   Chapter 17, Problem 48E

Chapter
Section
Textbook Problem

# For the reaction CS 2 ( g )   +   3 O 2 ( g )   →  CO 2 ( g )   +   2 SO 2 ( g ) ∆S° is equal to −143 J/K. Use this value and data from Appendix 4 to calculate the value of S° for CS2(g).

Interpretation Introduction

Interpretation: The reaction between CS2(g) , 3O2(g) and the value of ΔS for this reaction is given. The value of S is to be calculated for CS2(g) .

Concept introduction: Entropy (ΔS) is the measure of degree of disorder or randomness.

The expression for ΔS is,

ΔS=npS(product)nfS(reactant)

Explanation

Explanation

The stated reaction is,

CS2(g)+3O2(g)CO2(g)+2SO2(g)

The value of ΔS for this reaction is 143J/K .

Refer to Appendix 4 .

The value of S(J/Kmol) for the given reactant and product is,

 Molecules S∘(J/K⋅mol) CO2(g) 214 SO2(g) 248 O2(g) 205

The formula of ΔS is,

ΔS=npS(product)nfS(reactant)

Where,

• ΔS is the standard entropy of reaction.
• np is the number of moles of each product.
• nr is the number of moles each reactant.
• S(product) is the standard entropy of product at a pressure of 1atm .
• S(reactant) is the standard entropy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation

### 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 