The equilibrium concentration of X, Y and YX2 are 4, 2 and 2 moles respectively for the equilibrium — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution: Calculating Kc for the Equilibrium **Given Information:** - Equilibrium: $2X + Y \rightleftharpoons YX_2$ - $[X]_{eq} = 4$ mol - $[Y]_{eq} = 2$ mol - $[YX_2]_{eq} = 2$ mol **Step 1: Write the equilibrium constant expression** For the reaction $2X + Y \rightleftharpoons YX_2$: $$K_c = \frac{[YX_2]}{[X]^2[Y]}$$ **Step 2: Substitute equilibrium concentrations** $$K_c = \frac{2}{(4)^2 \times 2}$$ **Step 3: Calculate** $$K_c = \frac{2}{16 \times 2} = \frac{2}{32} = \frac{1}{16} = 0.0625$$ **Answer: Option B** — The value of $K_c = \frac{1}{16}$ or $0.0625$ The small value of $K_c$ indicates the equilibrium favors reactants, which is consistent with the relatively high concentrations of $X$ and $Y$ compared to the product $YX_2$.