Coupled Reactions Part A In nature, one common strategy to make thermodynamicall

Question

Coupled Reactions Part A 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 Consider these hypothetical chemical reactions 1. A-B, 2.B-C, 3, C D, G= 11.4 kJ/mol G=-28.8 kJ/mol G = 9.70 kJ/mol What is the free energy, G, for the overall reaction, A-D? Express your answer numerically in kilojoules per mole kJ/mol Submit Hints My Answers Give Up Review Part

Explanation / Answer

A)

1. A ---> B dG1 = 11.4

2. B ---> C dG2 = -28.8

3. C ---> D dG3 = 9.7

4. A ---> D dG4 = ??

we can see that

4 = 1 + 2 + 3

so

dG4 = dG1 + dG2 + dG3

dG4 = 11.4 -28.8 + 9.7

dG4 = -7.7 kJ/mol

so

free energy for the overall reaction is -7.7 kJ/mol

C)

dG for coupled reaction = dG of reaction1 + dG for reaction2

so

-8.1 = dG of reaction1 - 31.6

dG of reaction1 = 23.5 kJ/mol

now

we know that

dGo = -RT lnK

23.5 x1000 = -8.314 x 286 x lnK

K = 5.103 x 10-5

so

equilibrium constant for first reaction is 5.103 x 10-5

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