/*! 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 23 List three different ways to cal... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 23

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 different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), at 25掳C are: 1. Using the Gibbs-Helmholtz equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\) 2. Using the equilibrium constant: \(\Delta G^{\circ} = -RT \ln(K)\) 3. Using data tables and the equation: \(\Delta G^{\circ} = \sum \Delta G_f^{\circ}(\text{products}) - \sum \Delta G_f^{\circ}(\text{reactants})\) To estimate \(\Delta G^{\circ}\) at other temperatures, we can use the van't Hoff equation: \[\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Assumptions made while estimating \(\Delta G^{\circ}\) at a different temperature include: 1. The reaction is assumed to be at equilibrium. 2. The standard free energy change, enthalpy change, and entropy change are assumed to be constant over the temperature range being considered. 3. The van't Hoff equation assumes that the change in standard free energy and enthalpy with temperature is negligible.

Step by step solution

01

Method 1: Using Gibbs-Helmholtz equation

The standard free energy change, \(\Delta G^{\circ}\), can be calculated using the Gibbs-Helmholtz equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\] Here, we need to first calculate the standard enthalpy change, \(\Delta H^{\circ}\), and the standard entropy change, \(\Delta S^{\circ}\) of the reaction, then plug them into this equation to calculate the standard free energy change at a given temperature, T (in Kelvin).
02

Method 2: Using equilibrium constant

We can also calculate \(\Delta G^{\circ}\) using the relationship between the equilibrium constant, K, and the standard free energy change: \[\Delta G^{\circ} = -RT \ln(K)\] Here, R is the ideal gas constant (8.314 J/(mol路K)), and T is the temperature (in Kelvin). To use this method, we need to first determine the equilibrium constant, K, for the reaction at the desired temperature, and then plug it into this equation to calculate \(\Delta G^{\circ}\).
03

Method 3: Using the data tables

Another way to calculate \(\Delta G^{\circ}\) is by using the data tables that contain the standard free energy of formation, \(\Delta G_f^{\circ}\), for various substances. The standard free energy change for a reaction can be calculated using the equation: \[\Delta G^{\circ} = \sum \Delta G_f^{\circ}(\text{products}) - \sum \Delta G_f^{\circ}(\text{reactants})\] In this method, we need to look up the standard free energy of formation for each substance involved in the reaction and then use this equation to calculate \(\Delta G^{\circ}\).
04

Estimating \(\Delta G^{\circ}\) at other temperatures

To estimate the standard free energy change at a different temperature, we can use the following van't Hoff equation: \[\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\] Here, K1 and K2 are the equilibrium constants at the initial (T1) and final (T2) temperatures, respectively. First, we need to determine \(\Delta H^{\circ}\) and K1 for the reaction at the initial temperature using one of the methods mentioned above. Then, we can use this equation to calculate K2 at the desired temperature T2. Finally, we can use the relationship between K and \(\Delta G^{\circ}\) mentioned in Method 2 to calculate the standard free energy change at the new temperature.
05

Assumptions made

There are a few assumptions made while estimating \(\Delta G^{\circ}\) at a different temperature: 1. The reaction is assumed to be at equilibrium. 2. The standard free energy change, enthalpy change, and entropy change are assumed to be constant over the temperature range being considered. 3. The van't Hoff equation assumes that the change in standard free energy and enthalpy with temperature is negligible.

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-Helmholtz Equation
The Gibbs-Helmholtz equation is a fundamental tool in thermodynamics that allows us to calculate the standard free energy change (\( \Delta G^{\circ} \)) for a chemical reaction. This equation is expressed as:
\[ \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \]
where \( \Delta H^{\circ} \) is the standard enthalpy change, \( \Delta S^{\circ} \) is the standard entropy change, and \( T \) is the temperature in Kelvin.
This equation shows that the free energy change of a reaction is influenced by both the enthalpy and entropy changes.
  • Positive \( \Delta G^{\circ} \): This indicates a non-spontaneous reaction under standard conditions.
  • Negative \( \Delta G^{\circ} \): This means the reaction is spontaneous under standard conditions.
To use the Gibbs-Helmholtz equation, you need to determine \( \Delta H^{\circ} \) and \( \Delta S^{\circ} \) from experimental data or tables and substitute them into the equation along with the temperature in Kelvin. Understanding this equation helps predict whether a reaction will occur on its own or if it requires external energy input.
Equilibrium Constant
The equilibrium constant (\( K \)) is a number that gives insight into the position of balance in a reversible chemical reaction at a given temperature. It reflects the ratio of product concentrations to reactant concentrations when the system is at equilibrium:
  • \( K > 1 \): Products are favored at equilibrium.
  • \( K < 1 \): Reactants are favored at equilibrium.
\( \Delta G^{\circ} \) is related to the equilibrium constant through the relationship:
\[ \Delta G^{\circ} = -RT \ln(K) \]
Here, \( R \) is the ideal gas constant (8.314 J/(mol路K)), and \( T \) is the temperature in Kelvin. This equation indicates how a change in \( \Delta G^{\circ} \) affects \( K \):
  • If \( \Delta G^{\circ} \) is negative, \( K \) is greater than 1, suggesting a favorable formation of products.
  • If \( \Delta G^{\circ} \) is positive, \( K \) is less than 1, implying a reaction that does not favor product formation under standard conditions.
Understanding this relationship helps predict how a reaction will behave in equilibrium and the extent to which reactants are converted to products.
Free Energy of Formation
The free energy of formation, given by \( \Delta G_f^{\circ} \), is a measure of the energy change when one mole of a compound is formed from its constituent elements in their standard states.
This value provides crucial information for calculating the standard free energy change of a reaction using:
\[ \Delta G^{\circ} = \sum \Delta G_f^{\circ}(\text{products}) - \sum \Delta G_f^{\circ}(\text{reactants}) \]
By using tabulated values for \( \Delta G_f^{\circ} \), you can easily find \( \Delta G^{\circ} \) for reactions without direct experimentation.
  • Spontaneity: Negative \( \Delta G^{\circ} \) indicates spontaneous reactions.
  • Positive \( \Delta G^{\circ} \): Reactions are non-spontaneous and likely need energy input.
Understanding free energy of formation helps chemists and engineers design viable processes by predicting which reactions are thermodynamically favorably based solely on starting materials and products.
Temperature Dependence
The temperature dependence of a reaction determines how temperature variations influence the thermodynamic properties like free energy, entropy, and enthalpy during a reaction.
Standard free energy change (\( \Delta G^{\circ} \)) is particularly affected by temperature, as dictated by the Gibbs-Helmholtz equation.Unlike \( \Delta H^{\circ} \) and \( \Delta S^{\circ} \), which are often approximated to be constant over small temperature ranges, \( \Delta G^{\circ} \) has a more pronounced temperature dependency.
  • For exothermic reactions (\( \Delta H^{\circ} < 0 \)), increasing temperature can make \( \Delta G^{\circ} \) less negative, potentially reducing spontaneity.
  • For endothermic reactions (\( \Delta H^{\circ} > 0 \)), higher temperatures may make \( \Delta G^{\circ} \) more negative, increasing spontaneity.
Assessing temperature effects helps predict the feasibility of reactions under non-standard conditions, such as industrial temperatures, and guides adjustments needed to maintain desired reaction pathways.
Van't Hoff Equation
The Van't Hoff equation is pivotal in understanding how temperature affects the equilibrium constant (\( K \)) of a reaction. Given the relationship:
\[ \ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
this equation allows us to estimate \( K \) at a different temperature given a known \( K \) at another temperature and the standard enthalpy change \( \Delta H^{\circ} \).To use the Van't Hoff equation:
  • Calculate or obtain \( \Delta H^{\circ} \) from sources or experiments.
  • Use the known \( K_1 \) for an initial temperature \( T_1 \).
  • Substitute \( \Delta H^{\circ} \), \( T_1 \), and the desired \( T_2 \) into the equation to find \( K_2 \).
This is crucial for predicting how changes in temperature will shift the equilibrium of a reaction, which is essential for processes like chemical manufacturing and environmental assessments.

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

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C},\) the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

At \(100 .^{\circ} \mathrm{C}\) and \(1.00 \mathrm{atm}, \Delta H^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol}\) for the vaporization of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .^{\circ} \mathrm{C}\) and \(110 .^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) do not depend on temperature.

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point \(\left(78^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}},\) and \(\Delta S_{\mathrm{univ}}\) when 1.00 mole of ethanol is vaporized at \(78^{\circ} \mathrm{C}\) and 1.00 atm.

For mercury, the enthalpy of vaporization is 58.51 kJ/mol and the entropy of vaporization is 92.92 J/K ? mol. What is the normal boiling point of mercury?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

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.