   Chapter 17, Problem 36E

Chapter
Section
Textbook Problem

# At what temperatures will the following processes be spontaneous?a. ∆H = −18 kJ and ∆S = − 60. J/Kb. ∆H = +18 kJ and ∆S = + 60. J/Kc. ∆H = +18 kJ and ∆S = − 60. J/Kd. ∆H = −18 kJ and ∆S = +60. J/K

(a)

Interpretation Introduction

Interpretation:

The temperature at which the process is spontaneous is to be predicted for given options.

Concept introduction:

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is less than zero.

Explanation

Explanation

Given

The value of ΔH is 18kJ .

The value of ΔS is 60J/K .

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

1J=103kJ

Hence,

The conversion of 60J/K into joule is,

60J=(60×103)kJ=0.060kJ

Formula

The formula of ΔG is,

ΔG=ΔHTΔS

Where,

• ΔH is the enthalpy of reaction

(b)

Interpretation Introduction

Interpretation:

The temperature at which the process is spontaneous is to be predicted for given options.

Concept introduction:

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is less than zero.

(c)

Interpretation Introduction

Interpretation:

The temperature at which the process is spontaneous is to be predicted for given options.

Concept introduction:

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is less than zero.

(d)

Interpretation Introduction

Interpretation:

The temperature at which the process is spontaneous is to be predicted for given options.

Concept introduction:

ΔG=ΔHTΔS

A reaction is said to be spontaneous if the value of ΔG is less than zero.

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