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 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** $$K_c = \frac{[HI]^2}{[H_2][I_2]}$$ (Products raised to their stoichiometric coefficients divided by reactants) **Step 2: Substitute equilibrium concentrations** $$K_c = \frac{(0.80)^2}{(0.10)(0.10)}$$ **Step 3: Calculate** $$K_c = \frac{0.64}{0.01} = 64$$ **Answer: The equilibrium constant $K_c = 64$** This is correct because the equilibrium constant expression uses equilibrium concentrations with each substance raised to the power of its stoichiometric coefficient in the balanced equation.