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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

Using data from Appendix 4 calculate ∆H°,, and ∆ for the reaction

N 2 ( g ) + O 2 ( g ) 2NO ( g )

Why does NO form in an automobile engine but then does not readily decompose back to N2 and O2 in the atmosphere?

Interpretation Introduction

Interpretation: Reaction of commercial production of NO is given. The value of ΔH°,ΔG° and ΔS° is to be calculated for the given reaction. The explanation of the fact that NO formed in an automobile engine but does not readily decompose back to N2 and O2 in the atmosphere is to be stated.

Concept introduction: The expression to calculate ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

The expression to calculate ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

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

To determine: The value of ΔH°,ΔG° and ΔS° for the given reaction; the explanation of the fact that NO is formed in an automobile engine but does not readily decompose back to N2 and O2 in the atmosphere.

Explanation

Explanation

The value of ΔH° for the given reaction is 180kJ_ .

The stated reaction is,

N2(g)+O2(g)2NO(g)

Refer to Appendix 4 .

The value of ΔH°(kJ/mol) for the given reactant and product is,

Molecules ΔH°(kJ/mol)
NO(g) 90
N2(g) 0
O2(g) 0

The formula of ΔH° is,

ΔH°=npΔH°(product)nfΔH°(reactant)

Where,

  • ΔH° the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant.
  • ΔH°(product) is the standard enthalpy of product at a pressure of 1atm .
  • ΔH°(reactant) is the standard enthalpy of reactant at a pressure of 1atm .

Substitute all values from the table in the above equation.

ΔH°=npΔH°(product)nfΔH°(reactant)=[2(90){(0)+(0)}]kJ=180kJ_

The value of ΔS° for the given reaction is 25J/mol_ .

Refer to Appendix 4

The stated reaction is,

N2(g)+O2(g)2NO(g)

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

Molecules ΔS°(J/Kmol)
NO(g) 211
N2(g) 192
O2(g) 205

The formula of ΔS° is,

ΔS°=npΔS°(product)nfΔS°(reactant)

Where,

  • ΔS° is the standard enthalpy of reaction.
  • np is the number of moles of each product.
  • nr is the number of moles each reactant

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