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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: joinn46 on May 10, 2017, 12:43:14 PM

Title: Very simple question - law of mass action
Post by: joinn46 on May 10, 2017, 12:43:14 PM

Is the following reaction possible under the specified conditions?

Pb²⁺ + 1/2 O₂ + H₂O  :rarrow: PbO₂ (s) + 2 H⁺

pH = 7, c(Pb²⁺) = 10^-8 mol/L, c(O₂) = 0.261 mmol/L, K = 4 * 10^-7 mol^1/2/L^1/2

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

Title: Re: Very simple question - law of mass action
Post by: mjc123 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.