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. What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ}\), the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1 ?\)

Short Answer

Expert verified
The Gibbs free energy change (ΔG) provides information about the spontaneity and direction of a chemical reaction under given conditions, while the standard free energy change (ΔG°) provides similar information under standard conditions. The relationship between ΔG° and the equilibrium constant (K) is given by the equation: \(K = \exp(\frac{-\Delta G^\circ}{RT})\). To estimate the value of K at different temperatures, the Van't Hoff equation can be used. Not all reactions have a specific temperature where K=1; calculating the temperature for K=1 from the Van't Hoff equation may lead to impossible or unrealistic temperatures that lie outside the applicable domain of the equation.

Step by step solution

01

The Gibbs free energy change (ΔG) of a reaction is a thermodynamic quantity that combines enthalpy change and entropy change. It provides information about the spontaneity and direction of a chemical reaction under given conditions. If ΔG is negative, the reaction is spontaneous in the forward direction. If ΔG is positive, the reverse reaction is spontaneous. If ΔG is zero, the system is at equilibrium and there is no net change in the concentrations of reactants and products. #Step 2: Understanding standard free energy change, ΔG°#

The standard free energy change (ΔG°) is the Gibbs free energy change under standard conditions (1 atm pressure, 1 M concentration, and 25°C temperature). It provides information about the spontaneity and direction of a chemical reaction under standard conditions. Like ΔG, if ΔG° is negative, the reaction is spontaneous in the forward direction under standard conditions. If ΔG° is positive, the reverse reaction is spontaneous under standard conditions. If ΔG° is zero, the system is at equilibrium under standard conditions. #Step 3: Determining the equilibrium constant, K, from ΔG°#
02

The relationship between ΔG° and the equilibrium constant, K, for a reaction is given by the equation: \[\Delta G^\circ = -RT\ln K\] where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. This equation can be rearranged to calculate K as follows: \[K = \exp(\frac{-\Delta G^\circ}{RT})\] By calculating ΔG° for a reaction and knowing the temperature, one can determine the value of K using this equation. #Step 4: Estimating the value of K at different temperatures#

To estimate the value of K at temperatures other than 25°C, the Van't Hoff equation can be used: \[\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}(\frac{1}{T_2}-\frac{1}{T_1})\] This equation relates the change in the equilibrium constant with the change in temperature, considering the standard enthalpy change (ΔH°) for the reaction. By knowing the value of K at a specific temperature (e.g., 25°C), one can estimate K at a different temperature using this equation. #Step 5: Estimating the temperature where K=1#
03

At the temperature where K=1, the reaction will be at equilibrium, and the Gibbs free energy change, ΔG, will be zero. From the equation, ΔG = ΔG° + RT ln K, and knowing ΔG should be 0 at this temperature, we can write: \[0 = \Delta G^\circ + RT\ln 1\] Since ln(1) is 0, the equation becomes: \[0 = \Delta G^\circ\] #Step 6: Determining if all reactions have a specific temperature where K=1#

Not all reactions have a specific temperature where K=1. If a reaction has a ΔG° of zero at a specific temperature, then K would be equal to 1 at that temperature. However, this does not mean that all reactions must have such a temperature. In addition, calculating the temperature for K=1 from the Van't Hoff equation may lead to impossible or unrealistic temperatures that lie outside the applicable domain of the equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Free Energy Change
Understanding the standard free energy change, denoted as \(\Delta G^\circ\), is crucial for predicting the spontaneity of a reaction under standardized conditions, typically set at 1 atmosphere pressure, 1M concentration, and 25°C (298K). It reflects the maximum amount of work that can be extracted from a chemical process without external interventions. When \(\Delta G^\circ\) is negative, the reaction will proceed spontaneously in the forward direction. A positive \(\Delta G^\circ\) implies a non-spontaneous forward reaction, indicating that the reverse reaction tends to occur spontaneously. At \(\Delta G^\circ = 0\), the system is at equilibrium, indicating no net progression of reactants to products or vice versa under standard conditions.

Equilibrium Constant (K)
The equilibrium constant, represented as \(K\), is a dimensionless quantity that expresses the relative concentrations of reactants and products at equilibrium for a particular reaction. Using the equation \(\Delta G^\circ = -RT\ln K\), where \(R\) is the gas constant and \(T\) the temperature in Kelvin, we can derive \(K\) from the standard free energy change. This relationship signifies that the magnitude of \(K\) is directly influenced by the value of \(\Delta G^\circ\); a large \(K\) corresponds to a highly negative \(\Delta G^\circ\), implying a strong drive toward product formation under standard conditions.
  • If \(K > 1\), the products are favored at equilibrium.
  • If \(K < 1\), the reactants are favored at equilibrium.
  • If \(K = 1\), neither reactants nor products are favored, indicating a perfectly balanced equilibrium.

Spontaneity of Reactions
The spontaneity of a reaction is a measure of whether a reaction will proceed on its own under a given set of conditions. It is closely linked to the concept of Gibbs free energy change. A negative Gibbs free energy change (\(\Delta G\)) indicates that a reaction will occur spontaneously in the forward direction at those specific conditions. Alternatively, a positive \(\Delta G\) means that the reaction is spontaneous in the reverse direction. Spontaneity does not imply speed; a spontaneous process may occur rapidly or may require a considerable amount of time to proceed. Only at equilibrium (\(\Delta G = 0\)), there is no spontaneous change as the system's free energy is minimized.

Van't Hoff Equation
The Van't Hoff equation bridges thermodynamics and equilibrium by providing insight into how the equilibrium constant (\(K\)) changes with temperature. It is expressed as \[\ln\frac{K_2}{K_1} = -\frac{\Delta H^\circ}{R}(\frac{1}{T_2}-\frac{1}{T_1})\] This relationship stems from the dependency of reaction spontaneity on enthalpy changes at different temperatures. The Van't Hoff equation allows us to estimate \(K\) at a temperature other than the standard 25°C, assuming that the enthalpy change (\(\Delta H^\circ\)) is constant over the temperature range considered. This can be a powerful tool in predicting and understanding the effect of temperature on chemical equilibrium.

Thermodynamics in Chemistry
Thermodynamics serves as the framework for understanding energy changes in chemical reactions. It involves studying system properties like enthalpy, entropy, and Gibbs free energy to predict whether processes will occur spontaneously. Chemical thermodynamics particularly focuses on the flow of energy and the conversion of heat and work into each other during chemical reactions. A deep comprehension of these principles is essential for the practical application of chemistry across various fields including biochemical reactions, industrial processes, and environmental science. Key principles like the first and second laws of thermodynamics govern the energetics of chemical reactions and are crucial for setting environmental policies and creating more energy-efficient processes.

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Most popular questions from this chapter

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

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 \(78 ?\)

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(g)\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

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