## Clarification from lecture 1 - Jan 7th

105169446
Posts: 32
Joined: Fri Sep 28, 2018 12:16 am

### Clarification from lecture 1 - Jan 7th

I was absent on the day of the first lecture, so there is something in the notes I received that I don't quite understand.
When we are given the example of a forward reaction with N2 and H2 as reactants and NH3 as the product. We calculate the Kc for the forward reaction by replacing the Kc formula with the values given for each reactant and product, and get 61.0 as a result.
What I am confused about is why when we calculate the Kc for the reverse reaction, the values we put for [N2], [H2]^3 and [NH3]^2 are not the ones given at the beginning of the example (we put ([N2]*[H2]^3)/[NH3]^2=1/61.0). I don't understand where the 1 and the 61.0 are coming from.

Amy Lefley 1J
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### Re: Clarification from lecture 1 - Jan 7th

When you are finding the equilibrium constant of the reverse reaction, you simply take the inverse of the equilibrium constant for the forward reaction.

Danielle_Gallandt3I
Posts: 70
Joined: Fri Sep 28, 2018 12:24 am

### Re: Clarification from lecture 1 - Jan 7th

The K value of a reverse reaction is always the inverse of the K value for the forward reaction.

Searra Harding 4I
Posts: 68
Joined: Fri Sep 28, 2018 12:29 am

### Re: Clarification from lecture 1 - Jan 7th

When you know the forward reaction and are trying to find the backward reaction you can simply take the inverse of your answer.

### Who is online

Users browsing this forum: No registered users and 1 guest