After calculating the standard reduction potential Eo under standard biochemical

Question

After calculating the standard reduction potential Eo under standard biochemical conditions using table 13-7 for reactions below write the expected equilibrium constant K in the reaction below. succinate + FAD fumarate + FADH2 Do not forget to check the number of electron transferred in the reaction! You have to determine it yourself. Since biochemical reactions are conducted at constant pH [H+] is not present in the equilibrium constant. The answer should just a number like 0.000001 or 2400000000 Answer: 285763561.8

Explanation / Answer

The two half reactions are

FAD + 2H+ + 2e- ---> FADH2 Eo' = 0.035 V (1)

Fumarate + 2H+ + 2e- ---> Succinate Eo' = 0.031 V (2), reversing the reaction-2 gives Succinate ------> Fumarate+2H+2e-, Eo=-0.031V (2A)

equation-1+ Eq.2A gives FAD+Succinate ---------> FADH2+ fumarate, Eo=0.035-0.031=0.004V, n= no of electrons exchanged= 2, E= 96500 coulumbs/mole,

deltaG= Gibbs free energy change= -nFEo= -RT lnK. K= equilibrium constant

lnK= nFE0/RT= 2*96500*0.004/(8.314*298)=0.3115

K= 1.365

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