Question

Consider the reaction

A(g)+B(g)->C(g)

For which Kc=1.30*10²

Assume that .406 mole C(g) is placed in a cylinder. The temperature is 300 K and the barometirc pressure on the piston (which is assumed to be massless and frictionless) is constant at 1.00 atm. The original volume (before the C(g) begins to decompose) is 10.00 L. What is the volume of the container at equilibrium?

Answer 1

              A(g) +    B(g) -> C(g)

Initial       0           0             0.406

Change    x         x            0.406-x

Anyways, the concentration of C(g) at equil. would be (.406-n)/(10-V). The equil. concentrations of A and B would be n/v. Therefore, n^2(10-v)/v^2(.406-n)=130. Also, PV=nrT, so V=nR(300). Solve these two equations to get n and v, n being the moles and v being the volume. The answer would be 10-v.

So at equilibrium [C] = 0.406 - x / 10+V

[A] = x / 10+V= [B]

Kc = [C] / [A] [B]

Kc =   [0.406-x] X (10+V) / x^2 = 130

130x^2 = ( 0.406 - x ) (10+V)

total number of moles at equilibrium = x + x + 0.406 -x = 0.406 +x

volume occupied by 0.406 moles = 10 L

So volume occupied by 1 mole = 10 / 0.406 L

Volume occupied by 0.406 +x = 10 (0.406+x) / 0.406 L

The change in volume =