Calculate ΔG at 25°C for the precipitation of lead(II) iodide from mixing 100. mL of 0.10 M lead(II) nitrate solution with 100. mL of 0.10 M sodium iodide solution. The ΔG° for this precipitation reaction at 25°C = −46.2 kJ/mol.
dG = dGo + RTlnQ
Write the equation. Determine the limiting reagent. Q = (NaNO3)^2/[Pb(NO2)]
Determine the concentrations for Q and calculate. Post your work if you get stuck.
To calculate ΔG at 25°C for the precipitation of lead(II) iodide, we can use the equation:
ΔG = ΔG° + RTln(Q)
Where:
ΔG is the change in Gibbs free energy
ΔG° is the standard Gibbs free energy change
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
ln(Q) is the natural logarithm of the reaction quotient
First, let's convert the temperature from Celsius to Kelvin:
T = 25°C + 273.15 = 298.15 K
Now, let's calculate Q, the reaction quotient. The reaction can be represented by the balanced equation:
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)
The molar ratio between lead(II) nitrate and lead(II) iodide is 1:1, which means Q can be calculated as:
Q = [PbI2] / [Pb(NO3)2]
We need to determine the concentrations of lead(II) iodide and lead(II) nitrate. To do this, we'll use the volume and molarity of the solutions provided.
For the lead(II) iodide solution:
Volume = 100 mL = 0.100 L
Molarity = 0.10 M
Concentration of PbI2 = (0.10 M) * (0.100 L) = 0.010 mol
For the lead(II) nitrate solution:
Volume = 100 mL = 0.100 L
Molarity = 0.10 M
Concentration of Pb(NO3)2 = (0.10 M) * (0.100 L) = 0.010 mol
Substituting the values into the equation, we have:
Q = [0.010 mol] / [0.010 mol] = 1
Now we can calculate ΔG:
ΔG = ΔG° + RTln(Q)
= -46.2 kJ/mol + (8.314 J/(mol·K)) * (298.15 K) * ln(1)
Since ln(1) = 0, we can simplify the equation:
ΔG = -46.2 kJ/mol + 0 J/mol
= -46.2 kJ/mol
Therefore, the ΔG at 25°C for the precipitation of lead(II) iodide is -46.2 kJ/mol.