Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. because this light generation is an atp-requiring reaction, firefly luciferase can be used to test for the presence of atp. in this way, luciferase can test for the presence of life. the coupled reactions are 1.2.luciferin+o2atp⇌⇌oxyluciferin+l ightamp+ppiδg∘=−31.6 kj/mol if the overall δg∘ of the coupled reaction is -9.00 kj/mol , what is the equilibrium constant, k, of the first reaction at 26 ∘c ? express your answer numerically.

K = 0.0001

Explanation:

The change on the free Gibbs energy is related to the equilibrium constant K with the following equation when the reaction is in equilibrium:

ΔG° = -RTlnK  [Eq 1]

Where:

ΔG° = Gibbs free energy of the first equation

R= ideal gas constant in kJ/molK

T= temperature in kelvin (K)

K= equilibrium constant

To use this equation, we have to analyze the coupled reaction of luciferine to oxyluciferine (for this you can watch the first image attached)

Then we have to calculate the Gibbs free energy of the first equation:

ΔG°= (overal ΔG° of coupled reaction) - (ΔG° of second reaction)

ΔG°=-9 kJ/mol -(-31.6 kJ/mol) = 22.6  kJ/mol.

If you note, the first reaction isn't spontaneous, but the second reaction is spontaneous. Because the negative value of ΔG° means "spontaneous".

Then, you have to replace these values in equation 1 [Eq 1], using the next values:

T = 299.15 K

R = 8.314 J/molK = 0.008314 kJ/ molK

ΔG°= 22.6  kJ/mol.

ΔG° = -RTlnK

22.6  kJ/mol = - (0.008314 kJ/ molK)x299.15 Kx ln K

Note: Please, don't confuse kelvin unit (K), with equilibrium constant (K)!

From this equation we find lnK

lnK = 22.6 kJ(mol / ( 2.847 kJ/mol)

K = 0.0001

If you want to see the mathematic process with more details, please watch the document attached, where is the full exercise solved.