What is DG° at 298 K for the following equilibrium?
Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq); Kf = 1.7 × 107 at 298 K
-41 kJ
To calculate the standard Gibbs free energy change (ΔG°) at 298 K for the given equilibrium, you can use the equation:
ΔG° = -RT ln(K)
Where:
ΔG° is the standard Gibbs free energy change
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298 K)
K is the equilibrium constant
First, let's determine the value of ln(K) using the given equilibrium constant (Kf):
ln(K) = ln(1.7 × 10^7)
Using a natural logarithm calculator, the value of ln(1.7 × 10^7) is approximately 15.757
Now, we can substitute the known values into the equation for ΔG°:
ΔG° = - (8.314 J/(mol·K)) x (298 K) x (15.757)
Calculating this expression gives us:
ΔG° ≈ - 39207 J/mol
Therefore, the standard Gibbs free energy change (ΔG°) at 298 K for the given equilibrium is approximately -39207 J/mol.