Answer :
The equilibrium concentration of A is approximately 0.65 M, corresponding to option (a), calculated using the equilibrium constant expression and given equilibrium concentrations of B and C.
To find the equilibrium concentration of A in the reaction (2A + B ⇌ C), we'll use the equilibrium constant expression [tex]\(K_c = \frac{[C]}{[A]^2[B]}\)[/tex], where (K_c = 1.6), (B = 1.9 M), and (C = 0.52 M). Rearranging the equation to solve for ([A]), we get [tex]\([A]^2 = \frac{[C]}{K_c[B]}\)[/tex]. Substituting the given values, we find [tex]\([A]^2 = \frac{0.52}{1.6 \times 1.9}\)[/tex]. Solving for ([A]), we obtain approximately 0.65 M, corresponding to option (a). This concentration represents the equilibrium concentration of A, ensuring that the reaction reaches a state where the rates of the forward and reverse reactions are equal, achieving dynamic equilibrium.
