In a chemical reaction, substrate molecule \( A \) is broken down to form one molecule of product \( B \) and one molecule of product \( C \). The equilibrium constant, \( K \), for this reaction is 0.5 . If we start with a mixture containing only substrate \( A \) at a concentration of 1 M , what will be the concentration of \( A \) when the reaction reaches equilibrium? Choose one: \( \begin{array}{l}0.250 \mathrm{M} \\ 0.125 \mathrm{M} \\ 0.333 \mathrm{M} \\ 0.667 \mathrm{M} \\ 0.500 \mathrm{M}\end{array} \)

To find the equilibrium concentration of substrate A, we can set up the equilibrium expression based on the provided reaction. The reaction is: \[ A \rightleftharpoons B + C \] Given that \( K = 0.5 \), we can relate the concentrations at equilibrium: \[ K = \frac{[B][C]}{[A]} \] Let \( x \) be the concentration of A that reacts at equilibrium. Initially, we have 1 M of A, and as the reaction proceeds, A will decrease by \( x \) while B and C increase by \( x \): - At equilibrium: \( [A] = 1 - x \) - \( [B] = x \) - \( [C] = x \) Substituting these into the equilibrium expression gives us: \[ 0.5 = \frac{x \cdot x}{1 - x} = \frac{x^2}{1 - x} \] Rearranging this, we get: \[ 0.5(1 - x) = x^2 \] \[ 0.5 - 0.5x = x^2 \] \[ x^2 + 0.5x - 0.5 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 0.5, c = -0.5 \): \[ x = \frac{-0.5 \pm \sqrt{(0.5)^2 - 4(1)(-0.5)}}{2(1)} \] Calculating the discriminant: \[ (0.5)^2 + 2 = 0.25 + 2 = 2.25 \] So, \[ x = \frac{-0.5 \pm \sqrt{2.25}}{2} = \frac{-0.5 \pm 1.5}{2} \] This gives two possible \( x \) values: 1. \( x = \frac{1}{2} = 0.5 \) 2. \( x = \frac{-2}{2} = -1 \) (not a valid concentration) So, \( x = 0.5 \). Finally, we can find the concentration of A at equilibrium: \[ [A] = 1 - x = 1 - 0.5 = 0.5 \text{ M} \] Therefore, the concentration of A when the reaction reaches equilibrium is \( 0.500 \; \text{M} \).