Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 35

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P\) ? a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)

Short Answer

Expert verified
The spontaneous changes at constant \(T\) and \(P\) are: b. \( \Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K} \) c. \( \Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K} \) d. \( \Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K} \)

Step by step solution

01

Calculate Gibbs free energy change for choice a

For choice a: \( \Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 \mathrm{~K} \) First, convert \( \Delta H\) to J: \( \Delta H = 25,000 \mathrm{~J} \) Now, apply the Gibbs free energy equation: \( \Delta G = \Delta H - T \Delta S \) \( \Delta G = 25,000 - (300 \times 5) = 25,000 - 1,500 = 23,500 \mathrm{~J} \) Since \( \Delta G \) for choice a is positive, the process is not spontaneous.
02

Calculate Gibbs free energy change for choice b

For choice b: \( \Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K} \) First, convert \( \Delta H \) to J: \( \Delta H = 25,000 \mathrm{~J} \) Now, apply the Gibbs free energy equation: \( \Delta G = \Delta H - T \Delta S \) \( \Delta G = 25,000 - (300 \times 100) = 25,000 - 30,000 = -5,000 \mathrm{~J} \) Since \( \Delta G \) for choice b is negative, the process is spontaneous.
03

Calculate Gibbs free energy change for choice c

For choice c: \( \Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K} \) First, convert \( \Delta H \) to J: \( \Delta H = -10,000 \mathrm{~J} \) Now, apply the Gibbs free energy equation: \( \Delta G = \Delta H - T \Delta S \) \( \Delta G = -10,000 - (298 \times 5) = -10,000 - 1,490 = -11,490 \mathrm{~J} \) Since \( \Delta G \) for choice c is negative, the process is spontaneous.
04

Calculate Gibbs free energy change for choice d

For choice d: \( \Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K} \) First, convert \( \Delta H \) to J: \( \Delta H = -10,000 \mathrm{~J} \) Now, apply the Gibbs free energy equation: \( \Delta G = \Delta H - T \Delta S \) \( \Delta G = -10,000 - (200 \times -40) = -10,000 + 8,000 = -2,000 \mathrm{~J} \) Since \( \Delta G \) for choice d is negative, the process is spontaneous. The spontaneous changes at constant \( T \) and \( P \) are: b. \( \Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K} \) c. \( \Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K} \) d. \( \Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K} \)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) c. \(\Delta H=+18 \mathrm{~kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{~kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6\) \(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Calculate the value of \(\mathrm{K}\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

Using data from Appendix 4, calculate \(\Delta H^{\circ}, \Delta G^{\circ}\), and \(K(\) at \(298 \mathrm{~K}\) ) for the production of ozone from oxygen: $$ 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g) $$ At \(30 \mathrm{~km}\) above the surface of the earth, the temperature is about \(230 . \mathrm{K}\) and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{~km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)\) b. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\), using the following data: \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at 600. \(\mathrm{K}\) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at \(600 . \mathrm{K}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

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.

Sign-up for free