Question

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) The Δ G ° ′ of the reaction is − 9.130 kJ ⋅ mol − 1 . Calculate the equilibrium constant for the reaction at 25 °C. K ′ e q = What is Δ G for the reaction at body temperature (37.0 °C) if the concentration of A is 1.9 M and the concentration of B is 0.80 M ?

Δ G =

Answer 1

K = 39.85

ΔG= -6.9 kJ/mol

Step-by-step explanation:

Step 1: Data given

The ΔG°′ of the reaction is − 9.130 kJ/mol

Temperature = 25.0 °C = 298 K

Body temperature = 37.0 °C = 310K

the concentration of A is 1.9 M

the concentration of B is 0.80 M

Step 2: The reaction

A (aq) ⇌ B (aq)

Step 3:

ΔG° = -RT ln K

⇒with ΔG° = standard Gibbs free energy change = − 9.130 kJ/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 298 K

⇒with K = the equilibrium constant = TO BE DETERMINED

− 9130 J/mol = - 8.314 * 298 * ln K

ln K = 3.685

K = e^3.685

K = 39.85

Step 4: The reaction at body temperature

ΔG= ΔG° + RT ln [B]/[A]

⇒with ΔG° = Gibbs free energy change

⇒with ΔG° = standard Gibbs free energy change = − 9130 J/mol

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T = the temperature = 310 K

⇒with [A] = 1.9 M

⇒with [B] = 0.80 M

ΔG= -9130 J/mol + 8.314 J/mol*K * 310 K * ln (1.9/0.80)

ΔG= -9130 J/mol + 2229.4J/mol

ΔG=-6900.6 J/mol = -6.9 kJ/mol