Question

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 \(78 ?\)

Short answer

The equilibrium constant K for the hydrolysis of ATP at 25°C is approximately \(1.64 \times 10^{6}\). About 94 ATP molecules can be produced for every molecule of glucose. Approximately 13.8 moles of ATP need to be hydrolyzed to provide the energy required for the transport of 1 mol of K+ ions.

Step-by-step solution

Use the relationship between ΔG° and K

To find the equilibrium constant K, we can use the formula: \[\Delta G^{\circ} = -RT \ln K\] where ΔG° is the standard Gibbs free energy change (-30.5 kJ/mol), R is the gas constant (8.314 J/mol·K), T is the temperature (298 K), and K is the equilibrium constant we want to find. Note that we need to convert ΔG° to J/mol before plugging into the formula.

Calculate K

Rearrange the formula to solve for K: \[K = e^{-\frac{\Delta G^{\circ}}{RT}}\] Now plug in the values and convert ΔG° to J/mol: \[K = e^{-\frac{-30.5 \times 10^3\ \mathrm{J/mol}}{8.314\ \mathrm{J/mol\cdot K} \times 298\ \mathrm{K}}}\] \[K \approx 1.64 \times 10^{6}\] So the equilibrium constant K for the hydrolysis of ATP at 25°C is approximately \(1.64 \times 10^{6}\). b. Calculate the number of ATP molecules produced for every molecule of glucose

Determine the available energy from glucose metabolism

For glucose metabolism, the overall reaction is: \[C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)\] The ΔG° for the metabolism of one mole of glucose is -2880 kJ. This energy is used to form ATP from ADP.

Calculate the number of ATP molecules produced

Each mole of ATP uses -30.5 kJ of energy during hydrolysis. To calculate the number of ATP molecules produced, we can divide the total energy from glucose metabolism by the energy required to form one mole of ATP: \[\frac{-2880\ \mathrm{kJ/mol}}{-30.5\ \mathrm{kJ/mol}} \approx 94.4\] Thus, about 94 ATP molecules can be produced for every molecule of glucose. c. Calculate the amount of ATP required to transport 1 mol of K+ ions

Determine the energy required for transport

From Exercise 78, we know that the energy required to transport 1 mol of K+ ions from the blood to the inside of a muscle cell at 37°C is -420 kJ.

Calculate the amount of ATP required

Since the hydrolysis of 1 mole of ATP provides -30.5 kJ of energy, we can calculate the number of ATP moles needed for the transport as follows: \[\frac{-420\ \mathrm{kJ}}{-30.5\ \mathrm{kJ/mol}} \approx 13.8\] Therefore, approximately 13.8 moles of ATP need to be hydrolyzed to provide the energy required for the transport of 1 mol of K+ ions.

Key concepts

Gibbs free energy

Gibbs free energy is a crucial concept in chemistry that helps determine whether a chemical reaction will occur spontaneously. It combines enthalpy (the total energy within a system) and entropy (the measure of disorder or randomness) to provide a comprehensive picture. In simple terms, it tells us the balance between the energy needed to start a reaction and the extent to which the reaction increases disorder.
The formula for Gibbs free energy change is \(\Delta G = \Delta H - T\Delta S\), where:

• \(\Delta G\) is the change in Gibbs free energy
• \(\Delta H\) is the change in enthalpy
• \(T\) is the temperature in Kelvins
• \(\Delta S\) is the change in entropy

A negative \(\Delta G\) indicates a spontaneous reaction, meaning it can occur without any input of energy. In ATP hydrolysis, a negative Gibbs free energy (-30.5 kJ/mol) shows that breaking down ATP into ADP and phosphate is favored and releases energy that can be used by cells.
This energy is vital for cells as it powers numerous biological processes, from muscle contraction to neuron firing.

equilibrium constant

The equilibrium constant \(K\) is a number that provides insight into the direction and extent of a chemical reaction. It emerges from the balanced state of a reversible reaction where the concentrations of reactants and products reach a constant ratio.
The formula linking Gibbs free energy and the equilibrium constant is:\[\Delta G^{\circ} = -RT \ln K\]Here \(R\) is the gas constant (approx. 8.314 J/mol·K) and \(T\) is the temperature in Kelvin. By rearranging this formula, we have:\[K = e^{-\frac{\Delta G^{\circ}}{RT}}\]This means a large \(K\) indicates a high concentration of products at equilibrium compared to reactants.
In ATP hydrolysis, a high equilibrium constant (\(K \approx 1.64 \times 10^{6}\)) signifies a significant favoring towards the formation of ADP and phosphate ions. This balance is crucial since it helps ensure that the energy release from ATP hydrolysis can drive vital cellular reactions.

glucose metabolism

Glucose metabolism is the process by which cells break down glucose to release energy, essential for supporting cellular functions. This energy is captured in the form of molecules like ATP.

The complete breakdown of glucose, known as cellular respiration, includes:

• Glycolysis: breaking glucose into two molecules of pyruvate
• Citric Acid Cycle: further oxidation of pyruvate
• Electron Transport Chain: producing the bulk of ATP

The summarized reaction is:\[C_6H_{12}O_6(s) + 6 O_2(g) \rightarrow 6 CO_2(g) + 6 H_2O(l)\]This reaction yields a \(\Delta G^{\circ}\) of -2880 kJ. With this energy, cells produce ATP. For every molecule of glucose, approximately 94 ATP molecules are synthesized during respiration.
This stepwise process is efficient and allows organisms to extract as much usable energy as possible, which cells depend on for their most energy-intensive tasks.

energy transport in cells

Energy transport in cells involves the movement and conversion of energy derived from ATP hydrolysis to power cellular activities. The energy stored in ATP is released when it's converted to ADP and a phosphate group, a crucial role in biological systems.

• It powers active transport, moving molecules against their concentration gradient, crucial for nutrient uptake and waste removal.
• Facilitates mechanical work, such as muscle contraction and the movement of cilia.
• Drives chemical reactions necessary for anabolism and catabolism.

In scenarios requiring high energy, like muscle cells needing to pump potassium ions (\(K^{+}\)) across membranes, a large amount of ATP is necessary. For instance, hydrolyzing around 13.8 moles of ATP can drive the movement of 1 mole of \(K^{+}\), using the energy efficiently for cellular function at physiological temperatures.
Cells have evolved this ATP utilization mechanism to maintain homeostasis and support complex life processes.