On a given condition, the equilibrium concentration of HI, and are 0.80, 0.10 and 0.10 mole/litre. T — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution: Equilibrium Constant Calculation **Given Data:** - $[HI] = 0.80 \text{ mol/L}$ - $[H_2] = 0.10 \text{ mol/L}$ - $[I_2] = 0.10 \text{ mol/L}$ **Reaction:** $H_2 + I_2 \rightleftharpoons 2HI$ **Step 1:** Write the equilibrium constant expression. $$K_c = \frac{[\text{products}]}{[\text{reactants}]} = \frac{[HI]^2}{[H_2][I_2]}$$ **Step 2:** Substitute equilibrium concentrations. $$K_c = \frac{(0.80)^2}{(0.10)(0.10)}$$ **Step 3:** Calculate numerator and denominator. $$K_c = \frac{0.64}{0.01} = 64$$ **Answer: $K_c = 64$** (Option A) This large value indicates the equilibrium favors product formation, which is consistent with HI having a much higher concentration than the reactants.