Asked by Chemistry Chick
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?
Answers
Answered by
DrBob222
A ==> 2B
Kc = (B)^2/(A)
Kc = (0.154)^2/(0.381)
You do the math.
Kc = (B)^2/(A)
Kc = (0.154)^2/(0.381)
You do the math.
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