2A - B
At room temperature about 10% of A is converted to B. What is the value of equilibrium constant? (is this question right?, i mean, if we'll suppose A's concentration as 1M then Keq will be 0.06, if we'll suppose it as 2M then Keq will be 0.03, Does Keq depend on concentration for a specific reaction, or is it constant?)

Question

2A -> B
At room temperature about 10% of A is converted to B. What is the value of equilibrium constant? (is this question right?, i mean, if we'll suppose A's concentration as 1M then Keq will be 0.06, if we'll suppose it as 2M then Keq will be 0.03, Does Keq depend on concentration for a specific reaction, or is it constant?)

jfraser · Answer

I'd use the percentages themselves, rather than a set stoichiometric value. Suppose you start with 100% A , and 0% B. At equilibrium, you've lost 10% of the original A value, but formed only HALF that much in B, because if the stoichiometry
$$K_{EQ} = \frac{[B]}{[A]^2} = \frac{[0.05]}{[0.90]^2}$$