The equilibrium constant for the reaction Br2(g) <---> 2Br(g) is Kc = 2.64 * 10–4
at 1400K.
What will the concentration of bromine atoms be if 0.250 mol Br2 comes to equilibrium
in a 0.500-L reaction vessel at this temperature?

A. 1.15 * 10^–2M

B. 1.32 * 10^–2M

C. 8.10 * 10^–3M

D. 1.62 * 10^–2M

I know the correct answer is A, but I do not know how to solve the equation. This was from my 2nd exam earlier in the year, im currently going over this material and material from my third exam since today is when my final is. ( I've been studying all week, but mostly over new material that was covered last week, and stuff from exam 1)
Any help is appreciated.

I keep getting the wrong answer....

What I did was :


K= Product/reactants



K= [2Br]^2/[Br_2]
K= [2x]^2/[.250-x]
2.64*10^-4 *.250 (drop the x since K is so small..) = .000066
.000066 = 2x^2

I noticed this did not give me the correct answer.... so I tried doing a ice table:

--- Br_2(g) <---> 2Br(g)
I ----- .250 --------- 0.00 ----

C ------ -x ------------- +2x ----

E -- .250-x ------------ +2x -----


2.64*10^-4 = [2x]^2 / [.250-x]
2.64*10^-4 * (.250-x) = [2x]^2


Sqrt(2.64*10^-4 * (.250-x)) = 2x

x= (sqrt(2.64*10^-4 *(.250-x)))/2


x= 0.00402915





Can anyone please help me ?

Looks like you are doing it mostly correct, but I suggest you throw your .5L into that equation (Kc = CONCENTRATION of products / CONCENTRATION of reactants, not just moles)


Kc = [2x/.5]^2 / [(.25-x)/.5]

Solve for x and you get 2.856 * 10^-3 mol

2x / .5L = 1.1*10^-2 M Not sure why A is 1.15 but close enough I guess

This post was edited by RzChaos on Apr 29 2013 05:32pm