Answer:
ΔG° = -RT ln(K)
Where:
ΔG° is the standard Gibbs free energy change
R is the gas constant (8.3145 J/mol K)
T is the temperature in Kelvin
K is the equilibrium constant
To calculate the ΔG° in kJ/mol, we need to convert the temperature from Celsius to Kelvin. The formula to convert Celsius to Kelvin is:
T(K) = T(°C) + 273.15
Given that the temperature is approximately 25.00°C, we can convert it to Kelvin:
T(K) = 25.00°C + 273.15 = 298.15 K
Now we can substitute the values into the equation:
ΔG° = -RT ln(K)
= -(8.3145 J/mol K)(298.15 K) ln(1000.0)
= -(8.3145 J/mol K)(298.15 K) ln(10^3)
= -(8.3145 J/mol K)(298.15 K)(6.9078)
= -16.696 kJ/mol
Therefore, the ΔG° for the reaction at room temperature with a Keq = 1000.0 is approximately -16.696 kJ/mol