Entropy, Gibbs Function, and Gibbs Function of Formation

6. Entropy, Gibbs Function, and Gibbs Function of Formation

The discussion about the enthalpies of reacting species is useful for applying the First law of thermodynamics Now we will introduce methodologies for determining the entropy, Gibbs function and other species properties, which are useful in the application of the Second law.

Since for an ideal gas, ds = c p0 dT/T – R dP/P, then for a pure component

s 0 k (T,P) – s k (T ref ,1) = s k (T) – R ln (P/1), where (16) s T 0

k = ∫ T ( c /) ref T dT pk .

It is usual to select the conditions T ref = 0 K, P = 1 bar, and s k (T ref ,1) = 0 for an ideal gas. The Gibbs function for a pure component k is

g = h - T s k k . (17) For instance, the Gibbs function under these conditions for molecular hydrogen is

g = h –T s = 0 – 298 ×130.57 = –38910 kJ kmole –1 . of H 2 , g = h –T s = 0 – 298 ×130.57 = –38910 kJ kmole –1 . of H 2 ,

g = h –T s = –285830 – 298 ×69.95 = –306675 kJ kmole –1 . of H 2 O. Since combustion problems typically involve mixtures, the entropy of the k–th component in a

mixture must be first determined, e.g., following the relation ˆs o

k (T,P,X k )= s k (T) – R ln (p k /1), where p k =X k P. The Gibbs function of the k–th mixture component

ˆg k can be obtained by applying the ex- pression

ˆg k = ˆh k –T ˆs k . For ideal gases.

ˆh k = h k , and h k = h f,k + ∆ h k . Alternately, the values of g k (298K, 1 bar) for any compound can also be determined

by ascertaining the Gibbs function of formation g k,f under those conditions. The value of ¯g k,f is identical to the Gibbs energy of formation ∆G for a reaction that forms the compound from its elements that exist in a natural form. For instance, in the case of water,

g fHO o , 2 = ∆ G RHO , 2 = g HO 2 (298K,1 bar)– g H 2 (298 K,1 bar)– g O 2 (298 K,1 bar). (18) The value of ¯g k,f (298K, 1 bar) is assigned as zero for elements that exist in their

natural form. Table A-27A contains values of ¯g k,f (298K, 1 bar) of many substances.

e. Example 5 Determine the Gibbs energy for water at 25ºC and 1 bar. Under those conditions, also

determine the change in the Gibbs energy when H 2 O(l) is formed from its elements,

and the Gibbs energies of formation for H 2 O(l) and of H 2 O(g).

Solution

HO 2 () l = h HO 2 () l –T ¯s HO 2 () l = –285830 – 298 × 69.95 = –306675 kJ kmole . (A) For the chemical reaction

H 2 + 1/2 O 2 →H 2 O(l),

(B)

∆G = g HO 2 () l – g H 2 – 1/2 g O 2 , where,

(C)

g –1

H 2 = h H 2 –T s H 2 = 0 – 298 ×130.57 = –38910 kJ kmole , and (D)

g 2 = 0 – 298 ×205.04 = –61102 kJ kmole O –1 . (E) Applying Eqs. (A), (D), (E) in Eq. (C),

∆G = –306675 – (–38910)–0.5×(–61102) = –237,214 kJ kmole –1 . (F) Therefore, the Gibbs function of a kmole of H 2 O(l) is 237214 kJ lower than that of its

elements (when they exist in a natural form at 25ºC and 1 bar). As in the case of the enthalpy of formation, the Gibbs function of all elements in their natural form can be

arbitrarily set to zero (i.e., g f, H 2 = 0, g f,O2 = 0). Thereafter, the Gibbs energy of for- mation is arbitrarily set to zero (i.e., g f, H 2 = 0, g f,O2 = 0). Thereafter, the Gibbs energy of for- mation is

(H) If the vapor is assumed to be an ideal gas, then

g f, HOg 2 () (T,P) = h HOg 2 () (T,P) – T s HOg 2 () (T,P).

HOg 2 () (T,P) ≈ h HOg 2 () (T,P )= h HO 2 () l (T,P )+ h fg (T,P ). (I) Similarly,

h sat

sat

sat

s HOg 2 () (T,P) = s

HOg 2 () (T,P ) – R ln (P/P )

sat

sat

(T,P = sat )+h (T,P HO sat 2 () l fg )/T – R ln (P/P sat ). (J) Applying Eqs. (I) and (J) in Eq. (H), we obtain

g HOg 2 () (T,P) = h sat

HO 2 () l (T,P )–T s HO 2 () l (T,P )+ RT ln (P/P ). i.e.,

sat

sat

(K) If water is treated as an incompressible fluid, then,

HOg 2 () (T,P) = g HO 2 () l (T,P )+ R T ln (P/P ).

g sat

sat

dg HOg 2 () =v f dP.

Integrating this expression between the limits P sat and P

g HO 2 () l (T,P) – g HO 2 () l (T,P sat ) = –v f (P–P sat ), i.e.,

(L) Typically, v f (P–P sat ) ≈ 0 so that Eq. (L) assumes the form

g HOg 2 () (T,P) = (g HO 2 () l (T,P) – v f (P–P sat )) + RT ln (P/P sat ).

(M) for water at 298 K, P sat = 0.03169 bar. Therefore, at 1 bar water does not exist as gas.

g HOg 2 () (T,P) ≈g HO 2 () l (T,P) + RT ln(P/P sat ).

However, we can define a hypothetical state for water as an ideal gas at 25ºC and 1