GIBB'S FREE ENERGY AND CALCULATIONS

This is a measure of the useful work that a system is capable of doing (apart from volume changes). This is defined by the equation below:

 G = H – TS ………………………..1

Like entropy (H) and enthalpy (S), gibb’s free energy (G) depends on the initial and final state of the system undergoing the change (i.e. Gibb’s free energy is also a state function).

 ∆G = G_(PRODUCT) – G_(REACTANT)

G_(REACTANT) = H_(REACTANT) – TS_(REACTANT)

G_(PRODUCT) = H_(PRODUCT) – TS_(PRODUCT)

G_(PRODUCT) - G_(REACTANT)  = H_(PRODUCT) – TS_(REACTANT) – T(S_(PRODUCT) - S_(REACTANT))

  ∆G = ∆H - T∆S……………………………………..2

This quantity (i.e. ∆G) gives a direct condiion of spontaneity of reaction:

·         If ∆G is negative, (i.e. ∆G = -ve): the reaction is spontaneous, i.e. the reaction will occur as written.

·         If ∆G is positive, (i.e. ∆G = +ve): the reaction is not spontaneous in the forward reaction but spontaneous in the reversed direction.

·         If ∆G is zero, (i.e. ∆G = 0); the reaction is at equilibrium

When Gibb’s free energy is measurd at 25oC (298K) and 1atm (i.e. at standard temperature and pressure), it is referred to as standard gibb’s free energy change ∆G^Ѳ

Gibb’s free energy change is also related to the equilibrium constant (K) by the expression.

 ∆G = -RTInK

  ∆G = -2.303 RT log₁₀K………………………..3

Where R is the gas constant = 8.314J/mol/K

T is the temperature in Kelvin

K is the equilibrium constant.

ESTIMATION OF FREE ENERGY CHANGE

Equation 3 can be used to estimate values of ∆G and to predict how systems will behave at low and high temperatures.

·         At low temperatures: At low temperatures, the factor T∆S becomes small and so ∆H controls the sign of ∆G. therefore only exothermic reaction are feasible at low temperature because ∆G is negative. When ∆H is negative from the relation in equation 2.

·         At high temperatures: at high temperatures, the factor T∆S becomes large and more important

a.       ENDOTHERMIC REACTION: for the chemical reactions to be feasible (i.e. ∆G = -ve), the endothermic reactions must be accomplished by an increase in entropy, i.e. entropy change must be large enough for the product T∆S to exceed ∆H.

b.      EXOTHERMIC REACTION: exothermic reactions become explosive and favourable at high temperatures, for example, decomposition of ammonium dichromate below:

(NH₄)₂Cr₂O₇  ---  N_2(g) + Cr₂O_3(s) + 4H₂O_(l) (∆H = -ve)

WORKED EXAMPLES

Question 1: what is the standard gibb’s free energy change, ∆G^Ѳ for the following reactions at 25oC, given the following information :

                     N_2(g)      +      3H₂   --------    2NH₃

∆H_f (KJ) :          0                               0                   -45,9

S⁰ (J/K):           191.5                       130.6           193

SOLUTION:

∆H _(REACTION) = Ɛ n ∆H _(producr)  -  Ɛ m ∆H _(Reactant)

= [2∆H_f (NH3) ]  -  [∆H_f (N2)  +  3∆H_f (H2) ]

= [2 x (-45.9)]  -  [0+0]

∆H _(REACTION) = -91.8KJ

∆S^Ѳ = Ɛ n S^Ѳ _(producr)  -  Ɛ m S^Ѳ _(Reactant)

= [2X S^Ѳ (NH3)] – [S^Ѳ(N2) + 3 S^Ѳ(H2)]

= [(2 X 193)] – [ 191.5 + 3(130.6)]

= -197.3 J/K

= -0.1973KJ/K

Therefore,

∆G^Ѳ = ∆H^Ѳ - T∆S^Ѳ

= -91.8 KJ – (298K X (-0.1873 KJ/K)

= -91.8 + 58.80

∆G^Ѳ = -33 KJ

Question 2: calculate the standard gibb’s free energy change (∆G^Ѳ) at 25oC for the thermal decomposition of cadmium trioxocarbonate (IV).

CdCO_3(s)   --------   CdO_(s)  +  CO_2(g)

Given that:

∆H^Ѳ_(reaction) = 178KJ/mol

∆S^Ѳ = +161J/K

SOLUTION:

∆H^Ѳ_(reaction) = 178KJ/mol

∆S^Ѳ = +161/1000

        = 0.161  KJ/K

∆G^Ѳ = ∆H^Ѳ - T∆S^Ѳ

= 178 KJ – (298K X 0.161 KJ/K)

= + 130 KJ/mol

The reaction is not spontaneous because the ∆G^Ѳ value is positive.