Question

Suppose a 250. ml flask is filled with 0.50 mol of H, and 0.40 mol of HCl. The following reaction becomes possible: H2(g) + Cl2(g) = 2HCl (g) The equilibrium constant K for this reaction is 5.61 at the temperature of the flask. Calculate the equilibrium molarity of Cl. Round your answer to two decimal places. TOM xo?

Answer 1

According to law of mass action, the concentration of reactant is directly proportional to concentration of formation of product which gives the equilibrium constant, as follows: H, (g)+C1,(8)=2HCl(8) [HCl]a[H,][Cl] [HC1]a[Hz][Cl] [H2][Cl] K = [HC1] K [HCI] [H][cl] Given that equilibrium constant of the given reaction is K = 5.61 and according to given moles the initial concentrations of hydrogen and chlorine is as follows: Molarity (Concentration): The number of moles of solute per liter solution is defined as the molarity or concentration of solution. The unit of molarity is Mole per liter or M. Numbers of moles of solute Molarity = Solution in litre Molarity of H. - 0.50 mole 0.250L = 2M Similarly, the initial concentration of HCl is as follows:

Molarity of HCI – 0.40mole 0.250L =1.6M The ICE table for the above reaction is as follows: Initial Volume (M) Change in Volume At equilibrium H2(g) + Cl2(g) + 2HCl(g) 2 0 1.6 +x +x -2x (2+x) (x) (1.6- 2x) Substitute the given values in equation (1), as follows: 5.61=(1.6- 2r) (2+x)(x) x=0.14 Concentration of Cl, at equlibrium =(x) = 0.14M Therefore, the molarity of Cl, at equilibrium is 0.14M.