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
Question
On a given condition, the equilibrium concentration of HI, $H_2$ and $I_2$ are 0.80, 0.10 and 0.10 mole/litre. The equilibrium constant for the reaction $H_2$ + $I_2$ 2HI will be
💡 Solution & Explanation
# Solution: Equilibrium Constant Calculation **Given reaction:** $H_2 + I_2 \rightleftharpoons 2HI$ **Equilibrium concentrations:** - $[HI] = 0.80$ mol/L - $[H_2] = 0.10$ mol/L - $[I_2] = 0.10$ mol/L **Step 1: Write the equilibrium constant expression** $$K_c = \frac{[HI]^2}{[H_2][I_2]}$$ **Step 2: Substitute the equilibrium concentrations** $$K_c = \frac{(0.80)^2}{(0.10)(0.10)}$$ **Step 3: Calculate** $$K_c = \frac{0.64}{0.01} = 64$$ **Answer: $K_c = 64$ (Option A)** The equilibrium constant depends only on concentrations at equilibrium, not on how the system reached equilibrium. The large value indicates the reaction favors product formation at these conditions.