The Go, for the reaction ATP + H2O-ADP + Pi + H+ is-30.5 kJ mol. Other organopho
ID: 1031452 • Letter: T
Question
The Go, for the reaction ATP + H2O-ADP + Pi + H+ is-30.5 kJ mol. Other organophosphate species also undergo hydrolysis of the phosphate moiety via a similar reaction. Determine the value of G°, for the following reactions and indicate if the reaction will proceed spontaneously in the direction written if the reactants and products are initially in a 1:1 molar ratio. a. ATP+ Acetic acidADPAcetyl phosphate kJ mol- The reaction is The hydrolysis reaction for acetyl phosphate is Acetyl phosphateH20Acetic acid+PH G",--42.2 kJ mol-1.) b. ATP+GlucoseGlucose-6-phosphate ADP kJ mol The reaction is (The hydrolysis reaction for glucose-6-phosphate is Glucose-6-phosphateH2OGlucose PH AG 12.5 kJ mol1)Explanation / Answer
Ans. #a. Given,
Acetyl phosphate + H2O <------> Acetic acid + Pi + H+ ; dG0’= -42.2 kJ mol-1
# When the reaction is reversed, the numerical sign of dG0’ is also reversed.
So,
Acetic acid + Pi + H+ <------> Acetyl phosphate + H2O ; dG0’= +42.2 kJ mol-1
# The overall coupled reaction can be written as the sum of following reaction-
ATP + H2O <------> ADP + Pi + H+ ; dG0’= -30.5 kJ mol-1
(+) Acetic acid + Pi + H+ <------> Acetyl phosphate + H2O ; dG0’= +42.2 kJ mol-1
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ATP + Acetic acid <------> ADP + Acetyl phosphate ; dG0’net = ?
The dG0’net of the coupled reaction is given as the sum of dG0’ of the two reactions as follow-
dG0’net= (-30.5 kJ mol-1) + 42.2 kJ mol-1 = +11.7 kJ mol-1
#b. Given,
G-6-P + H2O <------> Glucose + Pi + H+ ; dG0’= -12.5 kJ mol-1
# When the reaction is reversed, the numerical sign of dG0’ is also reversed.
So,
Glucose + Pi + H+ <------> G-6-P + H2O ; dG0’= +12.5 kJ mol-1
# The overall coupled reaction can be written as the sum of following reaction-
ATP + H2O <------> ADP + Pi + H+ ; dG0’= -30.5 kJ mol-1
(+) Glucose + Pi + H+ <------> G-6-P + H2O ; dG0’= +12.5 kJ mol-1
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ATP + Glucose <------> ADP + G-6-P ; dG0’net = ?
The dG0’net of the coupled reaction is given as the sum of dG0’ of the two reactions as follow-
dG0’net= (-30.5 kJ mol-1) + 12.5 kJ mol-1 = -18.0 kJ mol-1
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