General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 15, Problem 15.106SP

a)

Interpretation Introduction

Interpretation:

The given equation is the prediction based on Le-Chatelier’s principle about the shift in equilibrium with temperature has to be interpreted.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

aA+bBcC+dDKeq=[C]c[D]d[A]a[B]bforaqueousKeq=(PC)c(PD)d(PA)a(PB)bforgases

Van’t Hoff equation: the change in equilibrium constant, Keq of a chemical reaction to the change in temperature, T given the standard enthalpy change ΔH for the process.

Clausius-Claypeyron equation:

lnP=-ΔHoRTwhere,PvarporpressureatTemperatureT(inK)ΔHoenthalpyofvaporizationforthesubstanceR=8.314J/mol.K

a)

Expert Solution
Check Mark

Explanation of Solution

Let’s write the Van’t Hoff equation at two different temperatures:

At T1lnK1=ΔHRT1+C------ (1)

At T2lnK2=ΔHRT2+C------ (2)

Taking the difference between two equations,

lnK1lnK2=ΔHRT1+C(ΔHRT2+C)=ΔHRT1+ΔHRT2ln(K1K2)=ΔHR(1T21T1)

Assuming an endothermic reaction, ΔH>0 and T2>T1. Then, ΔHR(1T21T1)<0 meaning that

ln(K1K2)<0 or K1<K2. A larger K2 indicates that there are more products at equilibrium as the temperature is raised. This agrees with Le-Chatelier’s principle that an increase in temperature favors the forward endothermic reaction. The opposite of the above discussion holds for an endothermic reaction.

b)

Interpretation Introduction

Interpretation:

The molar heat of vaporization of water has to be calculated.

Concept introduction:

Law of mass action: The rate of chemical reaction is directly proportional to the product of concentrations of reactant to products.

aA+bBcC+dDKeq=[C]c[D]d[A]a[B]bforaqueousKeq=(PC)c(PD)d(PA)a(PB)bforgases

Multiple equilibria: If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

A+   B    C + D   Kc'C+   D    E + F    Kc''_overallreaction:A+   B   E + FKc_

The equilibrium constant for two separate equilibrium constants are,

Kc'=[C][D][A][B]andKc''=[E][F][C][D]

For overall reaction, the equilibrium constant Kc is,

Kc'Kc''=[C][D][A][B]×[E][F][C][D]=[E][F][A][B]

Therefore, Kc=Kc'×Kc''

Van’t Hoff equation: the change in equilibrium constant, Keq of a chemical reaction to the change in temperature, T given the standard enthalpy change ΔH for the process.

lnKeq=ΔHRT+C

Clausius-Claypeyron equation:

lnP=-ΔHoRTwhere,PvarporpressureatTemperatureT(inK)ΔHoenthalpyofvaporizationforthesubstanceR=8.314J/mol.K

b)

Expert Solution
Check Mark

Explanation of Solution

For the given reaction,

     H2(l)H2O(g)ΔHvap=?

The given reaction is heterogeneous equilibrium, so, the equilibrium constant expression for gas is,

Rearranging KP expression:  KP=PH2O as PH2O of standard state of water (PH2O(liquid)) is one.

Given: T1= 30C+273=303K;KH2O=31.82mmHgT2= 50C+273=323K;KH2O=92.51mmHg

Substituting given value into the derived Van’t Hoff equation as,

ln(K1K2)=ΔHoR(1T2-1T1)ln(31.82mmHg92.51mmHg)=ΔHo8.314J/mol.K(1323K-1303K)-1.067=ΔHo(-2.458×10-5)ΔHo=4.34×104J/mol=43.4kJ/mol

Therefore, the molar heat of vaporization of water is 43.4kJ/mol

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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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