Использование статистических методов для описания химического равновесия в идеальных системах (решение задач)
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.
Для вращательной статсуммы при T < θrot zrot=1σ∑J−0∞(2J+1)exp(−h2J(J+1)8π2IkT)=12∑J−0∞(2J+1)exp(−h2J(J+1)4π2mr2kT)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@7B55@.
При T >> θrotzrot=2π2mr2kTh2
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWGYbGaam4BaiaadshaaeqaaOGaeyypa0ZaaSaaaeaacaaIYaGaeqiWda3aaWbaaSqabeaacaaIYaaaaOGaamyBaiaadkhadaahaaWcbeqaaiaaikdaaaGccaWGRbGaamivaaqaaiaadIgadaahaaWcbeqaaiaaikdaaaaaaaaa@4503@.
Для колебательной статсуммы
при T < θvib=hνk
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGObGaeqyVd4gabaGaam4Aaaaaaaa@3992@zvib=∑V−0∞exp(−VhνkT)=11−exp(−hνkT)
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaabCaeaaciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTiaadAfadaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaSqaaiaadAfacqGHsislcaaIWaaabaGaeyOhIukaniabggHiLdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaeyOeI0IaciyzaiaacIhacaGGWbWaaeWaaeaacqGHsisldaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaaaaaa@5929@
При T >> θvibzvib=Tθvib=kThν
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaSaaaeaacaWGubaabaGaeqiUde3aaSbaaSqaaiaadAhacaWGPbGaamOyaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGRbGaamivaaqaaiaadIgacqaH9oGBaaaaaa@4621@.
или lnKP≈−ΔERT+2,5lnΤ+const
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGOmaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C9@
или
lnKP≈−ΔERT+1,5lnΤ+const
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGymaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C8@
или
lnKP≈−ΔERT+0,5lnΤ+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