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### Topic: Zumdahl Challenge Problem  (Read 2107 times)

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#### narutoverse13

• Regular Member
•   • Posts: 14
• Mole Snacks: +1/-0 ##### Zumdahl Challenge Problem
« on: January 23, 2015, 06:30:12 PM »
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?

#### mjc123

• Chemist
• Sr. Member
• • Posts: 1751
• Mole Snacks: +245/-11 ##### Re: Zumdahl Challenge Problem
« Reply #1 on: January 24, 2015, 05:42:42 PM »
For a start, Kc is not [C]/[A]. Where does B come in?
As to the change in volume, well the volume at equilibrium is the number of moles (0.406+X) times the molar volume at 1 atm and 300K. The change in volume is this quantity minus the original volume, 10L. This is equal to X*mol vol because 0.406*mol vol = 10 L. You can verify this using PV=nRT, but it is obvious from the given fact that 0.406 mol at 1 atm and 300K occupied 10L.