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?

Step by step solution

01

Write down the formula for calculating the equilibrium constant

We will use the relation between the standard free energy change (∆G°) and the equilibrium constant (K) for a given reaction: \[ \Delta G^\circ = -RT \ln K, \] where R is the ideal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and ln is the natural logarithm.

02

Convert the temperature to Kelvin and substitute the given values

First, convert the temperature from Celsius to Kelvin: T = 25°C + 273.15 = 298.15 K. Now, substitute the given values of ∆G° (-30.5 kJ/mol) and R (8.314 J/mol·K) into the formula: \[ -30500 \text{ J/mol} = -(8.314 \text{ J/mol·K})(298.15 \text{ K}) \ln K \]

03

Solve for K

Divide both sides by (-8.314 J/mol·K)(298.15 K), and then take the exponent of the result to calculate K: \[ K = e^\frac{30500 \text{ J/mol}}{(8.314 \text{ J/mol·K})(298.15 \text{ K})} \approx 1.66 \times 10^{11} \] The equilibrium constant (K) for the ATP hydrolysis reaction at 25°C is approximately \(1.66 \times 10^{11}\). #b. Calculate the number of ATP molecules produced from one glucose molecule#

04

Calculate the free energy change for the glucose metabolism reaction

The free energy change for the glucose metabolism reaction (∆G_g) can be determined using the standard free energy of formation (∆G_f°) values for the reactants and products in the reaction: \[ \Delta G_g = \sum [\Delta G_f°(\text{products})] - \sum [\Delta G_f°(\text{reactants})] \] ∆G_f° values for the reactants and products are as follows: - Glucose (∆G_f°) = -910 kJ/mol - O2 (∆G_f°) = 0 kJ/mol - CO2 (∆G_f°) = -394 kJ/mol - H2O (∆G_f°) = -228 kJ/mol Using these values, calculate ∆G_g: \[ \Delta G_g = [6(-394 \text{ kJ/mol}) + 6(-228 \text{ kJ/mol})] - [-910 \text{ kJ/mol}] = -2808 \text{ kJ/mol} \] The free energy change for the glucose metabolism reaction is -2808 kJ/mol.

05

Calculate the number of ATP molecules formed from one glucose molecule

Assuming all the free energy obtained from glucose metabolism is used for the synthesis of ATP from ADP, we can calculate the number of ATP molecules formed by dividing the total free energy change of glucose metabolism by the free energy change of the ATP hydrolysis reaction: \[ \text{Number of ATP molecules} = \frac{\Delta G_g}{\Delta G_{ \text{ATP hydrolysis}}} = \frac{-2808 \text{ kJ/mol}}{-30.5 \text{ kJ/mol}} = 92 \text{ ATP molecules} \] Approximately 92 ATP molecules can be produced from the metabolism of one glucose molecule, assuming all the free energy obtained is used for ATP synthesis.

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