/*! 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 117 Which of the following condition... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Which of the following conditions may lead to a nonspontaneous change? (a) \(\Delta \mathrm{H}=-\mathrm{ve} ; \Delta \mathrm{S}=+\mathrm{ve}\) (b) \(\Delta \mathrm{H}=-\mathrm{ve} ; \Delta \mathrm{S}=-\mathrm{ve}\) (c) \(\Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) are both \(+\mathrm{ve}\) (d) \(\Delta \mathrm{H}=+\mathrm{ve} ; \Delta \mathrm{S}=-\mathrm{ve}\)

Short Answer

Expert verified
Condition (d) leads to a nonspontaneous change.

Step by step solution

01

Understand the Parameters

First, to determine the spontaneity of a reaction, we need to see how the Gibbs free energy change (G) is affected by the enthalpy change (H) and the entropy change (S). The formula is given by: \[\Delta G = \Delta H - T \Delta S\]where \(T\) is the temperature in Kelvin.
02

Analyzing Condition (a)

For condition (a):- \(\Delta H = -\text{ve}\) and \(\Delta S = +\text{ve}\) - Substituting these into the formula results in a negative \(\Delta G\), which typically means the process is spontaneous at all temperatures.
03

Analyzing Condition (b)

For condition (b):- \(\Delta H = -\text{ve}\) and \(\Delta S = -\text{ve}\) - Here, \(\Delta G = \Delta H - T \Delta S\) can be negative or positive depending on the size of \(T) \Delta S\) versus \(\Delta H\). It may be nonspontaneous at higher temperatures when \(T \Delta S\) is larger than \(\Delta H\).
04

Analyzing Condition (c)

For condition (c):- \(\Delta H = +\text{ve}\) and \(\Delta S = +\text{ve}\)- \(\Delta G\) can be negative when \(T \Delta S\) exceeds \(\Delta H\), typically spontaneous at high temperatures, but nonspontaneous at low temperatures.
05

Analyzing Condition (d)

For condition (d):- \(\Delta H = +\text{ve}\) and \(\Delta S = -\text{ve}\)- In this case, \(\Delta G\) will be positive as both terms contribute to a positive value, making it nonspontaneous at all temperatures.
06

Conclusion

The conditions in (d) inherently lead to a nonspontaneous change since both \(\Delta H\) is positive and \(\Delta S\) is negative, ensuring a positive \(\Delta G\) under all temperature conditions.

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 91Ó°ÊÓ!

Key Concepts

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

Enthalpy Change
What is enthalpy change, and why is it important? Enthalpy change, symbolized as \( \Delta H \), refers to the heat absorbed or released during a chemical reaction at constant pressure. It's crucial for understanding how energy moves in and out of a system.

Key points to consider:
  • A negative \( \Delta H \) (exothermic) indicates heat is released, which can favor spontaneity.
  • A positive \( \Delta H \) (endothermic) means heat is absorbed, often opposing spontaneity.
Enthalpy change helps predict whether a reaction might proceed on its own or needs external energy. Think of it as understanding the energy exchange within your cooking pot. When you heat water, you supply energy (positive \( \Delta H \)), and when it cools, the energy goes out (negative \( \Delta H \)).

Understanding how \( \Delta H \) plays a role in Gibbs Free Energy equations helps determine the favorability of reactions. It works alongside entropy in the formula \( \Delta G = \Delta H - T \Delta S \).
Entropy Change
Entropy, symbolized as \( \Delta S \), measures the disorder or randomness in a system. Imagine your room. A messy room with clothes everywhere has high entropy, while a tidy room has low entropy.

Here's why it matters:
  • If \( \Delta S \) is positive, the system becomes more disordered, often favoring spontaneity.
  • If \( \Delta S \) is negative, the system becomes more ordered, which may not favor spontaneity.
In chemical reactions, entropy change tells us about molecular freedom during the process. More disorder (positive \( \Delta S \)) usually means more possible microstates and higher probability of spontaneous reactions.

Entropy interacts with enthalpy in the Gibbs Free Energy equation to identify whether a reaction will happen by itself or not. It shows how changes in system organization affect the reaction's favorability.
Spontaneity of Reactions
Spontaneity of reactions refers to a reaction's ability to occur without external intervention. The Gibbs Free Energy equation \( \Delta G = \Delta H - T \Delta S \) helps determine this.

Consider these:
  • A negative \( \Delta G \) means the reaction is spontaneous, proceeding on its own.
  • A positive \( \Delta G \) indicates nonspontaneous reactions, needing energy input.
Example: Melting ice is spontaneous at room temperature because the system gains disorder (positive \( \Delta S \)) and the surroundings provide necessary heat despite the enthalpy change. Considerations like temperature can sway the direction of \( \Delta G \), emphasizing the critical role both \( \Delta H \) and \( \Delta S \) play.

It's a balancing act: enthalpy and entropy fight for favorability. This dynamic relationship elucidates why certain reactions spontaneously occur while others don't under specific 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

The enthalpy changes for the following processes are listed below. \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{C} 1(\mathrm{~g}) ; 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~g})=21(\mathrm{~g}) ; 151.0 \mathrm{kJmol}^{-1}\) \(\mathrm{ICl}(\mathrm{g})=\mathrm{I}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) ; 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\mathrm{I}_{2}(\mathrm{~s})=\mathrm{I}_{2}(\mathrm{~g}) ; 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Given that the standard states for iodine and chlorine are \(\mathrm{I}_{2}(\mathrm{~s})\) and \(\mathrm{Cl},(\mathrm{g})\), the standard enthalpy of formation for \(\mathrm{ICl}(\mathrm{g})\) is [2006] (a) \(-14.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+244.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

\(\Delta \mathrm{G}^{\circ}\) for the reaction, \(\mathrm{x}+\mathrm{y} \rightleftharpoons \mathrm{z}\) is \(-4.606 \mathrm{kcal}\). The value of equilibrium constant of the reaction at \(227^{\circ} \mathrm{C}\) is (a) \(0.01\) (b 100 (c) 2 (d) 10

When 1 mole gas is heated at constant volume tem perature is raised from 298 to \(308 \mathrm{~K}\). Heat supplied to the gas is \(500 \mathrm{~J}\). Then which of the following state ments is correct? (a) \(\mathrm{Q}=\mathrm{W}=500 \mathrm{~J}, \Delta \mathrm{U}=0\) (b) \(\mathrm{Q}=\Delta \mathrm{U}=500 \mathrm{~J}, \mathrm{~W}=0\) (c) \(Q=W=500 \mathrm{~J}, \Delta U=0\) (d) \(\Delta \mathrm{U}=0, \mathrm{Q}=\mathrm{W}=-500 \mathrm{~J}\)

2 moles of an ideal gas is expanded isothermally and reversibly from 1 litre of 10 litre at \(300 \mathrm{~K}\). The enthalpy change (in \(\mathrm{kJ}\) ) for the process is (a) \(11.4 \mathrm{~kJ}\) (b) \(-11.4 \mathrm{~kJ}\) (c) \(0 \mathrm{~kJ}\) (d) \(4.8 \mathrm{~kJ}\).

If the standard entropies of \(\mathrm{CH}_{4}, \mathrm{O}_{2}, \mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(186.2,205.3,213.6\) and \(69.96 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) respectively, then standard entropy change for the reaction \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}\) (1) is (a) \(-215.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (b) \(-243.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (c) \(-130.5 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(-85.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.