Post a New Question


posted by .

Consider the reversible reaction: A(g)-2B(g)At equilibrium, the concentration of A is 0.381 M and that of B is 0.154 M. What is the value of the equilibrium constant, Keq?

  • Chemistry -

    A ==> 2B
    Kc = (B)^2/(A)
    Kc = (0.154)^2/(0.381)
    You do the math.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question