Question

2 moles of each reactant A and B react as A(g) +B(g) <====> C (g) + D(g) at equilibrium the concentration of C is tripled to that of B calculate Kc for reaction.
let's see who can answer fast with full solution correct...

Answer 1

Initially we have :
A = 2 moles
B = 2 moles
C= 0 moles
D = 0 moles
At equilibrium we have : ( after having "x" moles of both the reactants dissociated )
A = 2-x moles
B = 2-x moles
C = x moles
D = x moles
___________
Let's suppose Volume of container be = 1 L .
It is given that at equilibrium Concentration of C is triple than that of B =>
x = 3×(2-x)
=> x = 3/2 moles
________________
Now the equilibrium concentrations of both reactants and products becomes :
A = 2-3/2 = 1/2 moles
=> [A] = 1/2 M
similarly ,
[B] = 1/2 M
[C] = 3/2 M
[D] = 3/2 M
_____________
Now;
Kc = {(3/2)×(3/2)}/{(1/2)×(1/2)}
=> Kc = 9 .
_________
hope it helps !