# Calculate Δ S °(universe) for the following reaction at 25.0 °C: C(graphite) + O 2 (g) → CO 2 (g) (a) −1317 J/K · mol-rxn (b) 3.1 J/K · mol-rxn (c) 4.4 J/K · mol-rxn (d) 1320 J/K · mol-rxn

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

Chapter
Section

### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18.5, Problem 2RC
Textbook Problem
1 views

## Calculate ΔS°(universe) for the following reaction at 25.0 °C:C(graphite) + O2(g) → CO2(g) (a) −1317 J/K · mol-rxn (b) 3.1 J/K · mol-rxn (c) 4.4 J/K · mol-rxn (d) 1320 J/K · mol-rxn

Interpretation Introduction

Interpretation: the value of ΔrS for the given reaction is to be identified from the given options.

Concept introduction:

In Spontaneous process, energy transfer from higher concentration to lower, or the energy is dispersed.

Entropy (S) is the quantity to measure the dispersal of energy in the spontaneous processes.

The equation for finding ΔS(system) is,

ΔrS=nS(products)nS(reactants)

ΔrH(system) is the change in enthalpy is calculated by sum of the product of stoichiometric coefficients and molar enthalpies of products minus sum of the product of stoichiometric coefficients and molar enthalpies of reactants. That is,

ΔrH(system)=nΔfH(products)nΔfH(reactants)

Equation for ΔS(surrounding) is,

ΔS(surrounding)=ΔfH(system)/T

Equation for ΔS(universe) ,

ΔS(universe)=ΔS(system)+ΔS(surrounding)

### Explanation of Solution

For the reaction,

Coefficient of product CO2 is 1 and its molar entropy is 213.74J/Kmol .

Coefficients of reactants C(graphite) and O2 are 1 and 1 respectively and their molar entropy are 5.7J/Kmol and 205.07J/Kmol respectively.

The equation for finding ΔS(system) is,

ΔS(system)=nS(products)nS(reactants)

Therefore,

The change in entropy is,

ΔS(system)=1×213.74J/Kmol(1×5.7J/Kmol+1×205.07J/Kmol)=2.9J/Kmol

ΔrH(system) is the change in enthalpy is calculated by sum of the product of stoichiometric coefficients and molar enthalpies of products minus sum of the product of stoichiometric coefficients and molar enthalpies of reactants. That is,

ΔfHΔrH(system)=nΔfH(products)nΔfH(reactants)

Therefore,

ΔrH(system)=1×(393.5kJ/mol)(1×0kJ/mol+1×0kJ/mol) =393

### 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
What are sister chromatids?

Human Heredity: Principles and Issues (MindTap Course List)

What is a gene?

Biology (MindTap Course List)

Is there a silicon dioxide compensation depth?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

Which of the foil owing is most commonly deficient in adolescents? folate zinc iron vitamin D

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Newton's law of universal gravitation is .represented by where F is the magnitude of the gravitational force ex...

Physics for Scientists and Engineers, Technology Update (No access codes included)