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Версия для печати | Главная > Образование > Кафедра физической химии > Химическая термодинамика > ... > Задачи по химической термодинамике > Решения некоторых задач

1. Основные понятия химической термодинамики

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2. Химическое равновесие в гомогенных системах

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3. Химическое равновесие в гетерогенных системах

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4. Химическое равновесие в системах, содержащих растворы электролитов

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5. Термодинамика дисперсных систем и поверхностных явлений

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6. Использование статистических методов для описания химического равновесия в идеальных системах

293. (1/3-97).* Используя методы статистической термодинамики, найти температурную зависимость константы равновесия Кр для газофазной диссоциации двухатомной молекулы. Привести график ожидаемого изменения Кр в широкой области температур.

Решение.

A2à 2A

K P =exp( ΔE RT ) ( kT P 0 V ) Δν i ( z i ' ) ν i = = kT P 0 V z trA 2 z tr A 2 z rot 1 z vib 1 ( 2 s A +1 ) 2 2 s A 2 +1 exp( ΔE RT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@890C@

z trA 2 z tr A 2 = ( 2πmkT h 2 ) 3 V 2 ( 4πmkT h 2 ) 3/2 V = ( πmkT h 2 ) 3/2 V MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@642B@ .

Для вращательной статсуммы при T < θrot
z rot = 1 σ J0 ( 2J+1 ) exp( h 2 J(J+1) 8 π 2 IkT )= 1 2 J0 ( 2J+1 ) exp( h 2 J(J+1) 4 π 2 m r 2 kT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7B55@ .

При T >> θrot     z rot = 2 π 2 m r 2 kT h 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWGYbGaam4BaiaadshaaeqaaOGaeyypa0ZaaSaaaeaacaaIYaGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGaamyBaiaadkhadaahaaWcbeqaaiaaikdaaaGccaWGRbGaamivaaqaaiaadIgadaahaaWcbeqaaiaaikdaaaaaaaaa@4503@ .

Для колебательной статсуммы

при T < θvib= hν k MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGObGaeqyVd4gabaGaam4Aaaaaaaa@3992@       z vib = V0 exp( V hν kT ) = 1 1exp( hν kT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaabCaeaaciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTiaadAfadaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaSqaaiaadAfacqGHsislcaaIWaaabaGaeyOhIukaniabggHiLdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaeyOeI0IaciyzaiaacIhacaGGWbWaaeWaaeaacqGHsisldaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaaaaaa@5929@

При T >> θvib      z vib = T θ vib = kT hν MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaSaaaeaacaWGubaabaGaeqiUde3aaSbaaSqaaiaadAhacaWGPbGaamOyaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGRbGaamivaaqaaiaadIgacqaH9oGBaaaaaa@4621@ .

Группируя всё выше, получаем:

при T < θrot

K P = 2 ( πm ) 3 2 h 3 ( 2 s A +1 ) 2 ( 2 s A 2 +1 ) exp( ΔE RT ) (kT) 5 2 P 0 ( 1exp( θ vib T ) ) J0 { ( 2J+1 )exp( θ rot J(J+1) T ) } constexp( ΔE RT ) (kT) 5 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@A7B5@

или ln K P ΔE RT + 2,5 lnΤ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGOmaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C9@

При θrot << T < θvib

K P = m π 1 r 2 h ( 2 s A +1 ) 2 ( 2 s A 2 +1 ) exp( ΔE RT ) (kT) 3 2 P 0 ( 1exp( θ vib T ) )constexp( ΔE RT ) (kT) 3 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@8904@

или         ln K P ΔE RT + 1,5 lnΤ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGymaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C8@

При T >> θvib

K P = m π ν vib r 2 ( 2 s A +1 ) 2 ( 2 s A 2 +1 ) exp( ΔE RT ) (kT) 1 2 P 0 =constexp( ΔE RT ) (kT) 1 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7D0D@

или      ln K P ΔE RT + 0,5 lnΤ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGimaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C7@

(сравните с задачей 67. (3/1-99), решенной выше:
для диссоциации Br2 ln KP = –23009/T + 0,663 lnT + 8,12)

311. (5/Э-04).* Природное содержание изотопа 13С составляет 1,12 % всего углерода, а дейтерия 2Н – 1,6 10–4 всего водорода. Определить минимальную работу, необходимую для выделения 1 моля “сверхтяжелого” метана 13СD4 в изобарном процессе при 300 К. Предполагается, что чистый метан доступен в неограниченных количествах.

Решение. Необходимо найти равновесную долю сверхтяжелого метана.
В равновесии доля mС(Dn)H4-n пропорциональна коэффициентам в произведении биномов Ньютона

(0,9888.12С + 0,0112.13С). (0,9998.H + 1,6.10–4.D)4

при соответствующих степенях изотопов, т. е. для 13СD4:

x(13СD4) = 0,0112.1,64.10–16 = 0,0112.6,55.10–16 = 7,34.10–18.

Минимальная работа по выделению 1 моля 13СD4 составит

δWmin = – RT ln(x(13СD4)) = 8,314.300.39,45 = 98,4 кДж моль–1



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