For a reaction, A+B 2 C 1.0 mole of A, 1.5 mole of B and 0.5 mole of C were taken in a 1 L vessel. At equilibrium, the concentration of C was 1.0 ~mol ~L ⁻¹ . The equilibrium constant for the reaction is x / 15 . The value of ' x ' is:

Step-by-step solution

The reaction is A + B 2C . Initial moles: [A]₀ = 1.0 mol L ⁻¹ [B]₀ = 1.5 mol L ⁻¹ [C]₀ = 0.5 mol L ⁻¹ Let the change in concentration of A be -y . Then the change in B is -y and the change in C is +2y . At equilibrium: [A]_ eq = 1.0 - y [B]_ eq = 1.5 - y [C]_ eq = 0.5 + 2y Given [C]_ eq = 1.0 mol L ⁻¹ , we have: 0.5 + 2y = 1.0 2y = 0.5 y = 0.25 . Calculating equilibrium concentrations: [A]_ eq = 1.0 - 0.25 = 0.75 = 3 4 [B]_ eq = 1.5 - 0.25 = 1.25 = 5 4 [C]_ eq = 1.0 The equilibrium constant K_c is given by: K_c = [