CHEMISTRY >CUSTOM<
CHEMISTRY >CUSTOM<
14th Edition
ISBN: 9781259137815
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 24, Problem 97AP
Interpretation Introduction

Interpretation:

The value of equilibrium constant at given temperatures is to be determined with given ΔH and ΔS

values.

Concept introduction:

The enthalpy of reaction is calculated by the expression as follows:

ΔH=ΔHf(products)ΔHf(reactants)

Here, ΔH is the standard change in enthalpy of reaction, ΔHf(reactants) is the sum of standard change in enthalpy of reactants, ΔHf(products) is the standard change in enthalpy of products.

The value of ΔS

for the given reaction is calculated by using the relation given below:

ΔS=S(product)S(reactant)

Here, ΔS

is the standard change in entropy,

The value of K is calculated by using the relation given below:

K=eΔGRT

Here K is the equilibrium constant, R is the gas constant, T is the temperature, and ΔG

is the standard change in energy.

The value of ΔG is calculated by using the relation given below:

ΔG=ΔH-TΔS

Expert Solution & Answer
Check Mark

Answer to Problem 97AP

Solution: K at 298 K

is 9.61×1022

and at 373 K

is 1.2×1015.

Explanation of Solution

Given information: The given reaction is C(s)+CO2(g)2CO(g).

The value of ΔHf(CO) from appendix 2

is -110.5 kJ/mol.

The value of ΔHf(CO2) from appendix 2

is 393.5 kJ/mol.

The value of S(C)

from appendix 2 is 5.69 J/Kmol.

The value of S(CO)

from appendix 2 is 197.9 J/Kmol.

The value of S(CO2)

from appendix 2 is 213.6 J/Kmol.

The value of ΔH

for the given reaction is calculated by using the relation given below:

ΔH=2ΔHf(CO)[ ΔHf(C)+ΔHf(CO2) ]

Here, ΔH

is the standard change in enthalpy, ΔHf(CO)

is the standard change in enthalpy of CO, ΔHf(C)

is the standard change in enthalpy of C

and ΔHf(CO2)

is the standard change in enthalpy of CO2.

Substitute 0

for ΔHf(C), 110.5kJ/mol for ΔHf(CO), and 393.5kJ/mol

for ΔHf(CO2) in the equation as follows:

ΔH=(2)(110.5kJ/mol)[ 0+(1)(393.5kJ/mol) ]=172.5 kJ/mol=172.5×103 J/mol

Therefore, the value of ΔH

for the given reaction is 172.5×103 J/mol.

The value of ΔS

for the given reaction is calculated by using the relation given below:

ΔS=2S(CO)[ S(C)+S(CO2) ]

Here, ΔS

is the standard change in entropy, S(CO)

is the standard entropy of CO, S(C)

is the standard entropy of C, and S(CO2)

is the standard entropy of CO2.

Substitute 5.69 J/Kmol

for S(C), 197.9 J/Kmol for S(CO), and 213.6 J/Kmol

for S(CO2) in the equation as follows:

ΔS=(2)(197.9 J/Kmol)[ 5.69 J/Kmol+213.6 J/Kmol ]=176.5 J/Kmol

Therefore, the value of ΔS

for the given reaction is 176.5 J/mol.

The value of ΔG is calculated by using the relation given below:

ΔG=ΔH-TΔS

The value of ΔG

at 298 K(25 C)

is as follows:

Substitute 176.5 J/mol

for ΔS, 172.5 ×103J/mol for ΔH, and 298 K

for T in the above equation as follows:

ΔG=172.5 ×103J/mol-(298 K)(176.5 J/mol)=1.99×105 J/mol

The value of K is calculated by using the relation given below:

K=eΔGRT

Here K is the equilibrium constant, R is the gas constant, T is the temperature, and ΔG

is the standard change in energy.

Substitute 1.99×105 J/mol

for ΔG, 8.314 J/mol.K for R, and 298 K

for T in the above equation as follows:

K=e1.99×105 J/mol(8.314 J/mol.K)(298 K)=e1.99×105 J/mol2477.57 J/mol=9.61×1022

The value of ΔG at 373 K(100 C) is as follows:

Substitute 176.5 J/mol

for ΔS, 172.5 ×103J/mol for ΔH, and 373 K

for T in the above equation as follows:

ΔG=172.5 ×103J/mol-(373 K)(176.5 J/mol)=1.07×105 J/mol

The value of K is calculated by using the relation given below:

K=eΔGRT

Here, K is the equilibrium constant, R is the gas constant, T is the temperature, and ΔG

is the standard change in energy.

Substitute 1.07×105 J/mol

for ΔG, 8.314 J/mol.K for R, and 373 K

for T in the above equation as follows:

K=e(1.07×105 J/mol)(8.314 J/mol.K)(373 K)=e1.07×105 J/mol3101.122 J/mol=1.2×1015

Conclusion

Therefore, the value of equilibrium constant K at 298 K is 9.61×1022 and at 373 K is 1.2×1015.

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Chapter 24 Solutions

CHEMISTRY >CUSTOM<

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