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 equilibrium concentrations:** - $[HI] = 0.80$ mol/L - $[H_2] = 0.10$ mol/L - $[I_2] = 0.10$ 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** $$K_c = \frac{0.64}{0.01} = 64$$ **Answer: $K_c = 64$** (Option A) This large $K_c$ value indicates the equilibrium lies heavily toward product formation, which is consistent with the high HI concentration (0.80 M) compared to reactants (0.10 M each).