One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium. A(g) B(g) . K_c for this reaction at 400 K is 4.0 . The partial pressures (in atm) of He and B(g) are respectively (at equilibrium) (Assume He, A(g) and B(g) behave as ideal gases) (Given: R = 0.082 L atm K ⁻¹ mol ⁻¹ )

Step-by-step solution

The given reaction is A(g) B(g) . Let the initial moles of A be 1 and at equilibrium, let x moles of A dissociate. Moles at equilibrium: n_A = 1 - x n_B = x Since n_g = 0 , the equilibrium constant K_c can be written in terms of moles: K_c = n_B n_A = x 1 - x = 4.0 x = 4 - 4x 5x = 4 x = 0.8 At equilibrium, the moles of the gases are: n_ He = 1 n_B = 0.8 Using the ideal gas equation P = nRT V , the partial pressures are calculated as follows: Partial pressure of He: P_ He = 1 0.082 400 10 = 3.28 atm Partial pressure