Consider the reaction of A(g)+B(g)yields C(g) for which Kc=130. assume 0.406 mole C(g) is placed in the cylinder below. the temperature is 300.0 K, and the barometric pressure on the piston (which is assumed to be massless and frictionless) is constant at 1.00 atm. the original volume [before the 0.406 mol C(g) begins to decompose] is 10.00 L. What is the volume of the cylinder at equilibrium?
This is what I've found & understood so far:
A(g) + B(g) <-> C(g)
Kc = [C]/[A] = 130
Starting with 0.406mol of C,
let X be mole of C decomposed.
let V be change in volume
Using Law of Equilibrium,
[(0.406-X)/(10.00 + V)]/[X/(10.00 + V)]2 = 130
(0.406-X)(10.00+V)/X2 = 130
(0.406-X)(10.00+V) = 130X2 (1)
total number of moles of gas in system
= moles of A + moles of B + moles of C
= X + X + (0.406-X)
= 0.406 + X
Change in volume, V
= (0.406 + X)(Molar Volume) - 10.00L
= X(Molar Volume) (2)
Solve (1) & (2) simultaneously will yield the required answer"
I don't understand how the last part ("Change in Volume, V) was found. Please can I get someone's help?