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Ch.18 - Free Energy and Thermodynamics
Chapter 18, Problem 91a

Living organisms use energy from the metabolism of food to create an energy-rich molecule called adenosine triphosphate (ATP). The ATP acts as an energy source for a variety of reactions that the living organism must carry out to survive. ATP provides energy through its hydrolysis, which can be symbolized as follows: ATP(aq) + H2O(l) → ADP(aq) + Pi(aq) ΔGrxn ° = -30.5 kJ where ADP represents adenosine diphosphate and Pi represents an inorganic phosphate group (such as HPO42-). a. Calculate the equilibrium constant, K, for the given reaction at 298 K.

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1. The first step is to understand the relationship between the standard Gibbs free energy change (ΔGrxn °) and the equilibrium constant (K). This relationship is given by the equation: ΔGrxn ° = -RT ln(K), where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin.
2. In this problem, we are given ΔGrxn ° = -30.5 kJ, but we need to convert this to J to match the units of R. So, ΔGrxn ° = -30.5 kJ * 1000 J/kJ = -30500 J.
3. We are also given the temperature T = 298 K. Now we can substitute these values into the equation ΔGrxn ° = -RT ln(K).
4. To solve for K, we need to isolate K on one side of the equation. We can do this by first dividing both sides of the equation by -RT, which gives us ln(K) = -ΔGrxn ° / RT.
5. Finally, to get K by itself, we need to get rid of the natural logarithm (ln) on the left side of the equation. We can do this by taking the exponent of both sides of the equation, which gives us K = e^(-ΔGrxn ° / RT). This is the equation you would use to calculate the equilibrium constant K for the given reaction at 298 K.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy and Equilibrium Constant

The Gibbs free energy change (ΔG) of a reaction is related to its equilibrium constant (K) by the equation ΔG° = -RT ln(K), where R is the universal gas constant and T is the temperature in Kelvin. A negative ΔG indicates that the reaction is spontaneous and favors the formation of products, which corresponds to a larger K value. Understanding this relationship is crucial for calculating K from the given ΔGrxn.
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Hydrolysis of ATP

ATP hydrolysis is a key biochemical reaction where ATP is converted into ADP and inorganic phosphate (Pi), releasing energy. This reaction is vital for cellular processes, as the energy released is harnessed for various biological functions. The standard Gibbs free energy change for this reaction is approximately -30.5 kJ, indicating that it is energetically favorable and drives many endergonic reactions in cells.
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Standard Conditions in Thermodynamics

Standard conditions in thermodynamics refer to a set of specific conditions (usually 1 atm pressure and 298 K temperature) under which thermodynamic measurements are made. These conditions allow for consistent comparisons of thermodynamic data, such as Gibbs free energy and equilibrium constants. When calculating K, it is essential to ensure that the reaction is evaluated under these standard conditions to obtain accurate and meaningful results.
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Related Practice
Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g) a. Show that the reaction is not spontaneous under standard conditions by calculating ΔG°rxn.

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Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2( g) b. If BaCO3 is placed in an evacuated flask, what is the partial pressure of CO2 when the reaction reaches equilibrium?

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Textbook Question

Consider this reaction occurring at 298 K: BaCO3(s) ⇌ BaO(s) + CO2(g) c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, at what temperature is the partial pressure of carbon dioxide 1.0 atm?

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Textbook Question

Living organisms use energy from the metabolism of food to create an energy-rich molecule called adenosine triphosphate (ATP). The ATP acts as an energy source for a variety of reactions that the living organism must carry out to survive. ATP provides energy through its hydrolysis, which can be symbolized as follows: ATP(aq) + H2O(l) → ADP(aq) + Pi(aq) ΔG°rxn = -30.5 kJ where ADP represents adenosine diphosphate and Pi represents an inorganic phosphate group (such as HPO42-). b. The free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the given reaction in reverse. Calculate the standard free energy change for the oxidation of glucose and estimate the maximum number of moles of ATP that can be formed by the oxidation of one mole of glucose.

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Open Question
The standard free energy change for the hydrolysis of ATP was given in Problem 91. In a particular cell, the concentrations of ATP, ADP, and Pi are 0.0031 M, 0.0014 M, and 0.0048 M, respectively. Calculate the free energy change for the hydrolysis of ATP under these conditions, assuming a temperature of 298 K.
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Textbook Question

These reactions are important in catalytic converters in automobiles. Calculate ΔG° for each at 298 K. Predict the effect of increasing temperature on the magnitude of ΔG°.

a. 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g)

b. 5 H2(g) + 2 NO(g) → 2 NH3(g) + 2 H2O(g)

c. 2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g)

d. 2 NH3(g) + 2 O2(g) → N2O(g) + 3 H2O(g)

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