Respuesta :
Answer:
a
The standard Gibbs free energy change for the reaction at 298 K is
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
b
The energetic (ΔH) is [tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is [tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Energetic is the dominant contribution
c
The equilibrium constant at 298 K is [tex]K = 2.53[/tex]
Explanation:
From the question we are told that
The chemical reaction is
[tex]CH_4 + 2 O_2 ----> 2 H_2 O + CO_2[/tex]
Generally ,
The free energy of formation of [tex]CH_4[/tex] is a constant with a value
[tex]\Delta G^o_f __{CH_4}} = -50.794 \ kJ / moles[/tex]
The free energy of formation of [tex]O_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The free energy of formation of [tex]H_2O[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -228.59 \ kJ / moles[/tex]
The free energy of formation of [tex]CO_2[/tex] is a constant with a value
[tex]\Delta G^o_f __{H_2O}} = -394.6 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CH_4[/tex] at standard condition i is a constant with a value
[tex]\Delta H^o_f __{CH_4}} = -74.848 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]CO_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{CO_2}} = -393.3 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]O_2[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{O_2}} = 0 \ kJ / moles[/tex]
The Enthalpy of formation of [tex]H_2O[/tex] at standard condition is a constant with a value
[tex]\Delta H^o_f __{H_2O}} = -241.83 \ kJ / moles[/tex]
The standard Gibbs free energy change for the reaction at 298 K is mathematically represented as
[tex]\Delta G^o_{re} = (\Delta G^o_f __{H_2O}} + (2 * \Delta G^o_f __{H_2O}} )) - ((\Delta G^o_f __{CH_4}} + (2 * \Delta G^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta G^o_{re} =\Delta G= ( (-394.6 ) + (2 * (-228.59)) ) - ((-50.794) +(2* 0))[/tex]
[tex]\Delta G^o_{re} = -800.99 kJ/moles[/tex]
The Enthalpy of formation of the reaction is
[tex]\Delta H^o _{re} =( \Delta H^o_f __{CH_4}} + (2 * (\Delta H^o_f __{H_2O}} ))) - ( \Delta H^o_f __{CH_4}} + (2 * \Delta H^o_f __{O_2}}))[/tex]
Substituting values
[tex]\Delta H^o _{re} = \Delta H = ((-393.3) + 2 * ( -241.83)) - ( -74.848 + (2 * 0))[/tex]
[tex]\Delta H^o _{re} = -802.112 \ kJ/mole[/tex]
The entropic contributions is mathematically represented as
[tex]T \Delta S = \Delta H -\Delta G[/tex]
Substituting values
[tex]T \Delta S =-802 .112-(-800.986)[/tex]
[tex]T \Delta S = -1.126 \ kJ/mole[/tex]
Comparing the values of [tex]T \Delta S \ and \ \Delta G[/tex] we see that energetic is the dominant contribution
The standard Gibbs free energy change for the reaction at 298 K can also be represented mathematically as
[tex]\Delta G = -RT lnK[/tex]
Where R is the gas constant with as value of [tex]R = 8.314 *10^{-3} kJ/mole[/tex]
K is the equilibrium constant
T is the temperature with a given value of [tex]T = 298K[/tex]
Making K the subject we have
[tex]K = e ^{- \frac{\Delta G }{RT} }[/tex]
Substituting values
[tex]K = e ^{- \frac{-800.99 }{(8.314 *10^{-3} ) * (298)} }[/tex]
[tex]K = 2.53[/tex]
