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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
Three ways to calculate the standard free energy change (ΔG°) at 25°C are: 1) using the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°) with the equation \(ΔG° = ΔH° - TΔS°\); 2) using the equilibrium constant (K) with the equation \(ΔG° = -RT \ lnK\); and 3) for redox reactions, using the standard reduction potentials (E°) with the equation \(ΔG° = -nFE°_{cell}\). To estimate ΔG° at different temperatures, the Van't Hoff equation \(\frac{d(lnK)}{dT} = \frac{ΔH°}{RT^2}\) can be used. Assumptions made in these calculations include the reaction being at equilibrium, standard states being well-defined, and the temperature-dependence of ΔH° and ΔS° being negligible or accounted for using empirical relationships.

Step by step solution

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1. Calculate ΔG° using ΔH° and ΔS°

One way to calculate the standard free energy change (ΔG°) for a reaction at 25°C is by using the standard enthalpy change (ΔH°) and the standard entropy change (ΔS°). This can be done using the following equation: \[ΔG° = ΔH° - TΔS°\] where T represents the temperature in Kelvin (298 K for 25°C).
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2. Calculate ΔG° using equilibrium constant

Another way to determine the standard free energy change is through the equilibrium constant (K) of the reaction. This can be done using the equation: \[ΔG° = -RT \ lnK\] where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K for 25°C).
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3. Calculate ΔG° using standard reduction potentials

For redox reactions, the standard free energy change can be calculated using the standard reduction potentials (E°) of the half-reactions involved: \[ΔG° = -nFE°_{cell}\] where n represents the number of moles of electrons transferred in the reaction, F is the Faraday constant (96,485 C/mol), and E°_{cell} is the standard cell potential, determined as the difference between the reduction potentials of the two half-reactions.
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Estimating ΔG° at other temperatures

The standard free energy change can be estimated at temperatures other than 25°C using the Van't Hoff equation: \[\frac{d(lnK)}{dT} = \frac{ΔH°}{RT^2}\] where K is the equilibrium constant, R is the gas constant, T is the temperature in Kelvin, and ΔH° is the standard enthalpy change. By integrating this equation over the desired temperature range and combining it with the relationship between ΔG° and K (ΔG° = -RT lnK), one can estimate ΔG° at different temperatures.
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Assumptions made

The main assumptions made in calculating the standard free energy change include: 1. The reaction is at equilibrium. 2. The standard states of the components are well-defined (e.g., ideal gases, pure solvents, etc.). 3. The temperature-dependence of ΔH° and ΔS° is either small enough to be negligible (over a limited temperature range) or can be accounted for using empirical relationships.

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Key Concepts

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

Standard Enthalpy Change (H°)
Understanding the role of standard enthalpy change (H°) in chemical reactions is crucial. It represents the total heat exchange under standard conditions—typically at 1 atm pressure and 25°C. When a reaction has a negative H°, it releases heat and is exothermic, while a positive H° indicates heat absorption and an endothermic process. The standard enthalpy change is the driving force behind the temperature-dependent behavior of a reaction, influencing the Gibbs free energy (G°) and the equilibrium constant (K).

Let's simplify this complex concept. Imagine you're at a park on a sunny day. The warmth you feel from the sunshine is analogous to an exothermic reaction, where the environment (the surroundings) gains heat from the sun (the system). Conversely, if you're absorbing coolness from a block of ice, you're experiencing an endothermic process, with the ice absorbing heat from the surroundings.
Standard Entropy Change (S°)
The standard entropy change (S°) is another thermodynamic term that describes the disorder or randomness within a system. It is a measurement of the possible arrangements of particles in a system at a constant temperature and pressure (standard conditions). It may seem a bit abstract, so let's use a real-world example.

Think of a classroom with neatly arranged desks as a system with low entropy. If students enter and scatter books and papers around, the entropy increases due to the added disorder. In chemical terms, when a solid turns into a liquid, entropy increases because the particles are more randomly arranged in the liquid state. Changes in S° contribute to the total G° and can predict whether or not a reaction will occur spontaneously.
Equilibrium Constant (K)
Our next concept is the equilibrium constant (K), an intriguing aspect of chemical reactions. It represents the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their stoichiometric coefficients. The value of K indicates the extent to which a reaction will proceed before reaching a state of balance. A large K means a reaction strongly favors product formation; a small K suggests the opposite, favoring reactants.

Imagine a tug-of-war match. The point at which neither team can pull the other is akin to reaction equilibrium. A high K indicates one team is much stronger (the products 'win'), while a low K means the other team is stronger (the reactants 'win').
Standard Reduction Potentials (E°)
Standard reduction potentials (E°) are integral to redox reactions, where electrons are transferred between chemical species. E°, measured in volts, indicates a substance's tendency to gain electrons (be reduced). The larger the positive value of E°, the greater the substance acts as an oxidizing agent, pulling electrons toward itself.

In our daily lives, battery function is a direct application of redox principles. A 9V battery has higher electrical potential energy compared to a 1.5V battery; similarly, substances with higher E° have a greater ability to drive electrical current. The standard reduction potentials enable the calculation of G°, thereby determining whether a redox reaction will occur spontaneously.
Van't Hoff Equation
The Van't Hoff equation is the bridge between thermodynamics and reaction kinetics. It reveals how the equilibrium constant (K) changes with temperature. This mathematical relation is derived from the integration of the entropy changes over temperature.

To visualize this, imagine a temperature dial controlling the brightness of a light bulb. As the temperature changes, the intensity of the light (the equilibrium position) adjusts, just as the K value changes with temperature in a chemical reaction. The Van't Hoff equation can be used to approximate G° for temperatures other than 25°C, enhancing our prediction capabilities for reaction behavior under varying conditions.
Reaction Equilibrium
Lastly, let's discuss reaction equilibrium—an essential concept in chemistry where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products. It's a state of dynamic balance, not to be mistaken for a static condition.

Think of a crowded mall where people enter and leave at the same rate. The total number of people inside the mall remains constant—this steady state is similar to chemical equilibrium. The equilibrium constant (K) quantifies this and helps predict the direction of the reaction. The principles that govern equilibrium, including standard free energy changes, enthalpy, and entropy, are fundamental to many applications in chemical engineering, environmental science, and even pharmacology.

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Most popular questions from this chapter

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

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?

You have a \(1.00\) - \(L\) sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\text {surr }}\) for this process. Assume the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\) over this temperature range, and the heat capacity of water is constant over this temperature range and equal to \(75.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ} $$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

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