March 28, 2024, 04:55:13 AM
Forum Rules: Read This Before Posting


Topic: Very simple question - law of mass action  (Read 1448 times)

0 Members and 1 Guest are viewing this topic.

Offline joinn46

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Very simple question - law of mass action
« on: May 10, 2017, 12:43:14 PM »
Is the following reaction possible under the specified conditions?

Pb2+ + 1/2 O2 + H2:rarrow: PbO2 (s) + 2 H+

pH = 7, c(Pb2+) = 10-8 mol/L, c(O2) = 0.261 mmol/L, K = 4 * 10-7 mol1/2/L1/2

I would begin with the law of mass action:

K = [PbO2] [H+]/([H2O][O2][Pb2+])

K = [H+]/([O2][Pb2+])

K = 10-pH/([O2][Pb2+])

But I have already an value for the equibilirium constant. I don't know what to calculate.  ???

I could also calculate the Gibbs Energy:

ΔG = - RT ln K = 36.5 kJ/mol.

I'm not sure if that's enough? Because I ignore the pH, etc.

Thanks for help,
Lena

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 2049
  • Mole Snacks: +296/-12
Re: Very simple question - law of mass action
« Reply #1 on: May 11, 2017, 04:41:24 AM »
First, your expression for K is wrong. What are the exponents of [H+] and [O2]?
Having got that right, calculate the reaction quotient Q, using the same expression as for K, but inserting the actual concentrations of the species.
If Q<K, the tendency is for the reaction to proceed in the forward direction, increasing Q until Q=K.
If Q>K, the tendency is for the reaction to proceed in the reverse direction.

Sponsored Links