A=0.200
Can not show work; it will not let me post it.
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
6 answers
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.
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
..........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?