eq. 6.8. The author seems to be using K_c for the equilibrium constant of both forward and backward reactions. This is very confusing. At equilbrium: forward and backward REACTION RATES are equal; equilibrium constants equal the ratio of REACTION RATE CONSTANTS for their respective direction; hence the forward and backward EQUILIBRIUM CONSTANTS are the inverses of EACH OTHER. This does not imply that they are each equal to their own inverse.

Seems just nuts to me, but hopefully @exchemist or @John Cuthber will be able to provide more expert details of the chemistry here.

Simulpost with @KJW : Same idea.

If you look at a clearer page

https://ncert.nic.in/textbook/pdf/kech106.pdf

The equation is K’=1/K, though it’s still not easy to discern the “prime” mark

(knowing the chapter and unit number would speed up the process of digging in to all this)