Question

Explore the relationship of AG and the equilibrium constant, K. Calculate the AG in kJ/mol for a reaction at room temperature a (~25.00°C) that has a Keq = 1000.0. Use R = 8.3145 J/mol K.

Answer 1

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