A system is represented by the following equation:
2A(aq) + B(aq) --> 3C(aq)
At equilibrium, [C]=0.6 M. Calculate the value of K if initially 1 mol of A and 2 mol of B were placed in a 1 L flask.
we will assume that the aqueous solutions fill the 1 L flask.
At start
1 mol A + 2 mol B and 0 mol C
At equilibrium
1-2x mol of A and 2-x mol of B and 3x mol of C
so x=0.2 M
K=x^3/(1-2x)^2(2-x)
K=(0.2)^3/(0.6)^2(1.8)
K=0.008/(0.36)(1.8)
K=0.012 (with no units)
I am having trouble with this question so I decided to check online. The question sheet in which I got this question from came with an answer sheet, and it says the answer for k = 0.33. Another online source gave an answer of 0.024. I'm not sure what values are needed in the change row.
2A + B <----> 3C
INITIAL 1 2 0
CHANGE +3X
EQUILIBRIUM 0.6 M
If 0.6 = 3x, x should equal 0.2? How is x affecting A + B?
To calculate the value of K, you need to use the equilibrium expression and the concentrations of the species involved. The equilibrium expression for the given equation is:
K = [C]^3 / ([A]^2 * [B])
Given that [C] = 0.6 M, we can substitute this value into the equilibrium expression:
K = (0.6 M)^3 / ([A]^2 * [B])
Now, we need to find the concentrations of A and B. Initially, 1 mol of A and 2 mol of B were placed in a 1 L flask, which means the initial concentration of A, [A], is 1 M, and the initial concentration of B, [B], is 2 M.
Substituting these values, we have:
K = (0.6 M)^3 / (1 M)^2 * (2 M)
Simplifying the equation:
K = 0.6^3 / 1 * 2
K = 0.216 / 2
K = 0.108
So, the value of K for the given system is 0.108.