hermodynamcs, Statstcal hermodynamcs, and Knetcs 4 th Edton,. Engel & P. ed Ch. 6 Part Answers to Selected Problems Q6.. Q6.4. If ξ =0. mole at equlbrum, the reacton s not ery far along. hus, there would be few products and many reactants, whch leads to a alue of K eq that s small ( K eq <). Usng the relaton ΔG = ln K eq, for K eq < we would obtan ΔG > 0. hus, snce ΔG s calculated as the dfference between standard Gbbs free energes of formaton for products and reactants, ΔG = ν ΔG f,, we predct that the standard Gbbs free energes of formaton of the products are larger than those of the reactants. he second reacton has stochometry that s twce that of the frst reacton. Snce the equlbrum constant N ν p n terms of partal pressures, K p, s defned as K p =, wth the partal pressure rased to the power of the stochometrc coeffcent, the equlbrum constant for the second reacton s the square of the. frst, K p = K p P N Q6.. For the reacton CO (g) + ½ O (g) D CO (g), the standard enthalpy of formaton s ΔH f = 83.0 kj/mol. As a result, we can wrte the equlbrum wth heat eoled, CO (g) + ½ O (g) D CO (g) + heat. If the temperature s ncreased (at constant pressure), the reacton wll shft to the left to alleate the stress of added heat. hus, the partal pressure of CO wll decrease. Q6.3. For the reacton CO (g) + ½ O (g) D CO (g), there are 3/ total moles of gas phase speces on the reactant sde and one mole on the product sde. If the pressure s ncreased (at constant temperature), the reacton wll shft to the rght to alleate the stress of added pressure. hus, the partal pressure of CO wll ncrease. Q6.4. In the case of addton of a non-reacte gas, t s helpful to dde the expresson for the equlbrum constant nto a pressure-ndependent porton and a pressure-dependent porton, K eq = K x P P, where K x s the mole fracton quotent, K x = x, and s the sum of the stochometrc coeffcents, =.
Q6.4. contnued Addton of an nert gas mpacts the mole fractons, so t s further helpful to consder that the mole fractons x may be wrtten as x = n n tot. Here, n s the number of moles of speces and n tot s the total number of gas phase moles, ncludng those of the added nert gas. Substtutng, the equlbrum constant expresson becomes K eq = P P n. n tot Snce the total moles s ndependent of the ndex, t may be moed outsde the product notaton to ge the equlbrum constant expresson as K eq = P n tot P n. For ths partcular example, CO (g) + ½ O (g) D CO (g), the sum of the stochometrc coeffcents s = CO + O + CO = + =. Snce the sum of the stochometrc coeffcents s negate, at constant pressure and temperature, f an nert P gas s added to the reacton essel, the term n tot P must ncrease because the total number of gas phase moles n tot ncreases (and t effectely appears n the numerator f s negate). In order for the oerall equlbrum constant K eq to mantan a constant alue, the expresson nolng the moles of each speces, n, must therefore decrease. hs means that products wll decrease and reactants ncrease, and the reacton wll shft to the left. herefore, the partal pressure of CO wll decrease. Q6.. For the reacton Cl (g) D Cl (g), there s one mole of gas phase speces on the reactant sde and two moles on the product sde. If the pressure s ncreased (at constant temperature), the reacton wll shft to the left, away from the sde wth more gas phase moles, n order to alleate the stress of added pressure. hus, the amount of Cl dssocated nto Cl atoms depends upon the total pressure een though the K eq alue s ndependent of pressure.
3 P6.6a. ΔG =.75 kj/mol P6.6b. K eq = 5.4 0 49 P6.6c. he process s not spontaneous. ( ) ( 3 ξ) ( ) 3 4 + ξ P6.7a. K eq = 3ξ P6.7b. p N = 0.5 bar p H = 0.0 bar p NH3 = 0.5 bar P6.7c. ξ = 0.67 mol P6.a. K eq = P6.b. K eq = ξ 3/.75+ 0.5ξ P P ( ) / (.75 ξ) P P ξ 3/ ( ) P 3/ P.75 / P6.c. ξ = 0.088 degree of dssocaton = 0.064 P6.d. ξ = 0.085 degree of dssocaton = 0.063 relate error n ξ =.0% / P6.3. K eq at 98 K =.9 0 6 K eq at 595 K =9. Snce the reacton s exothermc, ncreasng the temperature shfts the reacton to the left, reducng the equlbrum constant. So, as the temperature s ncreased further, the equlbrum constant s expected to become een smaller. P6.5a. K eq at 05 K = 0.8 P6.5b. K eq at 98 K =.58 0 8 P6.5c. ΔG ( 98 K) =00.4 kj/mol P6.6a. ΔG =.5 kj/mol P6.6b. K eq =.08 0 4 P6.6c. he proten s structurally stable under these condtons. P6.8a. K eq = 0.885 ΔG ( 50 K) =.0 kj/mol P6.8b. ΔG ( 98 K) = 7.9 kj/mol
4 P6.3a. ΔG = 3.0 kj/mol ΔH = 9.0 kj/mol ΔS =.6 J/mol-K P6.3b. x CO = 0.408 P6.3. ΔG = 09.7 kj/mol relate change = 9.7% P6.33. ΔG ( 98 K) = 363 kj/mol ( ) = 364 kj/mol ΔG r 30 K he calculaton of ΔG r at 30 K was made assumng that was ndependent of temperature and approxmatng the partal derate n the Gbbs-Helmholtz equaton as a fnte dfference. P6.35a. o dere the requested result, we start wth the equaton gen (a form of the an't Hoff equaton), dln K eq = K K. Expressng the -dependence of as gen n the problem, we hae ( ) = ( ) + ( ). Note that n ths expresson s assumed to be fxed (t corresponds to the ntal temperature) and s the arable oer whch the ntegraton takes place. Also, the term corresponds to the sum of the molar heat capactes tmes the respecte stochometrc coeffcents for all the speces n the reacton, = N ν C p,m,. Substtutng nto the ntegral (and ntegratng the left sde) yelds, ln K ln K = = ln K ln K = ( ) ( ) ( ) + ( ) +.
5 P6.35a. contnued In the last lne aboe, the terms that are constant (ncludng ) are pulled out of the ntegrals. Integratng leads to the expresson, ln K ln K = or P6.35b. p O =.7 0 6 bar ( ) + ln K = ln K + ΔH ( ) + ln. he equlbrum pressure of O s equal to K P ( K eq ) for the reacton because the other speces n the reacton are solds. herefore, the acttes for the solds are equal to and they drop out of the equlbrum constant expresson. P6.35c. p O = 4.3 0 6 bar P6.36a. For the reacton 3 O (g) D O 3 (g), there are three moles of gas phase speces on the reactant sde and two moles on the product sde. If the pressure s ncreased (at constant temperature), the reacton wll shft to the rght, toward products and away from the sde wth more gas phase moles, n order to alleate the stress of added pressure. P6.36b. For the reacton 3 O (g) D O 3 (g), the standard enthalpy of reacton s = 85.4 kj/mol. As a result, we can wrte the equlbrum wth heat absorbed, 3 O (g) + heat D O 3 (g). If the temperature s ncreased (at constant pressure), the reacton wll shft to the rght toward products to alleate the stress of added heat. P6.36c. K eq 800 K K eq ( ) =.5 0 6 ( 900 K) =.79 0 4 P6.36d. K x =.5 0 6 for P =.00 bar K x = 3.80 0 6 for P =.50 bar P6.37. ΔG =.68 kj/mol