At T ( K ), K _ c for the reaction. A ₂( ~g ) B ₂( ~g ) is 99.0 . Two moles of A₂( ~g ) was heated to T(K) in a 1 L closed flask to reach the above equilibrium. What are the concentrations (in mol L ⁻¹ ) of A ₂( ~g ) and B ₂( ~g ) respectively at equilibrium?

Step-by-step solution

given K_c=99.0 n =2 moles of A ₂ v =1 ~L aligned & K _ c = [ Product ] [ Reactant ] & K _ c = [ x / l ] [ 2- x 1 ] aligned aligned & 99= x 2-x & 198-99 x=x & x= 198 100 =1.98 aligned concentration of B ₂ at equilibrium is 1.98 ~mol ~L ⁻¹ concentration of A ₂ at equilibrium is (2-198)=0.02 ~mol ⁻¹