/*! 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 From the data given in the follo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 23

From the data given in the following table, determine \(\Delta S^{\circ} \quad\) for the reaction \(\quad \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) \longrightarrow\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) All data are at \(298 \mathrm{K}\) $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}, \mathrm{kJ} \mathrm{mol}^{-1} & \Delta G_{f,}^{\circ} \mathrm{kJ} \mathrm{mol}^{-1} \\ \hline \mathrm{NH}_{3}(\mathrm{g}) & -46.11 & -16.48 \\ \mathrm{HCl}(\mathrm{g}) & -92.31 & -95.30 \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & -314.4 & -202.9 \\ \hline \end{array}$$

Short Answer

Expert verified
\(\Delta S^{\circ} = -0.284 \ K^{-1} mol^{-1}\)

Step by step solution

01

Identify the relevant thermodynamic relation

The relationship between Gibbs Free Energy (\( \Delta G \)), Enthalpy (\( \Delta H \)), Temperature (T) and Entropy (\(\Delta S\)) is given by the equation: \[ \Delta G = \Delta H -T \Delta S \]Considering we are dealing with standard conditions (298 K), this can be adapted to:\[ \Delta G^{\circ} = \Delta H^{\circ} -T \Delta S^{\circ} \]To find \(\Delta S^{\circ}\), we rearrange the equation like so:\[ \Delta S^{\circ} = (\Delta H^{\circ} - \Delta G^{\circ}) / T \]
02

Calculate the change in enthalpy and Gibbs free energy of the reaction

Next, calculate the enthalpy (\(\Delta H_f^{\circ}\)) and Gibbs free energy (\(\Delta G_f^{\circ}\)) for the reaction. This can be done using the provided tables and the formulas:\[\Delta H^{\circ} = \sum \Delta H_f^{\circ} (products) - \sum \Delta H_f^{\circ} (reactants)\]\[\Delta G^{\circ} = \sum \Delta G_f^{\circ} (products) - \sum \Delta G_f^{\circ} (reactants)\]For our reaction:\[\Delta H^{\circ}= \Delta H_f^{\circ}(NH_{4}Cl) -(\Delta H_f^{\circ}(NH_{3}) + \Delta H_f^{\circ}(HCl)) = -314.4-(-46.11-92.31) = -175.98 \,kJ/mol \]\[\Delta G^{\circ}= \Delta G_f^{\circ}(NH_{4}Cl)- (\Delta G_f^{\circ}(NH_{3}) + \Delta G_f^{\circ}(HCl)) = -202.9 -(-16.48-95.3) = -91.12 \, kJ/mol \]
03

Determine the standard entropy change

Finally, calculate the standard entropy change (\(\Delta S^{\circ}\)) using the formula from Step 1:\[\Delta S^{\circ} = (\Delta H^{\circ} - \Delta G^{\circ}) / T = (-175.98 - (-91.12)) / 298 = -0.284 \ K^{-1} mol^{-1} \]Notice that the units of \(\Delta S^{\circ}\) are K^{-1}mol^{-1}, which are the proper units for entropy

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, denoted as \( \Delta G \), is a vital concept in thermodynamics that helps us understand the spontaneity of a reaction. It combines enthalpy, entropy, and temperature into a single value.
This concept is key because it predicts whether a reaction can occur without any external input. If \( \Delta G \) is negative, the reaction proceeds spontaneously.
  • Formula: \( \Delta G = \Delta H - T \Delta S \)
  • \( \Delta H \): Change in enthalpy (heat content)
  • \( T \): Absolute temperature in Kelvin
  • \( \Delta S \): Change in entropy (disorder)
In our example, substituting the values gives \( \Delta G^{\circ} = -91.12 \text{ kJ/mol} \).
This negative value indicates the formation of \( \text{NH}_4\text{Cl} \) from \( \text{NH}_3 \) and \( \text{HCl} \) is spontaneous.
Understanding Gibbs Free Energy can help you predict the feasibility of various chemical reactions in both industrial processes and natural phenomena.
Enthalpy
Enthalpy, represented as \( \Delta H \), is a measurement of heat content in a chemical system. It reflects the total energy change during a reaction, including heat absorbed or released.
Enthalpy change can be either positive (endothermic) or negative (exothermic).
  • Endothermic: Absorbs heat, \( \Delta H > 0 \)
  • Exothermic: Releases heat, \( \Delta H < 0 \)
For the reaction \( \text{NH}_3(\text{g}) + \text{HCl}(\text{g}) \rightarrow \text{NH}_4\text{Cl}(\text{s}) \), we calculated \( \Delta H^{\circ} = -175.98 \text{ kJ/mol} \).
This exothermic reaction releases heat, indicating that forming ammonium chloride from ammonia and hydrochloric acid is heat-releasing.
Enthalpy is crucial for understanding energy dynamics in reactions and helps in designing chemical processes where temperature control is essential.
Entropy
Entropy, denoted as \( \Delta S \), is a measure of disorder or randomness in a system. In chemistry, it helps us understand how energy disperses among molecules.
  • Increased entropy: Greater disorder, \( \Delta S > 0 \)
  • Decreased entropy: Greater order, \( \Delta S < 0 \)
For the given chemical reaction, we found \( \Delta S^{\circ} = -0.284 \text{ K}^{-1} \text{mol}^{-1} \).
This negative change suggests the system becomes more ordered as gases combine to form a solid (\( \text{NH}_4\text{Cl} \)).
Entropy is a fundamental concept because it affects the direction and feasibility of reactions. In combination with enthalpy, it helps chemists predict the behavior of chemical systems under different conditions.

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

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta S>0\) (c) \(\Delta G=\Delta H ;\) (d) \(\Delta G>0 ?\) Explain.

The standard molar entropy of solid hydrazine at its melting point of \(1.53^{\circ} \mathrm{C}\) is \(67.15 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(12.66 \mathrm{kJmol}^{-1} .\) For \(\mathrm{N}_{2} \mathrm{H}_{4}(1)\) in the interval from \(1.53^{\circ} \mathrm{C}\) to \(298.15 \mathrm{K}\), the molar heat capacity at constant pressure is given by the expression \(C_{p}=97.78+0.0586(T-280) .\) Determine the standard molar entropy of \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})\) at \(298.15 \mathrm{K}\). [Hint: The heat absorbed to produce an infinitesimal change in the temperature of a substance is \(d q_{\mathrm{rev}}=C_{p} d T\).

Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).

What must be the temperature if the following reaction has \(\Delta G^{\circ}=-45.5 \mathrm{kJ}, \Delta H^{\circ}=-24.8 \mathrm{kJ},\) and \(\Delta S^{\circ}=15.2 \mathrm{JK}^{-1} ?\) $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})$$
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

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.