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?

Question

Quizplus · Accepted Answer

The equilibrium constant, $K_{eq}$, for the reaction $A(g) \rightleftharpoons 2B(g)$ is given by the expression $K_{eq} = \frac{[B]^2}{[A]}$. Substituting the given concentrations into this expression gives $K_{eq} = \frac{(0.154)^2}{0.381} = 0.0622$.

Consider the Reversible Reaction: A(g)⇌ 2B(g) at Equilibrium,the Concentration of a Is