Cells use the hydrolysis of adenosine triphosphate,...

Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as $$\mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)$$ where ADP represents adenosine diphosphate. For this reaction, $\Delta G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol}$ a. Calculate $K$ at $25^{\circ} \mathrm{C}$. b. If all the free energy from the metabolism of glucose $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose? c. Much of the ATP formed from metabolic processes is used to provide energy for transport of cellular components. What amount (mol) of ATP must be hydrolyzed to provide the energy for the transport of $1.0 \mathrm{~mol} \mathrm{~K}^{+}$ from the blood to the inside of a muscle cell at $37^{\circ} \mathrm{C}$ as described in Exercise $78 ?$

Cells use the hydrolysis of adenosine triphosphate, abbreviated as ATP, as a source of energy. Symbolically, this reaction can be written as
$$\mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)$$
where ADP represents adenosine diphosphate. For this reaction, $\Delta G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol}$
a. Calculate $K$ at $25^{\circ} \mathrm{C}$.
b. If all the free energy from the metabolism of glucose
$$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$
goes into forming ATP from ADP, how many ATP molecules can be produced for every molecule of glucose?
c. Much of the ATP formed from metabolic processes is used to provide energy for transport of cellular components. What amount (mol) of ATP must be hydrolyzed to provide the energy for the transport of $1.0 \mathrm{~mol} \mathrm{~K}^{+}$ from the blood to the inside of a muscle cell at $37^{\circ} \mathrm{C}$ as described in Exercise $78 ?$

Key Concepts

Gibbs Free Energy

Gibbs free energy (?G) is a thermodynamic quantity that determines the spontaneity of a chemical reaction. It relates to the equilibrium constant (K) through the relationship ?G° = -RT ln K, where R is the gas constant and T is the temperature in Kelvin. This concept is central to understanding how reaction energetics drives biochemical processes, including the direction and extent of reactions in metabolic pathways.

ATP Hydrolysis

ATP hydrolysis involves the breakdown of adenosine triphosphate (ATP) into adenosine diphosphate (ADP) and an inorganic phosphate (Pi), releasing energy that cells use to perform work. This reaction is a primary example of how cells couple exergonic processes to drive endergonic processes, making ATP the universal energy currency in biological systems.

Metabolic Energy Conversion

Metabolic energy conversion refers to the process by which cells extract energy from nutrients, such as glucose, through pathways like cellular respiration. The chemical energy released in these reactions is used to synthesize ATP from ADP and phosphate, thus transforming an energy-rich molecule into a form that can be readily used by the cell for various energy-demanding processes.

Active Transport Energy Requirements

Active transport is the process by which cells move molecules against their concentration gradients, an energetically unfavorable process that requires the input of energy. This energy is typically supplied by ATP hydrolysis, which provides the necessary free energy to power transporter proteins that maintain concentration gradients crucial for cellular functions.

Transcript

00:01 Consider the reaction of adenosine triphosphate with water, and it produces adenosine diphosphate and dihydrogen phosphate ion.

00:11 We're told for this process, our free energy change is negative 30 .5 kilojoules per mole.

00:23 And we are asked to first calculate k at 25 degrees celsius.

00:30 Yes.

00:32 Ok, so my delta g.

00:35 Well, i'm just going to write my equation here.

00:38 My k will be equal to e to the negative g divided by rt.

00:45 T.

00:47 So k will be equal to negative, whoops, e to the negative negative 30 .5 divided by 8 .314 times 10 to the minus third kilojoules per kmol times 298 k.

01:24 Can't i get my eraser? there we go.

01:28 Okay, grabbing my calculator.

01:31 So i can see that's going to be positive.

01:33 I'm not going to bother to put that in there.

01:35 30 .5 divided by .0083, 8314 divided by 298, enter.

01:46 And i get one point, why am i getting a wrong number here? oh, i forgot my eight and my 314.

02:15 Okay, i forgot.

02:17 I was looking at my numbers i could tell it was a dumb number.

02:20 So 30 .5 divided by 0 .008314.

02:27 There we go.

02:28 Divided by 298.

02:30 And now i'm getting 222000 joules or 222000.