List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C}\) ? What assumptions are made?

Short Answer

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There are three ways to calculate ΔG° for a reaction at $25^{\circ}\mathrm{C}$: 1) Using the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) with the equation \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \); 2) Using the equilibrium constant (K) with the equation \( \Delta G^\circ = -RT \ln K \); 3) For redox reactions, using the standard electrode potential (E°) with the equation \( \Delta G^\circ = -nFE^\circ \). To estimate ΔG° at temperatures other than $25^{\circ}\mathrm{C}$, the van't Hoff equation \( \frac{\Delta G_2^\circ - \Delta G_1^\circ}{T_2 - T_1} = -\frac{\Delta H^\circ}{R}(\frac{1}{T_2} - \frac{1}{T_1}) \) can be used. The assumptions made are: 1) Standard state conditions are considered; 2) ΔH° and ΔS° are temperature-independent or their temperature-dependence is negligible; 3) The van't Hoff equation assumes ΔH° is constant over the temperature range considered.

Step by step solution

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1. Calculating ΔG° using the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°)

To calculate the standard free energy change (ΔG°) at 25°C using the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°), use the following equation: \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \) where T is the temperature in Kelvin (25°C = 298.15 K).
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2. Calculating ΔG° from the equilibrium constant (K)

The standard free energy change (ΔG°) at 25°C can also be calculated using the equilibrium constant (K) of the reaction using the following equation: \( \Delta G^\circ = -RT \ln K \) where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (25°C = 298.15 K).
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3. Calculating ΔG° using the relationship between the Gibbs free energy and the standard electrode potential (E°)

For redox reactions, the standard free energy change (ΔG°) can be calculated using the relationship between the Gibbs free energy change and the standard electrode potential (E°) using the following equation: \( \Delta G^\circ = -nFE^\circ \) where n is the number of electrons transferred in the redox reaction, F is Faraday's constant (96,485 C/mol or 96.485 kJ/mol·V), and E° is the standard electrode potential of the reaction.
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Estimating ΔG° at temperatures other than 25°C

At temperatures other than 25°C, the standard free energy change (ΔG°) can be estimated using the van't Hoff equation: \( \frac{\Delta G_2^\circ - \Delta G_1^\circ}{T_2 - T_1} = -\frac{\Delta H^\circ}{R}(\frac{1}{T_2} - \frac{1}{T_1}) \) where ΔG₁° and ΔG₂° are the standard free energy changes at temperatures T₁ and T₂ (in Kelvin) respectively, R is the universal gas constant, and ΔH° is the standard enthalpy change of the reaction.
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Assumptions made

The assumptions made in the calculation and estimation of the standard free energy change (ΔG°) are: 1. The standard state conditions, including a temperature of 25°C, 1 atm pressure, and 1 M concentration for each reactant and product, are considered. 2. The enthalpy and entropy changes (ΔH° and ΔS°) are temperature-independent or temperature-dependence is negligible. 3. The van't Hoff equation assumes that the standard enthalpy change (ΔH°) is constant over the temperature range considered.

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Most popular questions from this chapter

Consider the following reaction at \(800 . \mathrm{K}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm} .\) Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at \(600 . \mathrm{K}\) \(2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at 600. \(\mathrm{K}\)

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), cal- culate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})\) at \(1.00\) atm and \(1.00 \mathrm{~mol} \mathrm{~B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Calculate \(\Delta S_{\text {surr }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 atm. a. \(\mathrm{C}_{3} \mathrm{H}_{5}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(t) \Delta H^{\circ}=-2221 \mathrm{~kJ}\) b. \(2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H^{\circ}=112 \mathrm{~kJ}\)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

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