Answer : The concentration of B at equilibrium is 0.0262 M.
Solution :
The given equilibrium reaction is,
[tex]A(aq)\rightleftharpoons 3B(aq)[/tex]
Initially conc. 3.60 0
At eqm. (3.60-x) 3x
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[B]^3}{[A]}[/tex]
Now put all the value in this equation, we get:
[tex]4.99\times 10^{-6}=\frac{(3x)^3}{(3.60-x)}[/tex]
By solving the term, we get:
x = 0.00872 M
Thus, the concentration of B at equilibrium = 3x = 3(0.00872 M) = 0.0262 M
Therefore, the concentration of B at equilibrium is 0.0262 M.
