General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 15, Problem 15.101QP

(a)

Interpretation Introduction

Interpretation:

The mole fraction of the gases has to be calculated using the data.

Concept Introduction:

Mole fraction calculation:

Molefractionofsolutexinthemixtureofxandy=nxnx+ny

(a)

Expert Solution
Check Mark

Explanation of Solution

Given data:

  Molarmassofthereactingmixture=70.6g/mol

The reaction is given below:

  N2O4(g)2NO2(g)

The sum of the mole fractions in a mixture must equal one.

  XN2O4+XNO2=1XN2O4=1-XNO2

It is known that the average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.

(XNO2×46.01g)+(XN2O4×92.02g)=70.6g(XNO2×46.01g)+([1-XNO2]×92.02g)=70.6g(XNO2×46.01g)+92.02g-92.02g×XNO2=70.6g46.01gXNO2-92.02g×XNO2=70.6g-92.02g-46.01XNO2=-21.42XNO2=0.466XN2O4=1-XNO2=1-0.466=0.534

(b)

Interpretation Introduction

Interpretation:

The Kp for the reaction if the total pressure was 1.2atm has to be given.

Concept Introduction

  • Equilibrium constant at constant pressure:Kp

It is used to express the relationship between product pressures and reactant pressures.

  For a general reaction, aA+bBcC+dD

EquilibriumconstantKp=PCcPDdPAaPBb

  • Kp=Kc(RT)ΔnΔn-changeinnumberofmolesR-GasconstantT-Temperature

(b)

Expert Solution
Check Mark

Explanation of Solution

Given data:

  Ptotal=1.2atm

The reaction is given below:

  N2O4(g)2NO2(g)

Ptotal=PN2O4+PNO2=1.2atm

The partial pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

  PNO2=XNO2PtotalwhereXNO2isthemolefractionPN2O4=XN2O4PtotalwhereXN2O4isthemolefraction

PNO2=XNO2Ptotal=(0.466)(1.2atm)=0.56atm

PN2O4=XN2O4Ptotal=(0.534)(1.2atm)=0.64atm

  KP=P2NO2PN2O4=(0.56)20.64=0.49

(c)

Interpretation Introduction

Interpretation:

The mole fractions if the pressure were increased to 4.0atm by reducing the volume t the same temperature has to be determined.

Concept Introduction

The partial pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

  PA=XAPtotalwhereXAisthemolefractionofthecomponentA

(c)

Expert Solution
Check Mark

Explanation of Solution

Given data:

  Ptotal=4.0atm

The reaction is given below:

  N2O4(g)2NO2(g)

The partial pressure of a gas in a mixture is the product of the mole fraction of the gas and the total pressure.

PNO2=XNO2PtotalwhereXNO2isthemolefractionPN2O4=XN2O4PtotalwhereXN2O4isthemolefraction

The sum of the mole fractions will be equal to one.

  XNO2+XN2O4=1XN2O4=1-XNO2

  PN2O4=XN2O4Ptotal=(1-XNO2)Ptotal

Substituting the values in the equilibrium constant expression:

  KP=P2NO2PN2O4=(XNO2PT)2(1-XNO2)PT0.487=(XNO24.0)2(1-XNO2)4.00.487=16X2NO24.04.0XNO21.951.95XNO2=16X2NO20=16X2NO2+1.95XNO21.95

The value of XNO2 will be obtained by solving the quadratic equation.

XNO2=0.29XN2O4=1-XNO2=1-0.29=0.71

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Chapter 15 Solutions

General Chemistry

Ch. 15.4 - Practice Exercise At 430°C. the equilibrium...Ch. 15.4 - Prob. 2PECh. 15.4 - Prob. 1RCCh. 15.4 - Prob. 2RCCh. 15.4 - Prob. 3PECh. 15 - Prob. 15.1QPCh. 15 - 15.2 Explain the difference between physical...Ch. 15 - 15.3 Briefly describe the importance of...Ch. 15 - 15.4 Consider the equilibrium system 3A ⇌ B....Ch. 15 - 15.5 Define homogeneous equilibrium and...Ch. 15 - Prob. 15.6QPCh. 15 - 15.7 Write equilibrium constant expressions for...Ch. 15 - 15.8 Write the expressions for the equilibrium...Ch. 15 - 15.9 Write the equilibrium constant expressions...Ch. 15 - 15.10 Write the equation relating Kc and KP and...Ch. 15 - Prob. 15.11QPCh. 15 - Prob. 15.12QPCh. 15 - 15.13 The equilibrium constant (Kc) for the...Ch. 15 - Prob. 15.14QPCh. 15 - 15.15 What is the KP at 1273°C for the...Ch. 15 - 15.16 The equilibrium constant KP for the...Ch. 15 - Prob. 15.17QPCh. 15 - Prob. 15.18QPCh. 15 - Prob. 15.19QPCh. 15 - Prob. 15.20QPCh. 15 - Prob. 15.21QPCh. 15 - 15.22 Ammonium carbamate, NH4CO2NH2, decomposes...Ch. 15 - Prob. 15.23QPCh. 15 - 15.24 Pure phosgene gas (COCl2), 3.00 × 10−2...Ch. 15 - Prob. 15.25QPCh. 15 - 15.26 A 2.50-mol quantity of NOCl was initially...Ch. 15 - 15.27 Define reaction quotient. How does it...Ch. 15 - Prob. 15.28QPCh. 15 - Prob. 15.29QPCh. 15 - Prob. 15.30QPCh. 15 - Prob. 15.31QPCh. 15 - Prob. 15.32QPCh. 15 - Prob. 15.33QPCh. 15 - Prob. 15.34QPCh. 15 - Prob. 15.35QPCh. 15 - Prob. 15.36QPCh. 15 - Prob. 15.37QPCh. 15 - Prob. 15.38QPCh. 15 - Prob. 15.39QPCh. 15 - Prob. 15.40QPCh. 15 - Prob. 15.41QPCh. 15 - Prob. 15.42QPCh. 15 - Prob. 15.43QPCh. 15 - Prob. 15.44QPCh. 15 - Prob. 15.45QPCh. 15 - 15.46 What effect does an increase in pressure...Ch. 15 - Prob. 15.47QPCh. 15 - Prob. 15.48QPCh. 15 - 15.49 Consider the reaction Comment on the...Ch. 15 - Prob. 15.50QPCh. 15 - Prob. 15.51QPCh. 15 - Prob. 15.53QPCh. 15 - Prob. 15.54QPCh. 15 - Prob. 15.55QPCh. 15 - Prob. 15.56QPCh. 15 - Prob. 15.57QPCh. 15 - 15.58 Baking soda (sodium bicarbonate) undergoes...Ch. 15 - 15.59 Consider the following reaction at...Ch. 15 - Prob. 15.60QPCh. 15 - Prob. 15.61QPCh. 15 - Prob. 15.62QPCh. 15 - Prob. 15.64QPCh. 15 - Prob. 15.65QPCh. 15 - Prob. 15.66QPCh. 15 - Prob. 15.67QPCh. 15 - Prob. 15.68QPCh. 15 - Prob. 15.69QPCh. 15 - Prob. 15.70QPCh. 15 - Prob. 15.71QPCh. 15 - Prob. 15.72QPCh. 15 - Prob. 15.73QPCh. 15 - Prob. 15.74QPCh. 15 - Prob. 15.75QPCh. 15 - Prob. 15.76QPCh. 15 - Prob. 15.78QPCh. 15 - Prob. 15.79QPCh. 15 - Prob. 15.81QPCh. 15 - Prob. 15.82QPCh. 15 - Prob. 15.83QPCh. 15 - Prob. 15.84QPCh. 15 - Prob. 15.85QPCh. 15 - Prob. 15.86QPCh. 15 - Prob. 15.89QPCh. 15 - Prob. 15.90QPCh. 15 - Prob. 15.91QPCh. 15 - Prob. 15.92QPCh. 15 - Prob. 15.93QPCh. 15 - 15.94 Consider the decomposition of ammonium...Ch. 15 - Prob. 15.95QPCh. 15 - 15.96 In 1899 the German chemist Ludwig Mond...Ch. 15 - Prob. 15.98QPCh. 15 - Prob. 15.99QPCh. 15 - Prob. 15.100QPCh. 15 - Prob. 15.101QPCh. 15 - Prob. 15.102QPCh. 15 - Prob. 15.103SPCh. 15 - Prob. 15.104SPCh. 15 - Prob. 15.105SPCh. 15 - Prob. 15.106SPCh. 15 - Prob. 15.107SPCh. 15 - Prob. 15.110SPCh. 15 - Prob. 15.111SPCh. 15 - Prob. 15.112SP
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