/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 Derive the van't Hoff equation. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 85

Derive the van't Hoff equation. $$\frac{d \ln K}{d T}=\frac{\Delta H^{\circ}}{R T^{2}}$$ which gives the dependence of the equilibrium constant on temperature." Here \(\Delta H^{\circ}\) is the enthalpy change of the reaction, for pure substances in their standard states (1 bar pressure for gases). Notice that if \(\Delta H^{\circ}\) is positive (loosely speaking, if the reaction requires the absorption of heat), then higher temperature makes the reaction tend more to the right, as you might expect. Often you can neglect the temperature dependence of \(\Delta H^{\circ}\); solve the equation in this case to obtain $$\ln K\left(T_{2}\right)-\ln K\left(T_{1}\right)=\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$$

Short Answer

Expert verified
The van't Hoff equation is derived by integrating under the assumption of constant enthalpy, reflecting \( K \)'s temperature dependence.

Step by step solution

01

Write down the Gibbs Free Energy equation

The Gibbs Free Energy change \( \Delta G^\circ \) for a reaction is expressed as \( \Delta G^\circ = -RT \ln K \), where \( R \) is the universal gas constant and \( K \) is the equilibrium constant.
02

Use the relation between Gibbs Free Energy and Enthalpy

The relationship between Gibbs Free Energy change, Enthalpy change \( \Delta H^\circ \), and Entropy change \( \Delta S^\circ \) is given by \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \).
03

Differentiate the Gibbs Free Energy equation with respect to temperature

Differentiate the equation \( \Delta G^\circ = -RT \ln K \) with respect to temperature, \( T \), to get \(-R \frac{d (\ln K)}{dT} = \frac{d (\Delta G^\circ)}{dT} \).
04

Express \( \frac{d(\Delta G^\circ)}{dT} \) using \( \Delta H^\circ \) and \( \Delta S^\circ \)

By differentiating \( \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \) with respect to \( T \), using that \( \Delta G^\circ = 0 \) at equilibrium, we find \( \frac{d (\Delta G^\circ)}{dT} = \Delta S^\circ - \frac{d (T \Delta S^\circ)}{dT} = 0 - (\Delta H^\circ/T^2) = -\Delta H^\circ/T^2 \).
05

Combine equations to obtain van't Hoff equation

Equating the two expressions \(-R \frac{d (\ln K)}{dT} = \frac{d (\Delta G^\circ)}{dT} \) and \(\frac{d (\Delta G^\circ)}{dT} = \frac{\Delta H^\circ}{T^2} \) leads to \( \frac{d \ln K}{dT} = \frac{\Delta H^\circ}{R T^2} \).
06

Integrate the van't Hoff equation

Assume \( \Delta H^\circ \) is temperature independent. Integrate \( \frac{d \ln K}{dT} = \frac{\Delta H^\circ}{R T^2} \) to find \( \ln K(T) = -\frac{\Delta H^\circ}{R} \frac{1}{T} + C \), where \( C \) is the integration constant.
07

Determine change in \( \ln K \) between two temperatures

Consider temperatures \( T_1 \) and \( T_2 \), with \( \ln K(T_1) = -\frac{\Delta H^\circ}{R T_1} + C \) and \( \ln K(T_2) = -\frac{\Delta H^\circ}{R T_2} + C \). Subtracting these gives \( \ln K(T_2) - \ln K(T_1) = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy (\( \Delta G \)) is a thermodynamic quantity that offers insight into the spontaneity of a process. It combines enthalpy and entropy into a single value, providing a comprehensive view of the chemical reaction under a specific temperature pressure. The formula is\[\Delta G = \Delta H - T \Delta S\]where \( \Delta H \) is the enthalpy change, \( \Delta S \) is the entropy change, and \( T \) is the temperature in Kelvin.
- **Spontaneity**: A reaction is spontaneous when \( \Delta G \) is negative, indicating it might happen naturally without external energy input. - **\( \Delta G \) at Equilibrium**: At equilibrium, \( \Delta G \) is zero, as no net change occurs in the reactant and product concentrations.
Understanding \( \Delta G \) is crucial for predicting whether a reaction will occur and how it can be influenced by changing conditions like temperature.
Equilibrium Constant
The equilibrium constant (\( K \)) plays a vital role in chemistry, representing the ratio of the concentrations of products to reactants at equilibrium. It's a snapshot of the balance point of a reaction.
- **Equilibrium Expression**: For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium expression is:\[K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}}\]This expression quantifies how the equilibrium shifts toward products or reactants. A large \( K \) (greater than 1) indicates a product-favored reaction, whereas a small \( K \) (less than 1) suggests a reactant-favored reaction.
The Van't Hoff equation reveals how \( K \) changes with temperature, showing the dynamic nature of equilibrium.
Enthalpy Change
Enthalpy Change (\( \Delta H \)) measures the total energy change of a system during a reaction at constant pressure. It signifies whether a reaction absorbs or releases heat.- **Exothermic Reactions**: These reactions release heat (\( \Delta H \lt 0 \)), making them feel warm. Common examples include combustion and many synthesis reactions.- **Endothermic Reactions**: These absorb heat (\( \Delta H \gt 0 \)), typically requiring an input of energy to proceed.
Enthalpy change is an essential factor in thermodynamics, as it influences the equilibrium constant via temperature changes. In the Van't Hoff equation, \( \Delta H \) helps to illustrate how temperature impacts reaction spontaneity.
In a nutshell, \( \Delta H \) unveils the thermal characteristics of a reaction, guiding predictions about heat requirements or releases, thus impacting the equilibrium and favorability of reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4},\) readily dissociates into \(\mathrm{H}^{+}\) and \(\mathrm{HSO}_{4}^{-}\) ions: $$\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{H}^{+}+\mathrm{HSO}_{4}^{-}$$ The hydrogen sulfate ion, in turn, can dissociate again: $$\mathrm{HSO}_{4}^{-} \longmapsto \mathrm{H}^{+}+\mathrm{SO}_{4}^{2-}$$ 'The equilibrium constants for these reactions, in aqueous solutions at \(298 \mathrm{K},\) are approximately \(10^{2}\) and \(10^{-1.9}\), respectively. (For dissociation of acids it is usually more convenient to look up \(K\) than \(\Delta G^{\circ} .\) By the way, the negative base- 10 logarithm of \(K\) for such a reaction is called \(\mathbf{p K},\) in analogy to pH. So for the first reaction \(\mathrm{pK}=-2,\) while for the second reaction \(\mathrm{pK}=1.9 .2\) (a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated? (b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically \(5 \times 10^{-5} \mathrm{mol} / \mathrm{kg}\). The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of \(\mathrm{SO}_{4}^{2-} .\) What is the pH of this rainwater? (c) Explain why you can neglect dissociation of water into \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO \(_{4}^{-}\) and \(\mathrm{SO}_{4}^{2-} ?\)

What happens when you add salt to the ice bath in an ice cream maker? How is it possible for the temperature to spontaneously drop below \(0^{\circ} \mathrm{C} ?\) Explain in as much detail as you can.

Everything in this section assumes that the total pressure of the system is fixed. How would you expect the nitrogen-oxygen phase diagram to change if you increase or decrease the pressure? Justify your answer.

A mixture of one part nitrogen and three parts hydrogen is heated, in the presence of a suitable catalyst, to a temperature of \(500^{\circ} \mathrm{C}\). What fraction of the nitrogen (atom for atom) is converted to ammonia, if the final total pressure is 400 atm? Pretend for simplicity that the gases behave ideally despite the very high pressure. The equilibrium constant at \(500^{\circ} \mathrm{C}\) is \(6.9 \times 10^{-5}\). (Hint: You'll have to solve a quadratic equation.)

Seawater has a salinity of 3.5\%, meaning that if you boil away a kilogram of seawater, when you're finished you'll have 35 g of solids (mostly \(\mathrm{NaCl}\) ) left in the pot. When dissolved, sodium chloride dissociates into separate \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions. (a) Calculate the osmotic pressure difference between seawater and fresh water. Assume for simplicity that all the dissolved salts in seawater are \(\mathrm{NaCl}\). (b) If you apply a pressure difference greater than the osmotic pressure to a solution separated from pure solvent by a semipermeable membrane, you get reverse osmosis: a flow of solvent out of the solution. This process can be used to desalinate seawater. Calculate the minimum work required to desalinate one liter of seawater. Discuss some reasons why the actual work required would be greater than the minimum.
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Translational Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Linear Momentum

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.