Is there any way someone could help me complete this page? Still not sure what t

Question

Is there any way someone could help me complete this page? Still not sure what to do for B and C. Thanks for the help!

The combustion of glucose to CO2 and water is a major source of energy in aerobic organisms following equation may be useful: oretaninats a reaction favored mainly by a large negative enthalpy change. The CcH 1206 + 602 (g) 6C02 (g) +6H2O (1) Ho =-2816 kJ/mole AS = 181 J/mole a. At 37 C, what is the value of AGo? -2872.ll K1/rol In the overall reaction of aerobic metabolism of glucose, 38 moles of ATP are produced from ADP for every mole of glucose oxidized. Calculate the standard state free energy change for the overall reaction when glucose is coupled to the formation of ATP b. ATP + H2O ADP + Pi AG =-30.5 kJ/mole What is the overall efficiency of the process in terms of the percentage of the available free energy change captured in ATP? c.

Explanation / Answer

a)

we know that

dGo = dHo - TdSo

given

dHo = -2816 x 1000 J/mol

dSo = 181 J/mol K

temperature (T) = 37 C = 37 + 273 = 310 K

so

dGo = ( -2816 x 1000) - ( 310 x 181)

dGo = -2872110

dGo = -2872.11 x 1000

dGo = -2872.11 kJ/mol

b)

consider the given reaction

ATP + H20 ---> ADP + Pi dGo = -30.5 kJ/mol

now write the reverse reaction

ADP + Pi ----> ATP + H20 dGo = 30.5 kJ/mole

now

we need 38 moles of ATP

so

total energy required = 38 x 30.5

total energy required = 1159 kJ /mol of glucose

now

dGo for oxidation of glucose = -2872.11 kJ/mol

so

dGo for overall reaction = -2872.11 + 1159

dGo for overall reaction = -1713.11

so

dGo for overall reaction is -1713.11 kJ/mol

c)

now

given that

efficiecny = energy of ATP`s x 100 / amount of energy availble

now

energy of those 38 ATP = 1159

energr availbe = 2872.11

so

efficeincy = 1159 x 100 / 2872.11

efficiency = 40.53

so

the efficiency is 40.53 %

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.