2A(aq)-> B(aq) +C(aq)
Initial concentration of A and B is 1.00 M, with no C. Kc = 0.200
find equilibrium concentrations
A=0.200
Can not show work; it will not let me post it.
kc=0.200=C*B/A=C*(1.00 M)/(1.00 M)
C=0.200 M
A=1.00 M
B=1.00M
Sorry for the typo.
I don't buy Devron's solution.
..........2A ==> B + C
I........1.00...1.00..0
C........-2x.....x....x
E.....1.00-2x..1.00+x..x
Kc = 0.200 = (B)(C)/(A)^2
0.200 = )1.00+x)(x)/(1.00-2x)^2
Solve for x and evaluate 1-2x and 1+x
Read the problem wrong. The key words were, "initial concentration." I apologize, it was still early.
Devron, you're absolutely right and I thought that's what you did. I've made the same error but not recently.
You did the easy part... wheres the algebra to find x?
To find the equilibrium concentrations, we need to set up an ICE (Initial, Change, Equilibrium) table and use the given information.
The balanced chemical equation for the reaction is:
2A(aq) -> B(aq) + C(aq)
Let's set up the ICE table:
| A(aq) | B(aq) | C(aq) |
---------------------------------------
Initial | 1.00 M | 1.00 M | 0.00 M |
Change | -2x | +x | +x |
Equilibrium | 1.00 - 2x | 1.00 + x | x |
The "-2x" in the change column represents the decrease in the concentration of A, as two moles of A react to form one mole each of B and C. The "+x" in the change column represents the increase in the concentrations of B and C.
The equilibrium concentrations are given by the values obtained in the equilibrium column of the ICE table.
According to the given information, the value of Kc (the equilibrium constant) is 0.200. The expression for Kc can be written as:
Kc = [B] * [C] / [A]^2
Plugging in the equilibrium concentrations obtained from the ICE table:
0.200 = (1.00 + x) * (x) / (1.00 - 2x)^2
Now, solve this equation to find the value of x, which represents the equilibrium concentration of C. Once you find the value of x, you can substitute it back into the equilibrium column of the ICE table to calculate the equilibrium concentrations of A, B, and C.