   Chapter 17, Problem 35E

Chapter
Section
Textbook Problem

# Given the values of ∆H and ∆S, which of the following changes will be spontaneous at constant T and p?a. ∆H = + 25 kJ, ∆S = + 5.0 J/K, T = 300. Kb. ∆H = + 25 kJ, ∆S = + 100. J/K, T = 300. Kc. ∆H = − 10. kJ, ∆S = + 5.0 J/K, T= 298 Kd. ∆H = − 10.kJ, ∆S =−40.J/K, T = 200.K

(a)

Interpretation Introduction

Interpretation: The spontaneous change in each case is to be predicted at constant temperature and pressure.

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is negative.

Explanation

Explanation

Given

The value of ΔH is 25kJ .

The value of ΔS is 5.0J/K .

Temperature is 300K .

The conversion of kilo-joule (kJ) into joule (J) is done as,

1kJ=103J

Hence,

The conversion of 25kJ into joule is,

25kJ=(25×103)J=25×103J

Formula

The formula of ΔG is,

ΔG=ΔHTΔS

Where,

• ΔH is the enthalpy of reaction

(b)

Interpretation Introduction

Interpretation: The spontaneous change in each case is to be predicted at constant temperature and pressure.

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is negative.

(c)

Interpretation Introduction

Interpretation: The spontaneous change in each case is to be predicted at constant temperature and pressure.

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is negative.

(d)

Interpretation Introduction

Interpretation: The spontaneous change in each case is to be predicted at constant temperature and pressure.

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is negative.

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 