K _ c for the reaction, A ₂( ~g ) T ( K ) B ₂( ~g ) is 99.0 . In a 1 L closed flask two moles of B ₂( ~g ) is heated to T ( K ) . What is the concentration of B ₂( ~g ) (in mol L ⁻¹ ) at equilibrium?

Step-by-step solution

Initial concentration of A ₂=0 Initial concentration of B ₂=2 Let x be the concentration of A ₂ so the change in B ₂ would be -2 x Equilibrium concentration are [ A ₂ ]= x , [ B ₂ ]=2-2 x Putting the value in eq. (1) 99= (2-2 x)^2 x 99=4 (1-x)^2 x Let's simplify the equation - [( a - b )^2= a ^2-2 ab + b ^2 ]99= 4 (1-2 x+x^2 ) x 99 x =4-8 x +4 x ^2 4 x ^2-107 x +4=0 Solving the quadratic equation we use the Quadratic formula:- x = - b b ^2-4 ac 2 a a =4, ~b =-107, c =4x= -(-107) (-107)^2-4(4)(4) 2 4 x= 107 11449-64