44.7k views
2 votes
If Delta G initial = -125 kJ, calculate the value of the equilibrium constant at room temperature (298K).

User Amree
by
7.5k points

1 Answer

5 votes

Final answer:

To calculate the equilibrium constant (K) from a Gibbs free energy change (ΔG) of -125 kJ/mol at room temperature (298K), use the equation ΔG = -RTlnK. By substituting the values into the formula, we find that K is approximately 5.2 × 1021, showing a strong tendency towards product formation.

Step-by-step explanation:

The question pertains to the calculation of the equilibrium constant (K) from the Gibbs free energy change (ΔG) at room temperature, which is a common problem in thermodynamics. We can use the equation ΔG = -RTlnK, where R is the universal gas constant (8.314 J/K⋅mol), T is the temperature in kelvin, and K is the equilibrium constant. Given that at room temperature (298K), ΔG is -125 kJ/mol (or -125,000 J/mol), we can rearrange the equation to solve for K:

ΔG = -RTlnK

-125,000 J/mol = -(8.314 J/K⋅mol)(298K)lnK

lnK = ΔG / (-RT)

lnK = -125,000 J/mol / [-(8.314 J/K⋅mol)(298K)]

lnK = 50.198

K = e50.198

K ≈ 5.2 × 1021

Therefore, the equilibrium constant at room temperature is approximately 5.2 × 1021, indicating a reaction that strongly favors the formation of products.

User Eric Majerus
by
7.8k points