Thermodynamics

moles of the reactants and increases the number of moles of products if ∆nis positive, have the opposite effect if ∆n is negative, and have no effect if∆n is zero.6. When the stoichiometric coefficients are doubled, the value of K P issquared. Therefore, when one is using K P values from a table, the stoichiometriccoefficients (the n’s) used in a reaction must be exactly the sameones appearing in the table from which the K P values are selected. Multiplyingall the coefficients of a stoichiometric equation does not affect the massbalance, but it does affect the equilibrium constant calculations since thestoichiometric coefficients appear as exponents of partial pressures inEq. 16–13. For example,ForBut forH 2 1 2 O 2 ∆ H 2 O K P1 P H 2 OP H2P 1>22H 2 O 2 ∆ 2H 2 OK P27. Free electrons in the equilibrium composition can be treated as an idealgas. At high temperatures (usually above 2500 K), gas molecules start todissociate into unattached atoms (such as H 2 ∆ 2H), and at even highertemperatures atoms start to lose electrons and ionize, for example,H ∆ H e P 2 H 2 OP 2 H 2P O2 1K P12 2(16–16)The dissociation and ionization effects are more pronounced at low pressures.Ionization occurs to an appreciable extent only at very high temperatures,and the mixture of electrons, ions, and neutral atoms can be treated asan ideal gas. Therefore, the equilibrium composition of ionized gas mixturescan be determined from Eq. 16–15 (Fig. 16–11). This treatment may not beadequate in the presence of strong electric fields, however, since the electronsmay be at a different temperature than the ions in this case.8. Equilibrium calculations provide information on the equilibrium compositionof a reaction, not on the reaction rate. Sometimes it may even takeyears to achieve the indicated equilibrium composition. For example, theequilibrium constant of the reaction H 2 1 2O 2 ∆ H 2 O at 298 K is about10 40 , which suggests that a stoichiometric mixture of H 2 and O 2 at roomtemperature should react to form H 2 O, and the reaction should go to completion.However, the rate of this reaction is so slow that it practically doesnot occur. But when the right catalyst is used, the reaction goes to completionrather quickly to the predicted value.O 2nHN neH + N e –P∆nChapter 16 | 801H → H + + e –K P =nHN HwhereN total = N H + N H + + N e –∆ n= n H + + n e – – nH= 1 + 1 – 1= 1(N totalFIGURE 16–11Equilibrium-constant relation for theionization reaction of hydrogen.(EXAMPLE 16–3Equilibrium Compositionat a Specified TemperatureA mixture of 2 kmol of CO and 3 kmol of O 2 is heated to 2600 K at a pressureof 304 kPa. Determine the equilibrium composition, assuming the mixtureconsists of CO 2 , CO, and O 2 (Fig. 16–12).Initialcomposition2 kmol CO3 kmol O 2Equilibriumcomposition at2600 K, 304 kPax CO 2y COz O 2Solution A reactive gas mixture is heated to a high temperature. The equilibriumcomposition at that temperature is to be determined.FIGURE 16–12Schematic for Example 16–3.