A mixture initially contains A,B, and C in the following concentrations: A1 = 0.300 M, B1 = 1.15M, and C1 = 0.350 M. The following reaction occurs and equilibrium is established:
A+2B <---> C
At equilibrium, A2 = 0.190M and C2 = 0.460M.
Calculate the value of the equilibrium constant, Kc
I am not sure what to do:
This is what i did but i think its wrong:
SO since,
Kc= (C)^C/(A)^a*(B)^b
(0.350)/(0.300)(1.15)=(0.460/(0.190)(B)^2
=(0.350)/(0.345)/(0.460/(0.190)(B)^2
What i did is square root both sides to get rid of B2,then i solve for b i got 1.53.
Then i sub that in the equation:
Kc= 0.460/ (0.190)(1.53)
BUt the answer is wrong
2 answers
Never mind i got the right answer worked it out again
K = 2.80 is what I obtained when I worked the problem. But I think it is
Kc = 0.460/(0.190)(0.930)^2 = 2.80
Kc = 0.460/(0.190)(0.930)^2 = 2.80
2 answers