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(a) For a process that occurs at constant temperature, express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b) For a certain process that occurs at constant \(T\) and \(P\), the value of \(\Delta G\) is positive. What can you conclude? (c) What is the relationship between \(\Delta G\) for a process and the rate at which it occurs?

Short Answer

Expert verified
In summary, (a) the change in Gibbs free energy at constant temperature is given by \(\Delta G = \Delta H - T\Delta S\), (b) a positive \(\Delta G\) at constant \(T\) and \(P\) indicates a nonspontaneous process, and (c) while \(\Delta G\) provides information about the spontaneity of a process, it does not indicate the rate. The rate of a process depends on activation energy and rate constant, which are connected through the Arrhenius equation: \(k = Ae^{-\frac{E_a}{RT}}\).

Step by step solution

01

Part (a): Deriving the expression for change in Gibbs free energy at constant temperature

To derive the expression for the change in Gibbs free energy at constant temperature, we start with the definition of Gibbs free energy (\(G\)): \(G = H - TS\) Where, \(H\) is enthalpy, \(T\) is the temperature, and \(S\) is entropy. To find the change in Gibbs free energy (\(\Delta G\)) at constant temperature, we differentiate \(G\) with respect to some external parameter \(X\) (like pressure, volume, or composition) while holding temperature constant: \(\frac{dG}{dX} = \frac{dH}{dX} - T\frac{dS}{dX} - S\frac{dT}{dX}\) At constant temperature, \(\frac{dT}{dX} = 0\), so the expression reduces to: \(\frac{dG}{dX} = \frac{dH}{dX} - T\frac{dS}{dX}\) Integrating both sides with respect to \(X\), we get the expression for the change in Gibbs free energy at constant temperature: \(\Delta G = \Delta H - T\Delta S\)
02

Part (b): Positive \(\Delta G\) for a process at constant \(T\) and \(P\)

When a process occurs at constant \(T\) and \(P\), the value of \(\Delta G\) determines if the process is spontaneous or not. Specifically, the spontaneity criteria are: - If \(\Delta G < 0\), the process is spontaneous. - If \(\Delta G = 0\), the process is at equilibrium. - If \(\Delta G > 0\), the process is nonspontaneous. Given that the value of \(\Delta G\) is positive, we can conclude that the process is nonspontaneous. In other words, the process will not proceed in the forward direction on its own under the given conditions of constant temperature and pressure.
03

Part (c): Relationship between \(\Delta G\) and the rate of a process

While the value of \(\Delta G\) gives information about the spontaneity of a process, it does not provide any information about its rate (how fast the process occurs). A process with a negative \(\Delta G\) will be spontaneous, but it could occur very quickly or very slowly, depending on the reaction mechanism and activation energy. Activation energy is the minimum energy required to initiate a chemical or physical process, and it is related to the rate constant. The activation energy and the rate constant are connected through the Arrhenius equation: \(k = Ae^{-\frac{E_a}{RT}}\) where \(k\) is the rate constant, \(A\) is the pre-exponential factor (also called frequency factor), \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature. Thus, to determine the rate of a process, one should consider not only the thermodynamics (change in Gibbs free energy) but also the kinetics (activation energy and rate constant).

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Most popular questions from this chapter

(a) Express the second law of thermodynamics in words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a certain spontaneous process the system undergoes an entropy change, \(\Delta S=42 \mathrm{~J} / \mathrm{K}\). What can you conclude about \(\Delta S_{\text {surr }} ?\)

The volume of \(0.100 \mathrm{~mol}\) of helium gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from \(2.00 \mathrm{~L}\) to \(5.00 \mathrm{~L}\). Assuming the gas to be ideal, calculate the entropy change for the process.

Explain qualitatively how \(\Delta G\) changes for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: (a) \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) (c) \(2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\)

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