Answer:
Explanation:
The balanced equation is
I₂(g) + Br₂(g) ⇌ 2IBr(g)
Data:
Kc = 8.50 × 10⁻³
n(IBr) = 0.0600 mol
V = 1.0 L
1. Calculate [IBr]
2. Set up an ICE table.
3. Calculate [I₂]
4. Convert the temperature to kelvins
T = (150 + 273.15) K = 423.15 K
5. Calculate p(I₂)
Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.