The equilibrium constant of the reaction A <--> 2B + C is 25.6 mol^2/L^2 at an unspecified temperature. The equilibrium concentrations in this system are [A]= 0.110 M, [B]= 1.78 M and [C]= 0.890 M. The reaction system is slugged with an additional amount of reactant, equal to a concentration of 0.270 M, instantaneously raising the [A] concentration to 0.380 M. The system then undergoes a shift and reestablishes equilibrium. Calculate the new equilibrium concentrations.
I'm trying to learn about chemical equilibria, and Le Châtelier’s principle; I think if I have the solution to this problem, it will a little bit more clear.
2007-08-31 15:44:47 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
K = [B}^2[C]/[A]
2007-08-31 19:34:49 · answer #1 · answered by ag_iitkgp 7 · 0⤊ 1⤋
A <------> 2B + C
initial concentration
0.110 . . 1.78 .. 0.890
at new equilibrium
0.380-x. . . 1.78+2x. .. . 0.890+x
25.6 = ( 1.78+2x) ( 0.890+x) / 0.380 -x
solving by quadratic formula x = 0.225
[A] = 0.380 - 0.225 = 0.155 M
[B] = 1.78 + 2(0.225) = 2.23 M
[C] = 0.890 + 0.225 = 1.12 M
2007-09-01 05:41:57 · answer #2 · answered by Dr.A 7 · 0⤊ 0⤋