How do I know if this is spontaneous at 25 celsius and show work?
PbCl2 --->(equilibrium)--->Pb2+ + 2Cl-
a) when Pb2+=1M and Cl-=2M
b) when Pb2+=1X10^-5 and Cl-=2X10^-5
I don't understand your question. Is this in solution? In solution, of course PbCl2 will spontaneously dissolve although not completely. Do you mean will there be a precipitate? For that compare Qsp with Ksp. If Qsp > Ksp, yes. If Qsp < Ksp, no.
To determine if a reaction is spontaneous at a certain temperature, you can calculate the Gibbs free energy change (ΔG) using the equation:
ΔG = ΔH - TΔS
where:
ΔG is the change in Gibbs free energy,
ΔH is the change in enthalpy, and
ΔS is the change in entropy.
If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. If ΔG is zero, the reaction is at equilibrium.
For this particular reaction, we need to know the values of ΔH and ΔS at 25 degrees Celsius (298 K). Let's assume we have those values.
a) when Pb2+ = 1M and Cl- = 2M:
First, calculate the concentrations of the products and reactants:
[Pb2+] = 1M
[Cl-] = 2M
Next, substitute these values into the equation:
ΔG = ΔH - TΔS
Assuming you have the values of ΔH and ΔS, you can substitute them and calculate ΔG. If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. If ΔG is zero, the reaction is at equilibrium.
b) when Pb2+ = 1x10^-5 and Cl- = 2x10^-5:
Again, calculate the concentrations of the products and reactants:
[Pb2+] = 1x10^-5
[Cl-] = 2x10^-5
Substitute these values into the equation:
ΔG = ΔH - TΔS
Using the values of ΔH and ΔS, calculate ΔG. If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. If ΔG is zero, the reaction is at equilibrium.
Please note that the values of ΔH and ΔS can be determined experimentally or obtained from reference tables for specific reactions.