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 К_р для газофазной диссоциации двухатомной молекулы. Привести график ожидаемого изменения К_р 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 < θ_rot
z rot = 1 σ ∑ J−0 ∞ ( 2J+1 ) exp⁡(− h 2 J(J+1) 8 π 2 IkT )= 1 2 ∑ J−0 ∞ ( 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 >> θ_rot 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 < θ_vib= hν k MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGObGaeqyVd4gabaGaam4Aaaaaaaa@3992@ z vib = ∑ V−0 ∞ exp⁡( −V hν kT ) = 1 1−exp⁡( − hν kT ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBaaaleaacaWG2bGaamyAaiaadkgaaeqaaOGaeyypa0ZaaabCaeaaciGGLbGaaiiEaiaacchadaqadaqaaiabgkHiTiaadAfadaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaSqaaiaadAfacqGHsislcaaIWaaabaGaeyOhIukaniabggHiLdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaeyOeI0IaciyzaiaacIhacaGGWbWaaeWaaeaacqGHsisldaWcaaqaaiaadIgacqaH9oGBaeaacaWGRbGaamivaaaaaiaawIcacaGLPaaaaaaaaa@5929@

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

Группируя всё выше, получаем:

при T < θ_rot

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 ⋅( 1−exp⁡( − θ vib T ) ) ∑ J−0 ∞ { ( 2J+1 )exp⁡( − θ rot J(J+1) T ) } ≈const⋅exp⁡( − 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 ln⁡Τ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGOmaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C9@

При θ_rot << T < θ_vib

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 ⋅( 1−exp⁡( − θ vib T ) )≈const⋅exp⁡( − 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 ln⁡Τ + const MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaaMb8Uaam4samaaBaaaleaacaWGqbaabeaakiaadccacqGHijYUcaWGGaGaeyOeI0YaaSaaaeaacqqHuoarcaWGfbaabaGaamOuaiaadsfaaaGaeyiiaaIaey4kaSIaeyiiaaIaaGymaiaacYcacaaI1aGaeyiiaaIaciiBaiaac6gacaWGKoGaeyiiaaIaey4kaSIaeyiiaaIaam4yaiaad+gacaWGUbGaam4Caiaadshaaaa@52C8@

При T >> θ_vib

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 ⋅=const⋅exp⁡( − 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 ln⁡Τ + 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 K_P = –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(D_n)H_4-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^.1,6^4.10^–16= 0,0112^.6,55^.10^–16 = 7,34^.10^–18.

Минимальная работа по выделению 1 моля ¹³СD₄ составит

δW_min = – RT ln(x(¹³СD₄)) = 8,314^.300^.39,45 = 98,4 кДж моль^–1