## Quiz 2 Summer Workbook problem [ENDORSED]

Roya_N
Posts: 8
Joined: Fri Jun 17, 2016 11:28 am

### Quiz 2 Summer Workbook problem

In the last part the of Quiz 2 Preparation on question #9 asks:
"A researcher fills a 1.00L reaction vessel with 2.80x10^(-5) moles of BrCl gas and heats it to 500K. At equilibrium, 8.12% of the BrCl gas remains. Calculate the equilibrium constant, Kc, assuming the reaction: BrCl(g) <---> Br2(g) + Cl2(g)"

I am having a hard time understanding how to approach this problem, if anyone can briefly tell me how they solved this, it will be greatly appreciated.

Hope studying is going well for all of you!

-Roya Noori

Ana Ordonez 1G
Posts: 13
Joined: Fri Jun 17, 2016 11:28 am

### Re: Quiz 2 Summer Workbook problem  [ENDORSED]

Well what I did was that the initial concentration of BrCl is given which would be 2.80x10^-5 M. I ignored the 8.12% at first because it can get confusing so I just then I created the ICE box.
2BrCl Br2 Cl2
I (M) 2.80x10^-5 0 0
C (M) -2X +X +X
E (M) (2.80x10^-5)-2X X X

After I saw that they technically are giving us the concentration of BrCl at equilibrium which is that 8.12% remained of what was initial there.
So with that I put (2.80x10^-5)-2X= (2.80x10^-5)(0.0812) and solved for X. With that we now have the concentrations of BrCl, Br2, and Cl2 so we can solve for Kc.
Hopefully this helped you!
Goodluck studying

Roya_N
Posts: 8
Joined: Fri Jun 17, 2016 11:28 am

### Re: Quiz 2 Summer Workbook problem

Ana Ordonez 1E wrote:Well what I did was that the initial concentration of BrCl is given which would be 2.80x10^-5 M. I ignored the 8.12% at first because it can get confusing so I just then I created the ICE box.
2BrCl Br2 Cl2
I (M) 2.80x10^-5 0 0
C (M) -2X +X +X
E (M) (2.80x10^-5)-2X X X

After I saw that they technically are giving us the concentration of BrCl at equilibrium which is that 8.12% remained of what was initial there.
So with that I put (2.80x10^-5)-2X= (2.80x10^-5)(0.0812) and solved for X. With that we now have the concentrations of BrCl, Br2, and Cl2 so we can solve for Kc.
Hopefully this helped you!
Goodluck studying

Thank you so much!!