Question

Question is in the attachment

Answer 1

ans. option (c) 0.05 mol/L
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                 Equilibrium Constant  : Stoichiometry 

Chemical reaction :     A  + 2 B   ⇔  2 C

Initial state:
    Moles of A: 2.0.  Moles of B: 3.0.  Moles of C: 2.0.
    Volume of flask (constant): V = 2.0 L

Concentrations at equilibrium :
       [ C ] = 0.5 mol/L.
       Moles of C present = 0.5 * 2.0L = 1.0 mol

   The amount/concentration of C has reduced. It means that the reaction proceeded in the reverse direction. 1 mole of C has decomposed into A & B. [A] and [B] will increase to 0.50 mole of A and 1.0 mole of B as per :
                                     2 C  <==> A + 2 B

Concentrations at Equilibrium:
     [ A ] = (2.0 + 0.50) mol/2.0L = 1.25 Molar
     [ B ] = (3.0 + 1.0) mol/2.0L = 2.0 Molar

Equilibrium Constant for concentration forward direction: (Kc)
     Kc = [ C ]² / { [ A ] [ B ]² }
        = 0.5² / { 1.25 * 2.0² }
        = 0.05  L/mol

Answer 2

your answer is 0.05 mole/ litre option c