Question

Consider the reaction: 3A(g) + B(s) ↔ 5C(s) + 2D(g). In a 5.00 L container the reactants and products are at equilibrium. There are 0.18 mol A, 0.49 mol B, 1.41 mol C, and 0.15 mol D. What is the equilibrium constant?

Answer 1

K_c = concentration of product at equillibrium / concerntration of reactant at equillibrium

(Reactions containing pure solids and liquids results in heterogeneous reactions in which the concentrations of the solids and liquids are not considered when writing out the equilibrium constant expressions.)

3A(g) + B(s) ↔ 5C(s) + 2D(g)

at equillibrium

Kc = [a_C]⁵   [a_D]² / [a_A]³ [a_B]

(Remember that the activity, a, of any solid or liquid in a reaction is equal to 1. So, the activities of B and C will be set equal to 1. )

as the moles of gases are given we convert them to concerntration by dividing them by volume of vessel  

Kc = [D/5]² * 1/ [A/5 ]³ *1 = 0.15 ² * 5/ 0.18 ³ = 0.0225* 5 / 0.0058 = 0.1125 / 0.0058

= 19.3965 mole^-1L