Thermodynamics Chemical-reaction Equilibria Chapter 10 Chemical-reaction Equilibria

In a chemical reaction, both the rate and the equilibrium conversion are considered for commercial purposes. Equilibrium conversion represent the maximum possible conversion regardless the reaction rate or catalyst. This chapter focus on the effect of temperature, pressure, and the initial composition on the equilibrium conversions of chemical reactions.

10.1 The Reaction Coordinate The general chemical reaction: where |vi| is a stoichiometric coefficient and Ai stands for a chemical formula. For vi : Positive ( + ) for product Negative ( - ) for reactants

The stoichiometric numbers are : For example, CH4 + H2O → CO + 3H2 The stoichiometric numbers are : The stoichiometric number for an inert species is zero. Since:

This variable ε is called the reaction The differential change dni in the number of moles of a reacting species and dε is : dni = vi dε ( i = 1 , 2 ,….N) This variable ε is called the reaction coordinate, characterizes the extent or degree to which a reaction has taken place.

Summation over all species yields : gives ( i = 1 , 2 ,….N) Summation over all species yields : or n = n0 + vε where

Thus, the mole fractions yi of the species present are related to ε by :

Example 13.2 Consider a vessel which initially contains only n0 mol of water vapor. If decomposition occurs according to the reaction, find expressions which relate the number of moles and the mole fraction of each chemical species to the reaction coordinate.

Solution 13.2 For the given reaction Application of Eqs.(13.4) and (13.5) yields:

The fractional decomposition of water vapor is: Thus when no = 1, ε is identified with the fractional decomposition of the water vapor.

Multireaction Stoichometry ( i = 1 , 2 ,….N) After integration: Summing over all species yields : For the mole fraction of the species presents in particular reaction:

6.2 Application of Equilibrium Criteria To Chemical Reactions At the equilibrium state : (dGt)T , P = 0 The total Gibbs energy, Gt is a minimum. Its differential is zero.

6.3 The Standard Gibbs-Energy Change and The Equilibrium Constant The fundamental property relation for single-phase systems, provides an expression for the total differential of the Gibbs energy:

If changes in mole numbers ni occur as the result of a single chemical reaction in a closed system , then each dni may be replaced by the product vi dε. Since nG is a state function , the right side of this an exact differential expression :

For chemical-reaction equilibrium : Recall the definition of the fugacity of a species in solution : For a pure species i in its standard start at the same temperature : The difference between these equations:

For the equilibrium state of a chemical reaction: where

Alternative expression for K : where Called as standard Gibbs-energy change of reaction. K is dependence on temperature. In spite of its dependence on temperature, K is called the equilibrium constant for the reaction.

6.4 Effect of Temperature on the Equilibrium Constant The dependence of ΔG˚ on T : Then become: Equation above gives the effect of temperature on equilibrium constant and hence on the equilibrium conversion.

When ΔH˚ negative = exothermic reaction and the equilibrium constant, K decreases as the temp. increases. positive = endothermic reaction increases as the temp. increases. If ΔH˚ is assumed independent of T, integration from a reference T’ (at 298K) to random T, then:

In Figure 13.2, a plot of ln K vs 1/T for a number of common reactions, illustrates this near linearity.

Multiplication by vi and summation over all species gives: The effect of temperature on the equilibrium constant is based on the definition of the standard Gibbs energy: G˚i = H˚i -TSi˚ Multiplication by vi and summation over all species gives:

Hence, the standard property change of reaction; ΔG˚=ΔH˚ - TΔS˚ where the standard heat of reaction and standard entropy change is related to temperature:

However whence and recall,

K0 represents the equilibrium constant at reference temperature T0 : As an alternative, the preceding equation may be reorganized so as to factor K into three terms , each representing a basic contribution to its value : K = K0 K1 K2 K0 represents the equilibrium constant at reference temperature T0 :

6.5 Evaluation of Equilibrium Constant Example 13.4 Calculate the equilibrium constant for the vapor-phase hydration of ethylene at 145 and at 320°C from data given in App. C. Solution 13.4 First determine values for ΔA, ΔB, ΔC, and ΔD for the reaction: C2H4 (g) + H2O (g) → C2H5OH (g)

The meaning of Δ is indicated by: A = (C2H5 OH) - (C2H4) - (H2O) The meaning of Δ is indicated by: A = (C2H5 OH) - (C2H4) - (H2O). Thus, from the heat-capacity data of Table C.1: ΔA = 3.518 - 1.424 - 3.470 = - 1.376 ΔB = (20.001 - 14.394 - 1.450) x 10-3 = 4.157 x l0-3 ΔC = (-6.002 + 4.392 - 0.000) x 10-6 = -1.610 x 10-6 ΔD = (-0.000 - 0.000 - 0.121) x 105 = -0.121 x 105

Values of ΔH°298 and ΔG°298 at 298.15K for the hydration reaction are found from the heat-of-formation and Gibbs-energy-of-formation data of Table C.4: ΔH°298 = -235,100 - 52,510 - (-241,818) = -45,792 J mol-1 ΔG°298 = -168,490 - 68,460 - (-228,572) = -8,378 J mol-1 For T = 145 + 273.15 = 418.15 K, values of the integrals in Eq. (13.18) are:

Substitution of values into Eq. (13.18) for a reference temperature of 298.15 gives: For T = 320 + 273.15 = 593.15 K, values of the integrals in Eq. (13.18) are: Substitution of values into Eq. (13.18) for a reference temperature of 298.15 gives:

At 418.15K, : ln K = ln -1.9356 and K = 1.443 x 10-1 At 593.15K, : ln K = ln -5.8286 and K = 2.942x 10-3 Application of Eqs. (13.21), (13.22), and (13.24) provides an alternative solution to this example. By Eq. (13.21),

The following results are obtained: T/ K  K0 K1 K2 K 298.15 1 29.366 418.15 1.4025 4.985x10-3 0.9860 1.443x10-1 593.15 1.9894 1.023x10-4 0.9794 2.942x10-3 Clearly, the influence of K1, is far greater than that of K2. This is a typical result, and accounts for the fact that the lines on Fig. 13.2 are nearly linear.

6.6 Relation of Equilibrium Constant to Composition and Pressure Gas-Phase Reactions The standard state for gas is the ideal gas-state of the pure gas at the standard-state pressure P˚ of 1 bar. Since for ideal gas : f˚i = P˚ Thus : and where the constant K is a function of temp..

For a fixed temperature the composition at equilibrium must change with pressure in such a way that constant . Hence an equilibrium expression displaying pressure and composition; P˚ is the standard –state pressure of 1 bar.

For equilibrium mixture assumed to be in ideal solution, each is assumed to be the fugacity of pure species i : For pressure sufficiently low or temperature sufficiently high, the equilibrium behaves essentially as an ideal gas :

6.7 Equilibrium Conversions for Single Reactions Example 13.5 The water-gas-shift reaction, CO(g) + H2O(g) → CO2(g) + H2(g) is carried out under the different sets of conditions described below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an ideal gas.

Calculate if the conditions: The reactants consist of 1 mol of H2O vapor and 1 mol of CO. The temperature is 1,100 K and the pressure is 1 bar. (b) Same as (a) except that the pressure is 10 bar. (c) Same as (a) except that 2 mol of N2 is included in the reactants.

The reactants are 2 mol of H2O and 1 mol of CO The reactants are 2 mol of H2O and 1 mol of CO. Other conditions are the same as in (a). (e) The reactants are 1 mol of H2O and 2 mol of CO. Other conditions are the same as in (a). The initial mixture consists of 1 mol of H2O, 1 mol of CO, and 1 mol of CO2. Other conditions are the same as in (a) (g) Same as (a) except that the temperature is 1,650 K.

is an ideal gas, Eq. (13.28) applies, and here becomes: Solution 13.5 (a) For the given reaction at 1,100 K, 104/T = 9.05, and Fig. 13.2 provides the value. ln K0 or K = 1. For this reaction . Since the reaction mixture is an ideal gas, Eq. (13.28) applies, and here becomes: By Eq. (13.5):

Substitution of these values into Eq. (A) gives: or εe = 0.5 Therefore the fraction of the steam that reacts is 0.5. (b) Since v = 0 , the increase in pressure has no effect on the ideal-gas reaction, and εe is still 0.5.

(c) The N2 does not take part in the reaction, and serves only as a diluent. It does increase the initial number of moles no from 2 to 4, and the mole fractions are all reduced by a factor of 2. However, Eq. (A) is unchanged and reduces to the same expression as before. Therefore, εe is again 0.5.

In this case, the mole fractions at equilibrium are: and Eq. (A) becomes: or εe = 0.667 The fraction of steam that reacts is then 0.667/2 = 0.333

(f) In this case Eq. (A) becomes: Here the expressions for yCO and yH2O are interchanged , but this leaves the equilibrium equation the same as in (d). Therefore εe = 0.667, and the fraction of steam that reacts is 0.667. (f) In this case Eq. (A) becomes: or εe = 0.333 The fraction of steam reacted is 0.333.

At 1,650 K, 104/T = 6.06, and from Fig. 13.2, in K = -1.15 or K = 0.316. Therefore Eq. (A) becomes: or εe = 0.36 The reaction is exothermic, and conversion decreases with increasing temperature. (Must try: Examples 13.6, 13.7,13.8)

6.8 Phase Rule and Duhem’s Theorem for Reacting System This is phase rule for reacting systems. F = 2 – π + N –r where π is number of phases , N number of chemical species and r is number of independent chemical reactions at equilibrium within the system .

6.9 Multireaction Equilibria where j is the reaction index. For gas phase reaction : For the equilibrium mixture is an ideal-gas, (Must try: Examples 13.12, 13.13)

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