Calculate Δ G in two different ways for the combustion of benzene: 2 C 6 H 6 ( l ) + 15 O 2 ( g ) → 12 CO 2 ( g ) + 6 H 2 O ( l ) Are the two values equal?

Question

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Textbook Question

Chapter 4, Problem 4.16E

Calculate $Δ G$ in two different ways for the combustion of benzene:

$2 C 6 H 6 ( l ) + 15 O 2 ( g ) → 12 CO 2 ( g ) + 6 H 2 O ( l )$

Are the two values equal?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The value of $ΔG$ for the combustion reaction of benzene is to be calculated in two different ways and whether the two values are equal or not is to be predicted.

Concept introduction:

The standard Gibbs-energy gives the non-expansion work achieved from the system at a constant temperature and pressure for the reaction. The standard Gibbs-energy for the reaction is represented as $ΔG°$ .

Answer to Problem 4.16E

The value of $ΔG$ calculated by two different methods is $−6403.84 kJ$ and $−6404.26 kJ$ . The value of $ΔG$ calculated by two different methods is almost equal.

Explanation of Solution

The combustion reaction of benzene is shown below.

$2C6H6(l)+15O2(g)→12CO2(g)+6H2O(l)$

The value of $ΔG$ for the combustion reaction of benzene can be calculated by the formula as shown below.

$ΔG=[12ΔfG°(CO2)+6ΔfG°(H2O)]−[2ΔfG°(C6H6)+15ΔfG°(O2)]$ …(1)

Where,

• $ΔfG°(CO2)$ is the change in Gibbs free energy for the formation of $CO2$ .

• $ΔfG°(H2O)$ is the change in Gibbs free energy for the formation of $H2O$ .

• $ΔfG°(C6H6)$ is the change in Gibbs free energy for the formation of $C6H6$ .

• $ΔfG°(O2)$ is the change in Gibbs free energy for the formation of $O2$ .

The value of change in the Gibbs free energy for the formation of $CO2$ , $H2O$ , $C6H6$ , and $O2$ is $−394.35 kJ/mol$ , $−237.14 kJ/mol$ , $124.4 kJ/mol$ and $0 kJ/mol$ respectively.

Substitute the value of change in the Gibbs free energy for the formation of $CO2$ , $H2O$ , $C6H6$ , and $O2$ in equation (1).

$ΔG=[12×(−394.35 kJ/mol)+6×(−237.14 kJ/mol)]−[2×(124.4 kJ/mol)+15×0]=−4732.2 kJ−1422.84 kJ−248.8 kJ=−6403.84 kJ$

The value of $ΔfH° and ΔS°$ for $CO2$ is $−393.51 kJ/mol$ and $213.785 J/mol⋅K$ .

The value of $ΔfH° and ΔS°$ for $H2O$ is $−285.83 kJ/mol$ and $69.91 J/mol⋅K$ .

The value of $ΔfH° and ΔS°$ for $C6H6$ is $48.95 kJ/mol$ and $173.26 J/mol⋅K$ .

The value of $ΔfH° and ΔS°$ for $O2$ is $0 kJ/mol$ and $205.14 J/mol⋅K$ .

The value of $ΔG$ for the given reaction can be calculated by the formula as shown below.

$ΔG=ΔH−TΔS$ …(2)

The value of $ΔH$ for the given reaction is calculated by the formula as shown below.

$ΔH=[12ΔfH°(CO2)+6ΔfH°(H2O)]−[2ΔfH°(C6H6)+15ΔfH°(O2)]$ …(3)

Substitute the value of $ΔfH°$ of $CO2$ , $H2O$ , $C6H6$ , and $O2$ in equation (3).

$ΔH=[12×(−393.51 kJ/mol)+6×(−285.83 kJ/mol)]−[2×(48.95 kJ/mol)+15×0]=−4722.12 kJ−1714.98 kJ−97.9 kJ=−6535 kJ$

The value of $ΔS$ for the given reaction is calculated by the formula as shown below.

$ΔS=[12ΔS°(CO2)+6ΔS°(H2O)]−[2ΔS°(C6H6)+15ΔS°(O2)]$ …(4)

Substitute the value of $ΔS°$ of $CO2$ , $H2O$ , $C6H6$ and $O2$ in equation (4).

$ΔS=[(12×(213.785 J/mol⋅K)+6×(69.91 J/mol⋅K))−(2×(173.26 J/mol⋅K)+15×205.14 J/mol⋅K)]=2565.42 J/K+419.46 J/K−(346.52 J/K+3077.1 J/K)=−438.74 J/K$

Substitute the value of $ΔS$ and $ΔH$ in equation (2).

$ΔG=−6535 kJ−(298 K×(−438.74 J/K))=−6535 kJ+130.74 kJ=−6404.26 kJ$

The value of $ΔG$ calculated by two different methods is $−6403.84 kJ$ and $−6404.26 kJ$ . The value of $ΔG$ calculated by two different methods is almost equal.

Conclusion

The value of $ΔG$ calculated by two different methods is $−6403.84 kJ$ and $−6404.26 kJ$ . The value of $ΔG$ calculated by two different methods is almost equal.

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