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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: CrimpJiggler on January 14, 2012, 09:24:30 PM

Title: Calculating this equilibrium constant
Post by: CrimpJiggler on January 14, 2012, 09:24:30 PM
Heres the question:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg834.imageshack.us%2Fimg834%2F7974%2Fscreenshotbzu.png&hash=0afb794c43bd9e6e316a3096b7aeb8903502da43)
don't know why the hell I can't solve a simple equilibrium problem like this but I can't. I need to find out the equilibrium concentration of HBr but how do I do that when I don't know the starting concentration or the equilibrium constant? Also, if 0.5M of HBr is consumed, how does it only produce 0.13M of each product?
Title: Re: Calculating this equilibrium constant
Post by: Borek on January 15, 2012, 04:57:17 AM
Imagine you have 1 L of the mixture. If so, you started with 0.5 mole of HBr. If 0.13 mole of Br2 were produced, how much HBr was left? This is a simple stoichiometry.

Quote from: CrimpJiggler on January 14, 2012, 09:24:30 PM
Also, if 0.5M of HBr is consumed, how does it only produce 0.13M of each product?

It wasn't all consumed. 0.5M is the initial concentration, it doesn't mean there is no HBr left at equilibrium.
Title: Re: Calculating this equilibrium constant
Post by: Nisarg Shah on February 06, 2012, 01:43:48 AM
Quote from: CrimpJiggler on January 14, 2012, 09:24:30 PM
Heres the question:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fimg834.imageshack.us%2Fimg834%2F7974%2Fscreenshotbzu.png&hash=0afb794c43bd9e6e316a3096b7aeb8903502da43)
I need to find out the equilibrium concentration of HBr but how do I do that when I don't know the starting concentration or the equilibrium constant? Also, if 0.5M of HBr is consumed, how does it only produce 0.13M of each product?
                                           2HBr <--> Br2 + H2
At t=0s                                 0.5M        0M    0M
At t=tequilibrium                   (0.5-2x)M    xM    xM 

(Using stoichiometry, if 2x moles of Hbr are dissociated then x moles of Br2 and H2 will be formed at equilibrium. Assuming volume= 1 litre)

Since 0.13M of Br2 is formed at equilibrium, there x=0.13M.

K = [Br2] x [H2]/[HBr]2

The question doesn't mean that all of the HBr is consumed. HBr, H2 and Br2 all are present in the container in which the reaction has occurred. That is why they are said to be in equilibrium.
The reaction shifts backward or forward depending on various factors such as pressure, temperature, concentration of reactants and products, volume, etc.