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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: $$ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } $$ (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
(a) To determine ΔH° experimentally, measure K values at different temperatures (T) and plot \(\ln K\) against \(\frac{1}{T}\). The slope of the line equals \(\frac{-\Delta H^{\circ}}{R}\). Calculate ΔH° using \(\Delta H^{\circ} = -R \times (\text{slope})\). (b) The equation is derived using the van 't Hoff equation and integration. The constant C corresponds to the "constant" term in the given equation.

Step by step solution

01

(a) Using the equation to determine ΔH° experimentally

To determine ΔH° experimentally, we need to have equilibrium constants (K) at various temperatures (T). According to the given equation: \[ \ln K=\frac{-\Delta H^{\circ}}{R T}+\text { constant } \] This is a linear equation where the slope equals \(\frac{-\Delta H^{\circ}}{R}\) and the intercept is equal to the constant. By measuring K values at different temperatures (T) and making a plot of \(\ln K\) against \(\frac{1}{T}\), we can find the slope of the line corresponding to this dataset. Once we have the slope, we can calculate \(\Delta H^{\circ}\) using the formula: \[ \Delta H^{\circ} = -R \times (\text{slope}) \]
02

(b) Deriving the equation and identifying the constant

To derive the equation, we use the van 't Hoff equation: \[ \frac{\mathrm{d}(\ln K)}{\mathrm{d}T} = \frac{\Delta H^{\circ}}{RT^2} \] Now, we'll integrate this equation. Let's assume that the constant of integration is "C". Integrating both sides with respect to T, we get: \[ \int\frac{\mathrm{d}(\ln K)}{\mathrm{d}T} \mathrm{d}T = \int\frac{\Delta H^{\circ}}{RT^2} \mathrm{d}T \] \[ \ln K = -\frac{\Delta H^{\circ}}{R} \int\frac{1}{T^2} \mathrm{d}T + C \] Now, integrating the left side: \[ \int\frac{1}{T^2} \mathrm{d}T = \int T^{-2} \mathrm{d}T = -\frac{1}{T} \] Hence, the equation becomes: \[ \ln K = -\frac{\Delta H^{\circ}}{R}(-\frac{1}{T}) + C \] Simplifying the equation, we get: \[ \ln K = \frac{\Delta H^{\circ}}{RT} + C \] Comparing to the equation given in the exercise: \[ \ln K = \frac{-\Delta H^{\circ}}{R T}+\text { constant } \] We see that the constant C corresponds to the "constant" term in the given equation.

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Key Concepts

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

Standard Enthalpy Change
The concept of standard enthalpy change (\(\Delta H^{\circ}\)) refers to the heat absorbed or released during a chemical reaction at a constant pressure under standard conditions. These standard conditions typically include a pressure of 1 bar and a particular temperature (often 298.15 K), but for purposes of analyzing the van 't Hoff equation, any temperature can be considered.

In practical situations, knowing the standard enthalpy change of a reaction can tell us whether it is endothermic (absorbs heat) or exothermic (releases heat). This is fundamental for understanding the energy dynamics of a reaction. To find this experimentally, one would plot the natural logarithm of the equilibrium constant (\(\ln K\)) versus the reciprocal of the temperature (\(\frac{1}{T}\)) as specified in the original equation. The slope of this line is directly related to the standard enthalpy change by the relationship:
  • Slope = \(-\frac{\Delta H^{\circ}}{R}\).
Here, \(R\) is the universal gas constant. This gives a direct method for determining \(\Delta H^{\circ}\) from experimental data.

By understanding \(\Delta H^{\circ}\), we gain insights into the broader field of thermochemistry, guiding us towards potentially optimizing reactions for industrial or laboratory applications.
Equilibrium Constant
The equilibrium constant (\(K\)) is a value that expresses the ratio of concentrations of products to reactants at equilibrium for a given reversible reaction under constant temperature and pressure. It is a dimensionless number that provides significant insight into the position and feasibility of a chemical reaction.

When we integrate it into the van 't Hoff equation, \(\ln K = \frac{-\Delta H^{\circ}}{RT} + \text{constant} \), we can analyze how the equilibrium constant changes with temperature. This change embodies the reaction's energetic profile; specifically, how \(K\) adapts in response to temperature variations reflects directly on the Gibbs free energy change for the process.
  • If \(K > 1\), products are favored at equilibrium.
  • If \(K < 1\), reactants are favored.
  • When \(K = 1\), neither products nor reactants are favored.
This constant allows chemists to predict the extent of a reaction under certain conditions, thus optimizing conditions for desired outcomes in synthetic chemistry and industry.
Reaction Temperature
Reaction temperature is a fundamental parameter in the study of chemical kinetics and equilibrium. It influences reaction rates and equilibria shifts according to Le Chatelier's principle: increasing the temperature will favor the endothermic direction of a reaction.

In the context of the van 't Hoff equation, temperature (\(T\)) acts as a crucial variable when examining the dependency of the equilibrium constant (\(K\)) on temperature changes. The line described by plotting \(\ln K\) against \(\frac{1}{T}\) gives researchers the ability to discern the effect of temperature on the equilibrium constant. As temperature increases, the value of \(K\) can either increase or decrease based on the reaction's energy profile:
  • For exothermic reactions (\(\Delta H^{\circ} < 0\)), increasing temperature typically decreases \(K\)
  • For endothermic reactions (\(\Delta H^{\circ} > 0\)), increasing temperature generally increases \(K\)
This relationship highlights the balancing act between enthalpic and entropic contributions to a reaction's spontaneity. Understanding how reaction temperature shifts equilibrium positions is crucial for optimizing conditions in both experimental and industrial chemical processes.

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Most popular questions from this chapter

Using \(S^{\circ}\) values from Appendix C, calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{\circ} .\) (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. [Section 19.4\(]\)

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).

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{~kJ} /\) mol. (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00 \mathrm{~mol}\) of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8{ }^{\circ} \mathrm{C}\).

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