Exercise 6.4 15 of 17 The standard free energy for the reaction of oxygen bindin

Question

Exercise 6.4 15 of 17 The standard free energy for the reaction of oxygen binding to myoglobin Mb+029)MbO2 is ?/T -?30.0 kJ mol-1 at 298 K and pH = 7. The standard state of O2 is the dilute solution, molarity scale; therefore the concentration of O2 must be in M Part A What is the ratio Mb02/Mb in an aqueous solution at equilibrium with a partial pressure of oxygen po, gas and for the protein in solution. 400 Pa? Assume ideal behavior of O2 Express your answer using two significant figures. SubmitPre Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining

Explanation / Answer

for the reaction , Mb+ O2 ------->MbO2, deltaG= -30 Kj/mole

since 1000J= 1Kj, deltaG=-30KJ*1000J/Kj= -30000J/mole

deltaG is related to Equilibrium constant as deltaGo=-RT lnK

R= gas constant= 8.314 J/mole.K

lnK=-deltaGO/RT= 30000/(8.314*298)= 12.11

K= 181431

but K= [MbO2][/PO2*Mb]

PO2 is partial presure of O2 in atm, given PO2= 400Pa= 400 Pa*(1/1000)Kpa=0.4Kpa

101.3 Kpa= 1 atm, PO2=0.4/101.3 =0.004 atm

hence K= 181431= [MbO2]/0.004*[Mb]

[MbO2]/[Mb]=181431*0.004=725.7

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