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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

Expert verified
There are three ways to calculate 螖G掳 for a reaction at $25^{\circ}\mathrm{C}$: 1) Using the standard enthalpy change (螖H掳) and the standard entropy change (螖S掳) 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掳) with the equation \( \Delta G^\circ = -nFE^\circ \). To estimate 螖G掳 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) 螖H掳 and 螖S掳 are temperature-independent or their temperature-dependence is negligible; 3) The van't Hoff equation assumes 螖H掳 is constant over the temperature range considered.

Step by step solution

01

1. Calculating 螖G掳 using the standard enthalpy change (螖H掳) and the standard entropy change (螖S掳)

To calculate the standard free energy change (螖G掳) at 25掳C using the standard enthalpy change (螖H掳) and the standard entropy change (螖S掳), use the following equation: \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \) where T is the temperature in Kelvin (25掳C = 298.15 K).
02

2. Calculating 螖G掳 from the equilibrium constant (K)

The standard free energy change (螖G掳) at 25掳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路K) and T is the temperature in Kelvin (25掳C = 298.15 K).
03

3. Calculating 螖G掳 using the relationship between the Gibbs free energy and the standard electrode potential (E掳)

For redox reactions, the standard free energy change (螖G掳) can be calculated using the relationship between the Gibbs free energy change and the standard electrode potential (E掳) 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路V), and E掳 is the standard electrode potential of the reaction.
04

Estimating 螖G掳 at temperatures other than 25掳C

At temperatures other than 25掳C, the standard free energy change (螖G掳) 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 螖G鈧伮 and 螖G鈧偮 are the standard free energy changes at temperatures T鈧 and T鈧 (in Kelvin) respectively, R is the universal gas constant, and 螖H掳 is the standard enthalpy change of the reaction.
05

Assumptions made

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

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Key Concepts

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

Standard Enthalpy Change
The standard enthalpy change (螖H掳) is a measure of the heat absorbed or released during a chemical reaction under standard state conditions. It's important because it helps predict whether a reaction is endothermic (absorbs heat) or exothermic (releases heat).

When calculating Gibbs Free Energy (\(螖G^\) using 螖H掳 at 25掳C, we use the formula:
  • \(螖G^\circ = 螖H^\circ - T螖S^\ \circ\)
This formula helps in understanding the balance between enthalpy and entropy changes in determining the spontaneity of a reaction.

It is assumed that 螖H掳 remains constant with temperature, allowing us to use it across different calculations and conditions.
Standard Entropy Change
Standard entropy change (螖S掳) measures the disorder or randomness of a system when a reaction occurs under standard state conditions. A positive 螖S掳 indicates increased disorder, while a negative value suggests a decreased disorder.

Entropy is crucial in calculating the Gibbs Free Energy, particularly its role in determining whether a reaction is spontaneous. The equation, \(螖G^\circ = 螖H^\circ - T 螖S^\circ\), includes 螖S掳 to account for this disorder's impact. The temperature in Kelvin (T) amplifies any changes in 螖S掳, making it essential to consider even small entropy changes.
Equilibrium Constant
The equilibrium constant (K) measures a reaction's position at equilibrium under standard conditions. It is a ratio of the concentration of products to reactants, each raised to the power of their respective coefficients in the balanced equation.

For calculating Gibbs Free Energy from K, the equation \(螖G^\circ = -RT \ln K\) is used. Here, R is the universal gas constant, and T is the temperature in Kelvin. This relationship shows that reactions with large K values favor products and tend to be spontaneous, reflected by a negative 螖G掳 value.
Gibbs Free Energy
Gibbs Free Energy (螖G) is a measure of a system's maximum reversible work at constant temperature and pressure. It tells us whether a reaction is spontaneous or requires energy input. Negative 螖G values indicate spontaneous reactions, while positive values suggest non-spontaneity.

In redox reactions, the 螖G掳 can be calculated from the standard electrode potential (E掳) using the formula:
  • \(螖G^\circ = -nFE^\circ\)
Where 'n' is the number of electrons transferred, and 'F' is Faraday's constant. This equation connects electrochemical cell potentials with reaction spontaneity.
Standard Electrode Potential
Standard electrode potential (E掳) reflects the voltage associated with a redox reaction under standard conditions. It provides insight into the electron transfer tendency during a reaction. A positive E掳 indicates that a reaction is more likely to occur spontaneously.

You can utilize the E掳 to find the standard free energy change of a redox reaction through the equation \(螖G^\circ = -nFE^\circ\). This links the thermodynamic and electrochemical perspectives of reactions, showing how changes in electron flow relate to energy transformations.
Van't Hoff Equation
The Van鈥檛 Hoff equation helps estimate how equilibrium constants vary with temperature. The equation used in this context is:
  • \[\frac{\Delta G_2^\circ - \Delta G_1^\circ}{T_2 - T_1} = -\frac{\Delta H^\circ}{R}(\frac{1}{T_2} - \frac{1}{T_1})\]
This provides a means of estimating 螖G掳 for temperatures other than 25掳C. The Van鈥檛 Hoff equation assumes that 螖H掳 remains constant over the temperature range. This method is essential for reactions that do not strictly occur at standard temperature.
Standard State Conditions
Standard state conditions are a set of assumptions applied in thermodynamic calculations to have consistency across systems. These conditions include:
  • Temperature of 25掳C (298.15 K)
  • Pressure of 1 atm
  • Concentration of 1 M for all reactants and products
These standardized parameters provide a reference point to compare different reactions' thermodynamics. It establishes a common ground for evaluating changes in enthalpy, entropy, and free energy across various conditions. Moreover, calculations assume temperature-independent 螖H掳 and 螖S掳, simplifying the characterization of chemical processes.

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Most popular questions from this chapter

A green plant synthesizes glucose by photosynthesis, as shown in the reaction $$6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$$ Animals use glucose as a source of energy: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$$ If we were to assume that both these processes occur to the same extent in a cyclic process, what thermodynamic property must have a nonzero value?

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$

For a liquid, which would you expect to be larger, \(\Delta S_{\text {fusion }}\) or \(\Delta S_{\text {evaporation }}\) ? Why?

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.

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

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