Answer:
K = 9.620 × 10⁻⁶
Explanation:
From the given information:
Temperature T= 6° C
= (273 + 6)K
= 279 K
The correct and well presentation of the reactions are:
1. [tex]Luciferin + O_2[/tex] ⇆ oxyluciferin + light ΔG₁°
2. ATP ⇄ AMP + PP[tex]_i[/tex] ΔG₂° = -31.6 kJ/mol
The overall ΔG° = -4.80 kJ/mol
Let's first determine the ΔG₁° for the equation (1)
ΔG° = ΔG₁° + ΔG₂°
- ΔG₁° = - ΔG° + ΔG₂°
ΔG₁° = ΔG° - ΔG₂°
ΔG₁° = ( -4.80 - (-31.6) ) kJ/mol
ΔG₁° = 26.8 kJ/mol
Using the formula:
ΔG° = -RTIn K
[tex]In \ K =\dfrac{-\Delta G^0}{RT} \\ \\ log \ K = -\dfrac{\Delta G^0}{2.303RT}[/tex]
[tex]log \ K = -\dfrac{\Delta G_1^0}{2.303RT} \\ \\ log \ K = -\dfrac{26.8 \times 10^3 \ J/mol}{2.303\times 8.314 \ J/mol/K \times 279 \ K} \\ \\ log \ K =- 5.017[/tex]
K = antilog (-5.017)
K = 9.620 × 10⁻⁶
