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
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
To solve this problem and calculate the equilibrium constant, Kc, we need to use the concentrations of A, B, and C at equilibrium.
The reaction equation is: A + 2B ⇌ C
We are given the initial concentrations (A1, B1, and C1) and the equilibrium concentrations (A2 and C2):
A1 = 0.300 M, B1 = 1.15 M, C1 = 0.350 M
A2 = 0.190 M, C2 = 0.460 M
The equilibrium constant, Kc, can be calculated using the formula:
Kc = ([C]^c)/([A]^a * [B]^b)
where [C], [A], and [B] are the concentrations of C, A, and B at equilibrium, and c, a, and b are the respective coefficients of C, A, and B in the balanced equation.
In this case, we have: c = 1, a = 1, and b = 2.
Plugging in the given values, we have:
Kc = (0.460^1)/(0.190^1 * 1.15^2)
Kc = 0.460/(0.190 * 1.3225)
Kc = 0.460/0.251 = 1.833 M^-1
So, the value of the equilibrium constant, Kc, is 1.833 M^-1.