IDEAL-GAS REACTION EQUILIBRIUM

The equilibrium condition for the reaction 0 ^∆

in_iA_i(where n_iis the stoichiometric number of species A_i) is

in_im_i 0 [Eq. (4.98)]. We now specialize to the case where all reactants and products are ideal gases.

For the ideal-gas reaction

the equilibrium condition

in_im_i 0 is

Each chemical potential in an ideal gas mixture is given by Eq. (6.4) as m_i m°i  RT ln (P_i/P°), and substitution in the equilibrium condition gives

(6.5) Since m Gmfor a pure substance, the quantity on the left side of (6.5) is the stan-dard Gibbs energy change G°Tfor the reaction [Eq. (5.38)]

The equilibrium condition (6.5) becomes

(6.6) where the identities a ln x ln x^a, ln x ln y  ln xy, and ln x  ln y  ln (x/y) were used, and where the eq subscripts emphasize that these are partial pressures at equilib-rium. Defining the standard equilibrium constant K°_Pfor the ideal-gas reaction aA  bB : cC  dD as

(6.7) we have for Eq. (6.6)

We now repeat the derivation for the general ideal-gas reaction 0 :

in_iA_i. Substitution of the expression m_i m°i  RT ln (Pi/P°) for a component of an ideal gas mixture into the equilibrium condition

in_im_i 0 gives

(6.8) where the sum identities

i(a_i bi) 

ia_i

ib_iand

ica_i c

ia_i[Eq. (1.50)]

were used. We have m°_i(T)  G°m,T,i, Therefore

(6.9)

¢G°_T a

i

n_iG°_m,T,i a

i

n_im°_i1T2 a

i

n_im°_i1T2  RT a

i

n_i ln 1Pi,eq>P°2  0 a

i

n_im_i a

i

n_i3m°i  RT ln 1Pi,eq>P°2 4  0

¢G° RT ln K°P

K°_P⬅ 1PC,eq>P°2^c1PD,eq>P°2^d

1PA,eq>P°2^a1PB,eq>P°2^b, P° ⬅ 1 bar

¢G° RT ln1PC,eq>P°2^c1PD,eq>P°2^d 1PA,eq>P°2^a1PB,eq>P°2^b

¢G° RT3ln 1P^C>P°2^c ln 1P^D>P°2^d ln 1P^A>P°2^a ln 1P^B>P°2^b4

¢G°_T ⬅ a

i

n_iG°_{m, T,i}  a

i

n_im°_i1T2  cm°C dm°D am°A bm°B

RT3c ln 1PC>P°2  d ln 1PD>P°2  a ln 1PA>P°2  b ln 1PB>P°2 4 cm_C° dmD° am°A bm°B

 amA° aRT ln 1PA>P°2  bm°B bRT ln 1PB>P°2  0 cm_C° cRT ln 1PC>P°2  dm°D dRT ln 1PD>P°2

cm_C dmD amA bmB 0 am_A bmB cmC dmD

aA bB ∆ cC  dD

Section 6.2 Ideal-Gas Reaction Equilibrium 177 lev38627_ch06.qxd 3/3/08 10:07 AM Page 177

Chapter 6

Reaction Equilibrium in Ideal Gas Mixtures

178

and (6.8) becomes

(6.10) where k ln x ln x^kwas used. The sum of logarithms equals the log of the product:

where the large capital pi denotes a product:

(6.11)*

As with sums, the limits are often omitted when they are clear from the context. Use of

iln a_i ln ia_iin (6.10) gives

(6.12) We define K°_Pas the product that occurs in (6.12):

(6.13)*

Equation (6.12) becomes

(6.14)*

Recall that if y lnex, then x e^y[Eq. (1.67)]. Thus (6.14) can be written as (6.15) Equation (6.9) shows that G° depends only on T. It therefore follows from (6.15) that K°_Pfor a given ideal-gas reaction is a function of T only and is independent of the pres-sure, the volume, and the amounts of the reaction species present in the mixture: K°_P

 K°P(T). At a given temperature, K°_Pis a constant for a given reaction. K°_Pis the stan-dard equilibrium constant (or the stanstan-dard pressure equilibrium constant) for the ideal-gas reaction.

Summarizing, for the ideal-gas reaction 0 ^∆

in_iA_i, we started with the general condition for reaction equilibrium

in_im_i 0 (where the ni’s are the stoichiometric numbers); we replaced each m_i with the ideal-gas-mixture expression m_i  m°i  RT ln (P_i/P°) for the chemical potential m_i of component i and found that G° 

RT ln K°P. This equation relates the standard Gibbs energy change G° [defined by (6.9)] to the equilibrium constant K°_P[defined by (6.13)] for the ideal-gas reaction.

Because the stoichiometric numbers n_iare negative for reactants and positive for products, K°_Phas the products in the numerator and the reactants in the denominator.

Thus, for the ideal-gas reaction

(6.16)

we have so

(6.17)

(6.18) where the pressures are the equilibrium partial pressures of the gases in the reaction mixture. At any given temperature, the equilibrium partial pressures must be such as

K°_P 3P1NH32eq>P°4² 3P1N²2^eq>P°4 3P1H²2^eq>P°4³

K°P 3P1NH³2^eq>P°4²3P1N²2^eq>P°4^13P1H²2^eq>P°4^3 n_N

2 1, nH₂ 3, and nNH₃ 2,

N₂1g2  3H²1g2 S 2NH³1g2 K°_P e^¢G°>RT

¢G° RT ln K°P ideal-gas reaction equilib.

K°_P⬅ q

i

1Pi,eq>P°2ⁿⁱ ideal-gas reaction equilib.

¢G°_T RT ln c q

i

1Pi,eq>P°2ⁿⁱd q

n i1

a_i⬅ a1a₂p a_n a

n i1

ln a_i ln a1 ln a2 p  ln an ln 1a1a₂p an2  ln qⁿ

i1

a_i

¢G°_T RT a

i

n_i ln 1Pi,eq>P°2  RT a

i

ln 1Pi,eq>P°2ⁿⁱ

lev38627_ch06.qxd 3/3/08 10:07 AM Page 178

to satisfy (6.18). If the partial pressures do not satisfy (6.18), the system is not in re-action equilibrium and its composition will change until (6.18) is satisfied.

For the ideal-gas reaction aA  bB ^∆cC  dD, the standard (pressure) equilib-rium constant is given by (6.7).

Since P_i/P° in (6.13) is dimensionless, the standard equilibrium constant K°_P is dimensionless. In (6.14), the log of K°_Pis taken; one can take the log of a dimension-less number only. It is sometimes convenient to work with an equilibrium constant that omits the P° in (6.13). We define the equilibrium constant (or pressure equilibrium constant) K_Pas

(6.19) K_Phas dimensions of pressure raised to the change in mole numbers for the reaction as written. For example, for (6.16), K_Phas dimensions of pressure^2.

The existence of a standard equilibrium constant K°_Pthat depends only on T is a rigorous deduction from the laws of thermodynamics. The only assumption is that we have an ideal gas mixture. Our results are a good approximation for real gas mixtures at low densities.

EXAMPLE 6.1

Finding K°_Pand G° from the equilibrium composition

A mixture of 11.02 mmol (millimoles) of H₂S and 5.48 mmol of CH₄was placed in an empty container along with a Pt catalyst, and the equilibrium

(6.20) was established at 700°C and 762 torr. The reaction mixture was removed from the catalyst and rapidly cooled to room temperature, where the rates of the for-ward and reverse reactions are negligible. Analysis of the equilibrium mixture found 0.711 mmol of CS₂. Find K°_Pand G° for the reaction at 700°C.

Since 0.711 mmol of CS₂ was formed, 4(0.711 mmol)  2.84 mmol of H₂was formed. For CH₄, 0.711 mmol reacted, and 5.48 mmol  0.71 mmol  4.77 mmol was present at equilibrium. For H₂S, 2(0.711 mmol) reacted, and 11.02 mmol  1.42 mmol  9.60 mmol was present at equilibrium. To find K°_P, we need the partial pressures P_i. We have P_i⬅ xiP, where P 762 torr and the x_i’s are the mole fractions. Omitting the eq subscript to save writing, we have at equilibrium

The standard pressure P° in K°_Pis 1 bar ⬇ 750 torr [Eq. (1.12)], and (6.13) gives

The use of G°  RT ln K°P[Eq. (6.14)] at 700°C  973 K gives

¢G°₉₇₃ 38.314 J>1mol K2 4 1973 K2 ln 0.000331  64.8 kJ>mol

 0.000331

K°_P 1P^H2>P°2⁴1P^CS2>P°2

1PH2S>P°2²1PCH4>P°2  1120 torr>750 torr2⁴130.3 torr>750 torr2 1408 torr>750 torr2²1203 torr>750 torr2

P_H2 120 torr, PCS2 30.3 torr P_H2S 0.5361762 torr2  408 torr, PCH4 203 torr,

x_H2S 9.60>17.92  0.536, xCH4 0.266, xH2 0.158, xCS2 0.0397 n_H2S 9.60 mmol, nCH4 4.77 mmol, nH2 2.84 mmol, nCS2 0.711 mmol

2H₂S1g2  CH41g2 ∆ 4H21g2  CS21g2 K_P⬅ q

i

1Pi,eq2ⁿⁱ

Section 6.2 Ideal-Gas Reaction Equilibrium 179 lev38627_ch06.qxd 3/3/08 10:07 AM Page 179

Chapter 6

Reaction Equilibrium in Ideal Gas Mixtures

180

In working this problem, we assumed an ideal gas mixture, which is a good assumption at the T and P of the experiment.

Exercise

If 0.1500 mol of O₂(g) is placed in an empty container and equilibrium is reached at 3700 K and 895 torr, one finds that 0.1027 mol of O(g) is present.

Find K°_Pand G° for O2(g) ^∆2O(g) at 3700 K. Assume ideal gases. (Answers:

0.634, 14.0 kJ/mol.)

Exercise

If 0.1500 mol of O₂(g) is placed in an empty 32.80-L container and equilibrium is established at 4000 K, one finds the pressure is 2.175 atm. Find K°_P and

G° for O2(g) ^∆ 2O(g) at 4000 K. Assume ideal gases. (Answers: 2.22,

26.6 kJ/mol.)

Concentration and Mole-Fraction Equilibrium Constants

Gas-phase equilibrium constants are sometimes expressed using concentrations in-stead of partial pressures. For n_imoles of ideal gas i in a mixture of volume V, the par-tial pressure is P_i niRT/V [Eq. (1.24)]. Defining the (molar) concentration c_iof species i in the mixture as

(6.21)*

we have

(6.22) Use of (6.22) in (6.7) gives for the ideal-gas reaction aA  bB ^∆f F  dD

(6.23) where c°, defined as c°⬅ 1 mol/liter  1 mol/dm³, was introduced to make all frac-tions on the right side of (6.23) dimensionless. Note that c°RT has the same dimensions as P°. The quantity f d  a  b is the change in number of moles for the reaction as written, which we symbolize by n/mol ⬅ f  d  a  b. Since f  d  a  b is dimensionless and n has units of moles, n was divided by the unit “mole” in the definition. For N₂(g)  3H2(g) ^∆2NH₃(g), n/mol  2  1  3  2. Defining the standard concentration equilibrium constant K°_cas

(6.24) we have for (6.23)

(6.25) Knowing K°_P, we can find K°_cfrom (6.25). K°_cis, like K°_P, dimensionless. Since K°_P de-pends only on T, and c° and P° are constants, Eq. (6.25) shows that K°_cis a function of T only.

One can also define a mole-fraction equilibrium constant K_x:

(6.26) The relation between K_xand K°_Pis (Prob. 6.7)

(6.27) K°_P Kx1P>P°2^¢n>mol

K_x⬅ q

i

1xi,eq2ⁿⁱ K°_P K°c1RTc°>P°2^¢n>mol K°_c⬅ q

i

1ci,eq>c°2ⁿⁱ where c° ⬅ 1 mol>L ⬅ 1 mol>dm³ K°_P 1cF,eqRT>P°2^f1cD,eqRT>P°2^d

1cA,eqRT>P°2^a1cB,eqRT>P°2^b 1cF,eq>c°2^f1cD,eq>c°2^d

1cA,eq>c°2^a1cB,eq>c°2^b ac°RT

P° b^fdab P_i niRT>V  ciRT ideal gas mixture

c_i⬅ ni>V

lev38627_ch06.qxd 3/3/08 10:07 AM Page 180