Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a sou
ID: 913977 • Letter: C
Question
Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be represented as ATP(aq) + H2O(l) ADP(aq) + H2PO4-(aq) where ADP represents adenosine diphosphate. For this reaction G--30.5 kJ/mol. (a) Calculate K at 25°C. 0.9721 (b) If all the free energy from the metabolism of glucose C6H1206(s) + 6 O2(g) 6 CO2(g) + 6 H20(/) goes into the production of ATP, how many ATP molecules can be produced for every molecule of glucose? 32 X moleculesExplanation / Answer
Solution :-
a). Delta G = -30.5 kJ/mol * 1000 J / 1 kJ = -30500 J/mol
Delta G = - RT ln K
T= 25 C +273 = 298 K
-30500 J per mol = - 8.314 J per mol /k * 298 K * ln K
-30500 J per mol /(- 8.314 J per mol /k * 298 K) = ln K
12.31 = ln K
Anti ln 12.31 = K
2.22*10^5 = K
b). C6H12O6(s) + 6O2(g) ----- > 6CO2(g) + 6H2O(l)
lets calculate the delta G reaction using the standard deltaG values
Delta G rxn = sum of delta G product - sum of delta G reactant
=[(CO2*6)+(H2O*6)] –[C6H12O6*1)+(O2*6)]
= [(-394.4 *6)+(-237.2*6)] –[(-910.56*1)+(0*6)]
= -2879 kJ
Now lets calculate the moles of ATP that can be produced
-2879 kJ / -30.5 kJ per mol = 94 mol ATP
94.4 mol * 6.02*10^23 moleucles / 1 mol = 5.68*10^25 molecules of ATP
Autoionization of water
a). Kw = 1*10^-14 , T= 25 +273 = 298 K
delta G = - RT ln K
= -8.314 J per mol K * 298 K * ln 1*10^-14
= 7.99*10^4 J per mol
7.99*10^4 J per mol * 1k J / 1000 J = 79.9 kJ
b) kw = 2.09*10^-14 T = 35 C +273 = 308 K
delta G = - RT ln K
= -8.314 J per mol K * 308 K * ln 2.09*10^-14
= 8.066*10^4 J per mol
8.066*10^4 J per mol * 1k J / 1000 J = 80.7 kJ/mol
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