Question

Forty grams of Compound A (Molar Mass is 100 g/mol) is heated in a two liter reaction chamber where its temperature is set to 350oC in order for it to undergo thermal decomposition to form three gaseous products: Compound B, Compound C, and Compound D. Below is the balanced stoichiometry. A(s)->B(g)+C(g)+2D(g) , Kc= 5.00x10^-8 When the system has reached equilibrium, what must be the total pressure of the system? Once the system has reached equilibrium, how much of the reactant remains?

Answer 1

Moles of Compound A= 40/100 =0.4

Concentration of A =0.4/2 =0.2 Moles/L

The reaction is A--à B +C+ 2D

Let x= drop in concentration of A

At Equilibrium [B]=x , [C]= x, [D] =2x and [A] =0.4-x

Kc= [B][C] [D]²/ [A] =x*x*(2x)2/ (0.4-x)= 5*10^-8

4x³/ (0.4-x)= 5*10^-8, this is a equation which has to be solved by trial and error. Using solver and matching LHS and RHS, x= 0.00171

Concentrations [B]=[C]=[D] =0.00171 and [A] =0.4-0.00171=0.398 ( A remaining at equilibrium)

Total concentration = 0.00171+0.00171+0.00171+0.398=0.40313 M/L

From PV= nRT

P=(n/V)*RT, T =350 deg.c c=350+273.15= 623.15

P= 0.40313*0.08206Lit.atm/mole.K *623.15=20.61 atm