Question

Consider a general reaction A(aq) ⇌ B(aq) catalyzed by enzyme. The Δ°′ of the reaction is -5.590 kJ·mol⁻¹. Calculate the equilibrium constant for the reaction at 25 °C.

Answer 1

The equilibrium constant (K) for the reaction at 25°C is calculated using the standard free energy change (Δ°′) and is found to be approximately 9.58.

Step-by-step explanation:

The calculation of the equilibrium constant (K) for a reaction is related to the standard free energy change (Δ°′) by the equation ΔG°′ = -RT ln K, where T is the temperature in Kelvin, R is the ideal gas constant (8.314 J/mol·K), and K is the equilibrium constant.

Given that the Δ°′ for a reaction at 25°C (or 298.15 K) is -5.590 kJ·mol⁻¹, we can plug the values into the equation to find the equilibrium constant. Convert the given Δ°′ to joules (since 1 kJ = 1000 J) and then use the equation to solve for K:

ΔG°′ = -5.590 kJ/mol × 1000 J/kJ = -5590 J/mol

K = e^(-ΔG°′ / RT)

K = e^(-(-5590 J/mol) / (8.314 J/mol·K × 298.15 K))

K = e^(5590 / (8.314 × 298.15))
= e^(5590 / 2473.77)
= e^(2.260)
= 9.58 (rounded to two decimal places)

So, the equilibrium constant K for the reaction at 25°C is approximately 9.58.