Introduction To General, Organic, And Biochemistry
Introduction To General, Organic, And Biochemistry
12th Edition
ISBN: 9781337571357
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
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Chapter 5, Problem 89P
Interpretation Introduction

(a)

Interpretation:

The balanced chemical equation for decomposition of NH4NO3 should be determined.

Concept Introduction:

In a balanced chemical equation, all the constituents present in the reaction have equal number of atoms on both side of the reaction arrow.

Expert Solution
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Answer to Problem 89P

NH4NO3(s)  N2(g) + 2H2(g).

Explanation of Solution

Decommission of solid Ammonium Nitrate generates gaseous dinitrogen oxide along with the water vapour.

The balanced chemical equation of the reaction taking place is as depicted below:

NH4NO3(s)  N2(g) + 2H2(g)

From the above balanced reaction, for each mole of NH4 NO3 produces one mole of di-nitrogen oxide and 2 moles of water.

Interpretation Introduction

(b)

Interpretation:

The partial pressure of N2(g) and H2O(g) produced should be determined.

Concept Introduction:

To calculate the partial pressure of N2(g) and H2O(g) produced ideal gas equation is employed.

P=nRTV

Where,

R = Universal gas constant (0.08206 L atm K-1mol-1)T = Temperature (K)P = Pressure (atgm)n = Number of the molesV = Volume (l).

Expert Solution
Check Mark

Answer to Problem 89P

Partial pressure of N2(g) is 0.737 atm. Partial pressure of water is 1.47 atm.

Explanation of Solution

Calculate the number of moles of N2(g) as follows:

0.0312 mol NH4NO3× 2 mol N2OI mol NH4NO3 = 0.0312 mol N2O

Therefore, number of moles of N2(g) is 0.0312 mol N2 O.

Calculate the number of moles of water is as follows:

0.0312 mol NH4NO3× 2 mol H2OI mol NH4NO3 = 0.0624 mol H2O

Therefore, number of moles of water is 0.0624 mol H2 O.

From the ideal equation we have.

PV=nRT

Volume of tank (V) =1.75LNumber of moles of dinitrogen oxygen (n) = 0.0312 molesTemperature = 503 K.

To calculate the partial pressure of N2(g) substitute all the known value in the equation.

P=nRTV=(0.0312mol)(0.08206LatmKmol)(503K)(1.75L)=0.737atm

Therefore, the partial pressure of N2(g) = 0.737 atm.

Volume of tank v = 1.75 L.

Number of moles of water n = 0.0624 moles.

The temperature t = 503 K.

To calculate the partial pressure of water, substitute all the known value in the equation.

P=nRTV

=(0.0624mOl)(0.0821latmKmOl)(503K)(1.75l)

=1.47 atm

Therefore, partial pressure of water is 1.47 atm.

Interpretation Introduction

(c)

Interpretation:

The total gas pressure present in the flask at 2300 C should be determined.

Concept Introduction:

Dalton's law of partial pressure is state that the total pressure of a mixture gases is sum of the pressures that every gas would exert if it were present alone.

Expert Solution
Check Mark

Answer to Problem 89P

Total pressure present in flask is 2.21 atm.

Explanation of Solution

In the provided reaction the gases molecules are N2(g) and water vapor. Thus, total pressure present in the flask is the sum of the partial pressures of N2(g) and water vapor.

Ptotal= PN2O+ PH2O = 0.737 atm +1.47 atm= 2.21 atm

Therefore, total pressure present in flask is 2.21 atm.

Interpretation Introduction

(d)

Interpretation:

The three equivalent resonance structure for N2(g) should be drawn.

Concept Introduction:

The resonance structures show the arrangement of electrons and bonds in a molecule. The lone pair present on atom can show delocalization with pi electrons of double or triple bonds resulting formation of resonance structures. The position of atoms remains the same only position of bonds changes.

Expert Solution
Check Mark

Answer to Problem 89P

Introduction To General, Organic, And Biochemistry, Chapter 5, Problem 89P , additional homework tip  1

Explanation of Solution

The resonance structures of N2 O is shown in the following diagram.

Since, nitrogen has 3 valence electrons it can form three covalent bonds with other atoms. There are two valence electrons in oxygen thus, it can form one double or two single bonds with other atoms. Being more electronegative in nature, oxygen atom will be placed at the terminal position. Thus, there will be one double bond between two nitrogen atom and one double bond between nitrogen and oxygen atom resulting negative charge on one nitrogen atom and positive charge on other nitrogen atom.

The negative charge can delocalize with pi electrons of double bond resulting two resonance forms.

Introduction To General, Organic, And Biochemistry, Chapter 5, Problem 89P , additional homework tip  2

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Chapter 5 Solutions

Introduction To General, Organic, And Biochemistry

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