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The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: $$ \mathrm{n} K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (a) Explain how this equation can be used to determine \(\Delta H^{\circ}\) experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal?

Short Answer

Expert verified
The given equation represents the linear relationship between the equilibrium constant (K), reaction quotient (n), standard enthalpy change (∆H°), gas constant (R), and temperature (T). (a) To determine ∆H° experimentally, set up a reaction at different temperatures and measure the corresponding equilibrium constants (K). Plot a graph with K on the y-axis and \(\frac{1}{T}\) on the x-axis. Fit a straight line to the data points, calculate the slope (m), and use the formula \(m = \frac{-\Delta H^{\circ}}{R}\) to calculate ∆H°. (b) The equation can be derived starting from the relationships between Gibbs free energy, enthalpy, and entropy changes and the relationship between the equilibrium constant (K) and Gibbs free energy change (∆G°). The constant term in the equation equals \( n \frac{\Delta S^{\circ}}{R}\), the reaction quotient (n) multiplied by the ratio of the standard entropy change (∆S°) to the gas constant (R).

Step by step solution

01

Understanding the given equation

The given equation describes the relationship between equilibrium constant (K), reaction quotient (n), standard enthalpy change (∆H°), gas constant (R), and temperature (T), which can be written as: \[ n K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } \] We can see that this equation is of the form \( y = mx + c \), which implies it is a straight-line equation with K on the y-axis, T on the x-axis, and the slope of the line being \(\frac{-\Delta H^{\circ}}{R}\).
02

Part (a): Determining ∆H° experimentally

To determine ∆H° experimentally, the first step is to prepare a data set corresponding to several different temperatures and the corresponding equilibrium constants (K) at each temperature. 1. Set up the experiment: Perform the reaction under several different temperatures (\(T_1\), \(T_2\), \(T_3\), ...), and measure the equilibrium constants (\(K_1\), \(K_2\), \(K_3\), ...) for each reaction at these given temperatures. 2. Plot a graph: Plot a graph with K on the y-axis and \(\frac{1}{T}\) on the x-axis for the different temperature measurements. 3. Fit a straight line: Find the best-fit straight line for the plotted points using linear regression or a similar technique. 4. Calculate the slope: Determine the slope (m) of the regression line. 5. Determine ∆H°: Using the formula \(m = \frac{-\Delta H^{\circ}}{R}\), calculate the value of ∆H° using the obtained slope (m) and the gas constant (R). Now we will address part (b) of the question.
03

Part (b): Deriving the given equation

To derive the given equation, we will start with the relationship between the Gibbs free energy change (∆G°), standard enthalpy change (∆H°), and standard entropy change (∆S°) of a reaction, which can be written as: \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \] Also, the relationship between the equilibrium constant (K) and the Gibbs free energy change (∆G°) is given by: \[ \Delta G^{\circ} = -RT \ln K \] Now, we can substitute the first equation into the second equation, thus obtaining: \(-RT \ln K = \Delta H^{\circ} - T\Delta S^{\circ}\) Next, we divide both sides of the equation by RT, and multiply the term \(-\ln K\) by n: \[ nK = n \left(\frac{-\Delta H^{\circ} + T\Delta S^{\circ}}{RT}\right) \] Now, let's split the fraction: \[ n K = n \left(\frac{-\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}\right) \] Since \(\frac{\Delta S^{\circ}}{R}\) is a constant, we can simplify the equation by identifying the constant term: \[ n K = \frac{-\Delta H^{\circ}}{RT} + \text {constant} \]
04

Identifying the constant

According to the derived equation, the constant term in the equation is: \[ \text {constant} = n \frac{\Delta S^{\circ}}{R} \] Hence, the constant term in the given equation is equal to the reaction quotient (n) multiplied by the ratio of the standard entropy change (∆S°) to the gas constant (R).

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Key Concepts

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

Standard Enthalpy Change
When studying chemical reactions, the standard enthalpy change (ΔH°) is a crucial concept that represents the heat released or absorbed at constant pressure when reactants are transformed into products under standard conditions. It's an essential parameter in understanding how temperature influences a reaction's equilibrium.

For instance, if ΔH° is negative, the reaction is exothermic, meaning it releases heat; conversely, if ΔH° is positive, the reaction is endothermic and requires heat to proceed. By measuring the equilibrium constant at various temperatures, researchers can experimentally determine the standard enthalpy change for a reaction. This is done by plotting the logarithm of the equilibrium constant against the inverse of the temperature and calculating the slope of the resulting line. The slope is directly related to ΔH°, which can be extracted using the gas constant (R).
Gibbs Free Energy Change
The Gibbs free energy change (ΔG°) is an invaluable quantity that tells us whether a reaction is spontaneous under standard conditions. It combines the concepts of enthalpy (ΔH°) and entropy (ΔS°) to provide a fuller picture of a reaction's thermodynamics.

At equilibrium, ΔG° is zero, which allows for the derivation of a pivotal equation linking Gibbs free energy change to the equilibrium constant (K). This equation is given by ΔG° = -RT ln K, where R is the gas constant and T is the absolute temperature in Kelvins. Therefore, knowing the equilibrium constant enables the calculation of ΔG°. Moreover, through the relationship ΔG° = ΔH° - TΔS°, one can study how temperature influences the spontaneity of a reaction, as changes in temperature will affect ΔG° and the position of equilibrium.
Van 't Hoff Equation
The van 't Hoff equation is a powerful tool in physical chemistry that illustrates how equilibrium constants are affected by changes in temperature. It is based on the derivation of Gibbs free energy and the integration of this relationship with respect to temperature.

The integrated form of the van 't Hoff equation is frequently encountered in the following format: ln K = -ΔH°/(RT) + ΔS°/R, which shows the linear relationship between ln K and 1/T. By graphing this relationship, we can examine the reaction's thermodynamics and predict how equilibrium will shift with temperature changes. A more practical form of the van 't Hoff equation, sometimes used for data analysis, relates to the change in the equilibrium constant with temperature: dln K/dT = ΔH°/(RT²). This equation can provide insights into the temperature dependence of the standard enthalpy change.
Le Chatelier's Principle
Beyond mathematical equations, Le Chatelier's principle offers a qualitative guideline to predict how a system at equilibrium responds to external changes such as pressure, temperature, or concentration variations. According to this principle, if an external stress is applied to a system at equilibrium, the system will adjust itself to minimize that stress.

For example, in the context of temperature, if heat is added to an exothermic reaction (one that releases heat), the equilibrium will shift to consume that added heat, moving towards the reactants. Conversely, for an endothermic reaction (one that absorbs heat), adding heat would shift the equilibrium towards the products. Understanding this principle allows one to qualitatively predict the direction of shift in a reaction's equilibrium without relying solely on quantitative calculations, although both approaches often complement each other.

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Most popular questions from this chapter

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b}\). (b) By using the value of \(K_{b r}\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}+\right]=\left[\mathrm{H}^{+}\right]=1.5 \times 10^{-8} \mathrm{M}\) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]=5.5 \times 10^{-4} \mathrm{M}\), and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.120 \mathrm{M} ?\)

(a) What do you expect for the sign of \(\Delta S\) in a chemical reaction in which two moles of gaseous reactants are converted to three moles of gaseous products? (b) For which of the processes in Exercise \(19.9\) does the entropy of the system increase?

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K\), at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\). (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s\), graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about \(88 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) (a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{\circ}\) for \(\mathrm{Br}_{2}\) using data from Appendix \(C\). Assume that \(\Delta H_{\text {vap }}^{\circ}\) remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of \(\mathrm{Br}_{2}\) in a chemistry handbook or at the WebElements web site (www.webelements.com).

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