Question

The complete combustion of methane is: CH4 + 2O2 ! 2H2O + CO2 a. Calculate the standard Gibbs free energy change for the reaction at 298 K (i.e. ). b. Calculate the energetic (ΔH) and entropic contributions (TΔS) to the favorable standard Gibbs free energy change at 298 K and determine which is the dominant contribution,? c. Estimate the equilibrium constant at 298 K.

Answer 1

a

The standard Gibbs free energy change for the reaction at 298 K is

`\Delta G^o_(re) = -800.99 kJ/moles`

b

The energetic (ΔH) is
`\Delta H^o _(re) = -802.112 \ kJ/mole`

The entropic contributions is
`T \Delta S = -1.126 \ kJ/mole`

Energetic is the dominant contribution

c

The equilibrium constant at 298 K is
`K = 2.53`

Step-by-step explanation:

From the question we are told that

The chemical reaction is

`CH_4 + 2 O_2 ----> 2 H_2 O + CO_2`

Generally ,

The free energy of formation of
`CH_4` is a constant with a value

`\Delta G^o_f __(CH_4)} = -50.794 \ kJ / moles`

The free energy of formation of
`O_2` is a constant with a value

`\Delta G^o_f __(O_2)} = 0 \ kJ / moles`

The free energy of formation of
`H_2O` is a constant with a value

`\Delta G^o_f __(H_2O)} = -228.59 \ kJ / moles`

The free energy of formation of
`CO_2` is a constant with a value

`\Delta G^o_f __(H_2O)} = -394.6 \ kJ / moles`

The Enthalpy of formation of
`CH_4` at standard condition i is a constant with a value

`\Delta H^o_f __(CH_4)} = -74.848 \ kJ / moles`

The Enthalpy of formation of
`CO_2` at standard condition is a constant with a value

`\Delta H^o_f __(CO_2)} = -393.3 \ kJ / moles`

The Enthalpy of formation of
`O_2` at standard condition is a constant with a value

`\Delta H^o_f __(O_2)} = 0 \ kJ / moles`

The Enthalpy of formation of
`H_2O` at standard condition is a constant with a value

`\Delta H^o_f __(H_2O)} = -241.83 \ kJ / moles`

The standard Gibbs free energy change for the reaction at 298 K is mathematically represented as

`\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)}))`

Substituting values

`\Delta G^o_(re) =\Delta G= ( (-394.6 ) + (2 * (-228.59)) ) - ((-50.794) +(2* 0))`

`\Delta G^o_(re) = -800.99 kJ/moles`

The Enthalpy of formation of the reaction is

`\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)}))`

Substituting values

`\Delta H^o _(re) = \Delta H = ((-393.3) + 2 * ( -241.83)) - ( -74.848 + (2 * 0))`

`\Delta H^o _(re) = -802.112 \ kJ/mole`

The entropic contributions is mathematically represented as

`T \Delta S = \Delta H -\Delta G`

Substituting values

`T \Delta S =-802 .112-(-800.986)`

`T \Delta S = -1.126 \ kJ/mole`

Comparing the values of
`T \Delta S \ and \ \Delta G` 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

`\Delta G = -RT lnK`

Where R is the gas constant with as value of
`R = 8.314 *10^(-3) kJ/mole`

K is the equilibrium constant

T is the temperature with a given value of
`T = 298K`

Making K the subject we have

`K = e ^{- (\Delta G )/(RT) }`

Substituting values

`K = e ^{- (-800.99 )/((8.314 *10^(-3) ) * (298)) }`

`K = 2.53`