Answer:
K = 39.85
ΔG= -6.9 kJ/mol
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
