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Methane, a component in natural gas, can be used as a fuel in combustion reactions. What is the value for ΔGnon (in kJ) for the following reaction under the given conditions at 293 K? CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) where ΔHorxn = -803 kJ, ΔSorxn = -4.05 J/K, and [CH4] = 14.51 M, [O2] = 9.27 M, [CO2] = 3.83 M, and [H2O] = 6.41 M. (in kJ)

Respuesta :

Answer: The [tex]\Delta G[/tex] for the reaction is -806.86 kJ

Explanation:

We are given:

[tex]\Delta H^o_{rxn}=-803kJ=-803000J[/tex]      (Conversion factor:  1 kJ = 1000)

[tex]\Delta S^o_{rxn}=-4.05J/K[/tex]

Temperature of the reaction = 293 K

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

[tex]\Delta G^o_{rxn}=\Delat H^o_{rxn}-T\Delta S^o_{rxn}[/tex]

Putting values in above equation, we get:

[tex]\Delta G^o_{rxn}=-803000J-[(293K)\times (-4.05J/K)]=-801813.35J[/tex]

For the given chemical equation:

[tex]CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(g)[/tex]

The expression for [tex]K_c[/tex] is given as:

[tex]K_{c}=\frac{[H_2O]^2[CO_2]}{[CH_4][O_2]^2}[/tex]

We are given:

[tex][H_2O]=6.41M[/tex]

[tex][CO_2]=3.83M[/tex]

[tex][CH_4]=14.51M[/tex]

[tex][O_2]=9.27M[/tex]

Putting values in above equation, we get:

[tex]K_c=\frac{(6.41)^2\times 3.83}{14.51\times (9.27)^2}[/tex]

[tex]K_c=0.126[/tex]

To calculate the Gibbs free energy of the reaction, we use the equation:

[tex]\Delta G=\Delta G^o+RT\ln K_c[/tex]

where,

[tex]\Delta G[/tex] = Gibbs' free energy of the reaction = ?

[tex]\Delta G^o[/tex] = Standard gibbs' free energy change of the reaction = -801813.35 J

R = Gas constant = [tex]8.314J/K mol[/tex]

T = Temperature = 293 K

[tex]K_c[/tex] = equilibrium constant in terms of concentration = 0.126

Putting values in above equation, we get:

[tex]\Delta G=-801813.35J+(8.314J/K.mol\times 293K\times \ln(0.126))[/tex]

[tex]\Delta G=-806859.46J=-806.86kJ[/tex]

Hence, the [tex]\Delta G[/tex] for the reaction is -806.86 kJ

-806.859 kJ is the value of Gibb's free energy of the given reaction.

How we calculate the Gibb's free energy change?

We calculate the ΔG for the given reaction is as:

ΔG = ΔG° + RTlnKc, where

R = universal gas constant = 8.314 J/K.mole

ΔG° = standard Gibb's free energy which can be calculated as:

ΔG° = ΔH° - TΔS°, where

ΔH° = Change in enthalpy = -803 kJ (given) = -803000J

T = temperature = 293K (given)

ΔS° = Change in entropy = -4.05 J/K (given)

On putting all these values in the above equation, we get

ΔG° = -803000 - 293 × (-4.05) = -801813.35J

Also we have to calculate the value of equilibrium constant for the given reaction:

CH₄ (g) + 2O₂ (g) ⇄ CO₂ (g) + 2H₂O (g)

Kc = [CO₂][H₂O]² / [CH₄][O₂]²

Given that:

Concentration of CO₂ = 3.83 M

Concentration of H₂O = 6.41 M

Concentration of CH₄ = 14.51 M

Concentration of O₂ = 9.27 M

On putting all these values in the above equation, we get

Kc = (3.83)(6.41)² / (14.51)(9.27)² = 0.126

Now we put all the calculated values in the first equation of Gibb's free energy, we get

ΔG = -801813.35 + 8.314×293 ln(0.126)

ΔG = -806859.46J = -806.859 kJ

Hence, -806.859 kJ is the value of ΔG.

To know more about Gibb's free energy, visit the below link:

https://brainly.com/question/10012881

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