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mihalych1998 [28]
2 years ago
6

In nature, one common strategy to make thermodynamically unfavorable reactions proceed is to couple them chemically to reactions

that are thermodynamically favorable. As long as the overall reaction is thermodynamically favorable, even the unfavorable reaction will proceed.
Part A
Consider these hypothetical chemical reactions:
A⇌B,ΔG= 14.8 kJ/mol
B⇌C,ΔG= -29.7 kJ/mol
C⇌D,ΔG= 8.10 kJ/mol
What is the free energy, ΔG, for the overall reaction, A⇌D?
Part B
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
luciferin+O2ATP⇌⇌oxyluciferin+lightAMP+PPi
If the overall ΔG of the coupled reaction is -7.50 kJ/mol , what is the equilibrium constant, K, of the first reactions at 11 ∘C ? The ΔG for the hydrolysis of ATP to AMP is −31.6 kJ/mol.
Chemistry
1 answer:
padilas [110]2 years ago
5 0

Answer:

\triangle G= -6.7 KJ/mol

Explanation:

From the question we are told that:

Chemical Reactions:

X=A⇌B,ΔG= 14.8 kJ/mol

Y=B⇌C,ΔG= -29.7 kJ/mol

Z=C⇌D,ΔG= 8.10 kJ/mol

Since

Hess Law

The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.

Therefore

Generally the equation for the Reaction is mathematically given by

T = +1 * X +1 * Y +1 *Z

Therefore the free energy, ΔG is

\triangle G=1 * \triangle G*X +1 * \triangle G*Y +1 * \triangle G *Z

\triangle G= +1 * (14.9) +1 * (-29.7) +1 * (8.10)

\triangle G= -6.7 KJ/mol

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We will use this two reaction equation:

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we will use the ICE table for the first equation:

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initial     0.025                        0            0

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1.3 x 10^-2 = X^2 / (0.025 - X) by solving for X

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and when [HSO3-] = X

∴[HSO3-] = 0.0127 M

and when [H2SO3] = 0.025 - X

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when Kw = [OH-][H3O+]

and Kw = 1.1 x 10^-14 / 0.0127

∴[OH-] = 1.1 x 10^-14 / 0.0127

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- by using the ICE table for the second equation:

              HSO3- + H2O ↔ H3O+         + SO3 2-

initial    0.0127                      0.0127            0

change    -X                            +X                +X

Equ      (0.0127-X)                (0.0127+X)        X


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by substitution:

6.3 x 10^-8 = X(0.0127+X) / (0.0127-X) 

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6.3 x 10^-8 = 0.0127X /0.0127

∴X = 6.3 x 10^-8

when [SO3 2-] = X 

∴[SO32-] = 6.3 x 10^-8
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Explanation:

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