At equilibrium for reaction 1, in which substrate A is converted to product B, there is...

At equilibrium for reaction 1, in which substrate A is converted to product B, there is 4 times as many molecules of A as there are molecules of B. At equilibrium for reaction 2, in which substrate R is converted into product S, there is 8 times as many molecules of S as there are molecules of R. If these two reactions are coupled, what is the net standard free energy change?

Expert Solution

A ----------------------> B

1 4

K1 = 4 /1 = 4

Go = -RT lnK

= - 8.314 x 10^-3 x 298 x ln 4

= - 3.43 kJ/mol

R -----------------------> S

8 1

K2 = 1/8 = 1/ 8

Go = -RT lnK = -8.314 x 10^-3 x 298 x ln 1/8 = 5.15 kJ/mol

for coupled reaction

Go = -3.43 + 5.15

Go = 1.72 kJ/mol

queen_honey_blossom answered 2 years ago

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