One reaction that occurs in human metabolism is For this reaction ∆G° = 14 kJ at 25°c. a. Calculate K for this reaction at 25°C. b. In a living cell this reaction is coupled with the hydrolysis of ATP. (See Exercise 75.) Calculate ∆G° and K at 25°C for the following reaction: Glutamic acid ( a q ) + ATP ( a q ) + NH 3 ( a q ) ⇌ Glutamine ( a q ) + ADP ( a q ) + H 2 PO 4 − ( a q )
One reaction that occurs in human metabolism is For this reaction ∆G° = 14 kJ at 25°c. a. Calculate K for this reaction at 25°C. b. In a living cell this reaction is coupled with the hydrolysis of ATP. (See Exercise 75.) Calculate ∆G° and K at 25°C for the following reaction: Glutamic acid ( a q ) + ATP ( a q ) + NH 3 ( a q ) ⇌ Glutamine ( a q ) + ADP ( a q ) + H 2 PO 4 − ( a q )
Solution Summary: The author explains the value of Delta G° and K for the reaction of human metabolism.
b. In a living cell this reaction is coupled with the hydrolysis of ATP. (See Exercise 75.) Calculate ∆G° and K at 25°C for the following reaction:
Glutamic acid
(
a
q
)
+
ATP
(
a
q
)
+
NH
3
(
a
q
)
⇌
Glutamine
(
a
q
)
+
ADP
(
a
q
)
+
H
2
PO
4
−
(
a
q
)
Chemical pathways by which living things function, especially those that provide cellular energy, such as the transformation of energy from food into the energy of ATP. Metabolism also focuses on chemical pathways involving the synthesis of new biomolecules and the elimination of waste.
In the electrode Pt, H2(1 atm) | H+(a=1), if the electrode balance potential is -0.118 V and the interface potential difference is +5 mV. The current voltage will be 0.005 - (-0.118) = 0.123 V ¿Correcto?
In the electrode Pt, H2(1 atm) | H+(a=1) at 298K is 0.79 mA cm-2. If the balance potential of the electrode is -0.118 V and the potential difference of the interface is +5 mV. Determine its potential.
In one electrode: Pt, H2(1 atm) | H+(a=1), the interchange current density at 298K is 0.79 mA·cm-2. If the voltage difference of the interface is +5 mV. What will be the correct intensity at pH = 2?. Maximum transfer voltage and beta = 0.5.
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