Question

Coupled Reactions.

Let us assume that 2 reactions are connected as stated below;

A + Pi <------> BPi Keq at 37 C is 0.02 and

S<----->C + Pi Keq at 37 C is 1000

a) Determine the Standard state energy value (ΔG0) for each reaction then determine the value for the coupled reaction. (hint – you know the Keq value – how would the related equation be set up?)

b) Do you think this reaction would proceed spontaneously at STP conditions? Why or Why not?

c) Would a coupled reaction proceed faster in the presence of an enzyme versus a situation where the reactants alone were dissolved in water in a beaker? Explain at a molecular level. Pictures would help. Would the ΔG be different in these two situations?

Answer 1

Spontaneous reaction is one which occur on their own without any need of external agency. For spontaneous reactions ∆G is negative.

1) ∆G = - RT ln K

R = 8.313 J/K mol T = 37°C = 37+273 K = 310K

K =0.02

∆G = -8.314×310 ln0.02

∆G = -2577.34 × (-3.91)

∆G= 10083.39 J

For second reaction

K = 1000

∆G = -8.314×310 ln1000

∆G =-2577.34×(6.9)

∆G = -17803.63 J

For Coupled reaction

∆G = -17803.63+10083.39

∆G = -7720.24 J

2) Since for first reaction ∆G is positive it is not spontaneous and for second reaction and coupled reaction it is negative thus they are spontaneous.

3) In presence of enzyme, coupled reaction proceed faster as compared to in absence of it because enzyme lowers the activation energy due to which more reactant molecules gat convert into product.

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