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Part C 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+lightAMP+PPiΔG∘=−31.6 kJ/mol If the overall ΔG∘ of the coupled reaction is -4.80 kJ/mol , what is the equilibrium constant, K, of the first reaction at 6 ∘C ?

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ajeigbeibraheem

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

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