Question

For the reaction
2A(g) + B(s) ⇌ C(g) + D(g)
at 700 °C, Kc = 0.0389. Calculate the number of moles of C present at equilibrium if a mixture of 3.55 moles of C and 3.55 moles of D is heated to 700 °C in a 5.85-L container.

Answer 1

concentratinos : C = 3.55/5.85=0.61M, D= 3.55/5..85=0.61M

So there are more number of moles of C and D, the reaction shift in the forward direction so more A and more B is formed.

let x =drop in concentration of C as well as D

at Equilibrium [C] =0.61-x =[D]

A= 2x and B=x

Kc= (0.61-x)²/{ 4x²) =0.0389

(0.61-x)²/x²= 0.0389*2=0.1556

(0.61-x)/x= 0.3944

0.61-x =0.3944x

1.3944x= 0.61, x= 0.61/1.3944=0.437

At Equilibrium [C]= [D]= 0.61-0.437=0.173 [B] =0.437 and [A] =0.437*2= 0.874M

x=