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: Finding Kc for the Equilibrium **Given:** - Equilibrium: $2X + Y \rightleftharpoons YX_2$ - $[X]_{eq} = 4$ mol - $[Y]_{eq} = 2$ mol - $[YX_2]_{eq} = 2$ mol **Step 1: Write the expression for Kc** For the reaction $2X + Y \rightleftharpoons YX_2$: $$K_c = \frac{[YX_2]}{[X]^2[Y]}$$ **Step 2: Substitute equilibrium concentrations** $$K_c = \frac{(2)}{(4)^2(2)}$$ **Step 3: Calculate** $$K_c = \frac{2}{16 \times 2} = \frac{2}{32} = \frac{1}{16} = 0.0625$$ Or expressed as a fraction: $K_c = \frac{1}{16}$ **Answer: B** (which corresponds to $K_c = 0.0625$ or $\frac{1}{16}$) The key is correctly applying the equilibrium constant expression with stoichiometric coefficients as exponents: the concentration of $X$ is raised to the power of 2 (its stoichiometric coefficient).