Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 5, Problem 5.27E

Nitrogen dioxide, NO 2 , dimerizes easily to form dinitrogen tetroxide, N 2 O 4 :

2 NO 2 ( g ) N 2 O 4 ( g )

(a) Using data in Appendix 2, calculate Δ rxn G ° and K for this equilibrium.

(b) Calculate ξ for this equilibrium if 1.00 mol NO 2 were present initially and allowed to come to equilibrium with the dimer in a 20.0 -L system.

(a)

Expert Solution
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Interpretation Introduction

Interpretation:

The values of ΔrxnG° and K for the given equilibrium are to be calculated.

Concept introduction:

Standard Gibbs free energy of a reaction is calculated by subtracting the standard Gibbs free energy of formation of reactants from standard Gibbs free energy of formation of products. The formula for standard Gibbs free energy of reaction is given as,

ΔrxnG°=ΔfG°(products)ΔfG°(reactants)

Where,

ΔfG°(products) represents the standard Gibbs free energy of formation of products.

ΔfG°(reactants) represents the standard Gibbs free energy of formation of reactants.

The relation between equilibrium constant and standard Gibbs free energy of reaction is given as,

ΔrxnG°=RTlnK

Where,

R represents the gas constant with a value of 8.314J/molK.

T represents the temperature.

K represents the equilibrium constant

Answer to Problem 5.27E

The values of ΔrxnG° and K for the given equilibrium are 4.81kJ/mol and 6.96 respectively.

Explanation of Solution

From Appendix 2

The standard Gibbs free energy of formation of NO2 is 51.30kJ/mol.

The standard Gibbs free energy of formation of N2O4 is 97.79kJ/mol.

Temperature at the equilibrium is 298 K.

The given reaction is represented as,

2NO2(g)N2O4(g)

The standard Gibbs free energy of given reaction is given as,

ΔrxnG°=ΔfG°(N2O4)2ΔfG°(NO2)

Where,

ΔfG°(NO2) represents the standard Gibbs free energy of formation of NO2.

ΔfG°(N2O4) represents the standard Gibbs free energy of formation of N2O4.

Substitute the value of ΔfG°(NO2) and ΔfG°(N2O4) in the above equation.

ΔrxnG°=97.79kJ/mol(2)(51.30kJ/mol)=4.81kJ/mol

Therefore, the value ΔrxnG° at equilibrium is 4.81kJ/mol.

The relation between equilibrium constant and standard Gibbs free energy of reaction is given as,

ΔrxnG°=RTlnK

Where,

R represents the gas constant with a value of 8.314J/molK.

T represents the temperature.

K represents the equilibrium constant

Rearrange the above equation form the value of K.

K=exp(ΔrxnG°RT)

Substitute the value of R, T and ΔrxnG° in the above equation.

K=exp((4.81kJ/mol)(1000J1kJ)(8.314J/molK)(298K))=exp(1.6871)=6.96

Therefore, the value K at equilibrium is 6.96.

Conclusion

The values of ΔrxnG° and K for the given equilibrium are 4.81kJ/mol and 6.97 respectively.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of ξ for the given equilibrium reaction is to be calculated.

Concept introduction:

The equilibrium constant of a reaction is expressed as the ratio of partial pressure of products and reactants each raised to the power of their stoichiometric coefficients. A typical equilibrium reaction is represented as,

aA+bBcC+dD

The algebraic form of equilibrium constant for the above chemical reaction is expressed as,

K=(pC)c(pD)d(pA)a(pB)b

Where,

(pA) represents the partial pressure of reactant A.

(pB) represents the partial pressure of reactant B.

(pC) represents the partial pressure of product C.

(pD) represents the partial pressure of product D.

a represents the stoichiometric coefficient of reactant A.

b represents the stoichiometric coefficient of reactant B.

c represents the stoichiometric coefficient of product C.

d represents the stoichiometric coefficient of product D.

Answer to Problem 5.27E

The value of ξ for the given equilibrium reaction is 0.4 mol.

Explanation of Solution

The initial number of moles of NO2 is 1.00mol.

The volume of the reaction system is 20.0 L.

The temperature of the reaction system is 298K.

The value K at equilibrium is 6.96.

The ideal gas equation is given as,

PV=nRT (1)

Where,

V represents the volume occupied by the ideal gas.

P represents the pressure of the ideal gas.

n represents the number of moles of the ideal gas.

T represents the temperature of the ideal gas.

R represents the ideal gas constant with value 0.08206Latm/Kmol.

Rearrange the above equation for the value of P.

P=nRTV

Substitute the value of V, T, n and R in the above equation.

P=(1.00mol)(0.08206Latm/Kmol)(298K)20.0 L=1.2227atm

The table for initial and equilibrium amounts of the substances involved in the reaction is represented as,

2NO2(g)N2O4(g)Pressure(atm)2NO2(g)N2O4(g)Initial1.2227atm0atmequilibrium(1.22272x)atmxatm

The expression forequilibrium constant for the given equilibrium reaction is represented as,

K=(pN2O4)(pNO2)2

Substitute the value of pN2O4, pNO2 and K in the above expression.

6.96=(xatm)((1.22272x)atm)2

Rearrange the equation to form a quadratic equation.

27.84x218.02x+10.4052=0

Solve the quadratic equation form the value of x=0.481 atm.

Rearrange equation (1) form the value of n.

n=PVRT

Substitute the value of V, T, P and R in the above equation.

n=(0.481atm)(20.0 L)(0.08206Latm/Kmol)(298K)=0.3934 mol

The number of mole of N2O4 at equilibrium is 0.3934 mol.

The initial number of mole of N2O4 is 0 mol.

The expression for extent of reaction for N2O4 is given as,

ξ=nN2O4,fnN2O4,ivN2O4

Where,

nN2O4,f represent the number of moles of N2O4 at equilibrium.

nN2O4,f represent the initial number of moles of N2O4.

vN2O4 represent the stoichiometric coefficient of N2O4.

Substitute the value of nN2O4,f, nN2O4,f and vN2O4 in the above expression.

ξ=0.3934 mol0 mol1=0.3934 mol0.4 mol

Therefore, the extent of the reaction is 0.4 mol.

Conclusion

The extent of the reaction is 0.4 mol.

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Chapter 5 Solutions

Physical Chemistry

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