COMEDK2021ChemistryChemical Equilibrium
The value of G^ for the phosphorylation of glucose in glycolysis is 13.8 ~kJ / mol . The value of K_ C at 298 ~K is
Options
- A7.72 10⁻⁴
- B5.62 10⁻⁴
- C4.81 10⁻³
- D3.81 10⁻³
Correct answer
D. 3.81 10⁻³
Step-by-step solution
The relationship between the standard Gibbs free energy change G^ and the equilibrium constant K_ C is given by the equation G^ = -RT K_ C . Given values are G^ = 13.8 kJ/mol = 13800 J/mol , R = 8.314 J/(mol K) , and T = 298 K . Substituting these values into the equation: 13800 = -(8.314) (298) K_ C K_ C = - 13800 8.314 298 K_ C = - 13800 2477.572 -5.570 K_ C = e^ -5.570 K_ C 3.81 10⁻³ Answer: 3.81 10⁻³