52. (4/1-97).Для процесса диссоциации идеального газа А₂ = 2А выразить в явном виде зависимость константы равновесия K_P от степени диссоциации a , измеряемой в изобарном и изохорном процессах. При каком начальном давлении P^o(A₂) будет достигаться a = 0,5 в этих случаях, если K_P = 1 бар?

Решение.

При проведении процесса в изобарных условиях:

K_P = P A 2 P A 2 = P Σ (2ξ) 2 ( n 0 −ξ)( n 0 +ξ) = P Σ (2ξ) 2 ( n 0 ) 2 − ξ 2 = P Σ (2α) 2 1− α 2 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6C4E@ ,

откуда при K_P = 1 и α = 0,5, P⁰(A₂) = PS = 0,75 бар.

Если процесс вести изохорно, то в равновесном состоянии

K_P = P A 2 P A 2 = RT V (2ξ) 2 ( n 0 −ξ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGqbWaa0baaSqaaiaadgeaaeaacaaIYaaaaaGcbaGaamiuamaaBaaaleaacaWGbbWaaSbaaWqaaiaaikdaaeqaaaWcbeaaaaGccqGH9aqpdaWcaaqaaiaadkfacaWGubaabaGaamOvaaaadaWcaaqaaiaacIcacaaIYaGaeqOVdGNaaiykamaaCaaaleqabaGaaGOmaaaaaOqaaiaacIcacaWGUbWaaWbaaSqabeaacaaIWaaaaOGaeyOeI0IaeqOVdGNaaiykaaaaaaa@49B9@ = n 0 RT V (2α) 2 (1−α) = P A 2 нач (2α) 2 (1−α) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGUbWaaWbaaSqabeaacaaIWaaaaOGaamOuaiaadsfaaeaacaWGwbaaamaalaaabaGaaiikaiaaikdacqaHXoqycaGGPaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaiikaiaaigdacqGHsislcqaHXoqycaGGPaaaaiabg2da9iaadcfadaqhaaWcbaGaamyqamaaBaaameaacaaIYaaabeaaaSqaaiaad2dbcaWGWqGaam4reaaakmaalaaabaGaaiikaiaaikdacqaHXoqycaGGPaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaiikaiaaigdacqGHsislcqaHXoqycaGGPaaaaaaa@5335@ .

После подстановки K_P = 1 и α = 0,5 находим: P⁰(A₂) = 0,5 бар.

67. (3/1-99)*. Для реакции диссоциации Br₂ (газ) = 2Br (газ)

зависимость константы равновесия от температуры в единицах СИ имеет вид: ln K_P = –23009/T + 0,663 lnT + 8,12

Оценить энергию связи в молекуле Br₂.

Решение. K P =exp⁡( − Δ r G 0 RT )=exp⁡( − Δ r H 0 (T) RT )exp⁡( Δ r S 0 (T) R ) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4samaaBaaaleaacaWGqbaabeaakiabg2da9iGacwgacaGG4bGaaiiCamaabmaabaGaeyOeI0YaaSaaaeaacqqHuoardaWgaaWcbaGaamOCaaqabaGccaWGhbWaaWbaaSqabeaacaaIWaaaaaGcbaGaamOuaiaadsfaaaaacaGLOaGaayzkaaGaeyypa0JaciyzaiaacIhacaGGWbWaaeWaaeaacqGHsisldaWcaaqaaiabfs5aenaaBaaaleaacaWGYbaabeaakiaadIeadaahaaWcbeqaaiaaicdaaaGccaGGOaGaamivaiaacMcaaeaacaWGsbGaamivaaaaaiaawIcacaGLPaaaciGGLbGaaiiEaiaacchadaqadaqaamaalaaabaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4uamaaCaaaleqabaGaaGimaaaakiaacIcacaWGubGaaiykaaqaaiaadkfaaaaacaGLOaGaayzkaaaaaa@5EA2@

ln⁡ K P =− Δ r H 0 (T) RT + Δ r S 0 (T) R =− Δ r H 298 0 + Δ r c p ⋅( T−298 ) RT + Δ r S 298 0 + Δ r c p ⋅ln⁡ T 298 R =    =− Δ r H 298 0 −298⋅ Δ r c p RT + Δ r c p R ln⁡T+( Δ r c p ( 1−ln⁡298 )+ Δ r S 298 0 R ). MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@B3E8@

Δ r H 298 0 ≈(23009+298⋅0,663)R=192,94 кДж/моль. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamisamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaaIWaaaaOGaeyisISRaaiikaiaaikdacaaIZaGaaGimaiaaicdacaaI5aGaey4kaSIaaGOmaiaaiMdacaaI4aGaeyyXICTaaGimaiaacYcacaaI2aGaaGOnaiaaiodacaGGPaGaamOuaiabg2da9iaaigdacaaI5aGaaGOmaiaacYcacaaI5aGaaGinaiaaykW7caqG6qGaaeifeiaabAdbcaqGVaGaaeipeiaab6dbcaqG7qGaaeiteiaab6caaaa@5AB9@

Энергию связи в современной справочной литературе определяют как Δ r U 298 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamyvamaaDaaaleaacaaIYaGaaGyoaiaaiIdaaeaacaaIWaaaaaaa@3C82@ (для стандартного состояния вещества):

Δ r U 298 0 = Δ r H 298 0 − Δ r ( PV ) 298 = Δ r H 298 0 −298R=190,46  кДж/моль; MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6819@

либо как Δ r U 0K 0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaamyvamaaDaaaleaacaaIWaGaam4saaqaaiaaicdaaaaaaa@3BCB@ = 23009 R = 191,3 кДж/моль

72. (4/1-08).Газообразный углеводород A участвует в двух реакциях, приводящих к получению изомеров В и С:

A ↔ B и A ↔ С.

Значения стандартных энтальпий, энтропий и потенциалов Гиббса образования указанных веществ при 1000 К приведены в таблице

Определите равновесный состав при 1000 К. Какой из изомеров будет преобладать при последующем повышении температуры?

Решение.

K P1 =exp⁡( − Δ r1 G 0 RT )=exp⁡( − −740500+734700 8,31⋅1000 )=exp⁡(+0,698)=2. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6B14@

K P2 =exp⁡( − Δ r2 G 0 RT )=exp⁡( − −737000+734700 8,31⋅1000 )=exp⁡(+0,277)=1,32. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6D38@

Состав смеси:

x A = P A P A + P B + P C = 1 1+ K P1 + K P2 =0,23 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaWGbbaabeaakiabg2da9maalaaabaGaamiuamaaBaaaleaacaWGbbaabeaaaOqaaiaadcfadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGqbWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaamiuamaaBaaaleaacaWGdbaabeaaaaGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaey4kaSIaam4samaaBaaaleaacaWGqbGaaGymaaqabaGccqGHRaWkcaWGlbWaaSbaaSqaaiaadcfacaaIYaaabeaaaaGccqGH9aqpcaaIWaGaaiilaiaaikdacaaIZaaaaa@4F6D@

x B = P B P A + P B + P C = K P1 1+ K P1 + K P2 =0,46 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaWGcbaabeaakiabg2da9maalaaabaGaamiuamaaBaaaleaacaWGcbaabeaaaOqaaiaadcfadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGqbWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaamiuamaaBaaaleaacaWGdbaabeaaaaGccqGH9aqpdaWcaaqaaiaadUeadaWgaaWcbaGaamiuaiaaigdaaeqaaaGcbaGaaGymaiabgUcaRiaadUeadaWgaaWcbaGaamiuaiaaigdaaeqaaOGaey4kaSIaam4samaaBaaaleaacaWGqbGaaGOmaaqabaaaaOGaeyypa0JaaGimaiaacYcacaaI0aGaaGOnaaaa@514F@

x C = P C P A + P B + P C = K P2 1+ K P1 + K P2 =0,3 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBaaaleaacaWGdbaabeaakiabg2da9maalaaabaGaamiuamaaBaaaleaacaWGdbaabeaaaOqaaiaadcfadaWgaaWcbaGaamyqaaqabaGccqGHRaWkcaWGqbWaaSbaaSqaaiaadkeaaeqaaOGaey4kaSIaamiuamaaBaaaleaacaWGdbaabeaaaaGccqGH9aqpdaWcaaqaaiaadUeadaWgaaWcbaGaamiuaiaaikdaaeqaaaGcbaGaaGymaiabgUcaRiaadUeadaWgaaWcbaGaamiuaiaaigdaaeqaaOGaey4kaSIaam4samaaBaaaleaacaWGqbGaaGOmaaqabaaaaOGaeyypa0JaaGimaiaacYcacaaIZaaaaa@5091@

При увеличении температуры:

∂ ∂T ( x B x A )= ∂ ∂T ( P B P A )= ∂ K P1 ∂T = K P1 Δ r1 H 0 R T 2 =0. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5E43@

∂ ∂T ( x C x A )= ∂ ∂T ( P C P A )= ∂ K P2 ∂T = K P2 Δ r2 H 0 R T 2 = K P2 ⋅1,1⋅ 10 −3  > 0. MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F0F@

В равновесной смеси при повышении температуры будет повышаться мольная доля вещества С, соотношение B:A будет постоянным. Однако рост K_P2 очень “медленный” – порядка 0,1 % на 1 K, поэтому преобладать будет изомер B.

96. (4/2-00).Для систем

CuSO_4× 5H₂O = CuSO_4× 3H₂O + 2H₂O (1)

CuSO_4× 3H₂O = CuSO_4× H₂O + 2H₂O (2)

CuSO_4× H₂O = CuSO₄ + H₂O (3)

давление насыщенного пара при 50 °С равно соответственно 47, 30 и 4,4 торр. В герметичный сосуд небольшого объема поместили 1 моль безводного сульфата меди, откачали и затем стали вводить пары воды при данной температуре. Нарисовать зависимость давления паров воды в системе от количества введенных молей воды. Пояснить приведенный рисунок.

Решение. При давлении паров воды менее 4,4 торр, насыщение парами воды не достигается ни д ля одной реакции. Давление паров воды растет пропорционально количеству введенной воды. После достижения давления 4,4 торр (n_H2O = δ) равновесно сосуществуют моногидрат сульфата меди и безводный сульфат, соотношение этих фаз определяется количеством введенной воды, но, пока сосуществуют обе фазы, давление паров воды составляет 4,4 торр. Максимальное количество воды, при вводе которого Р_H2O = 4,4 торр, составяет δ + 1 моль.

Дальнейшее добавление в систему воды приведет к дальнейшему росту давления паров воды, пока не будет достигнуто давление 30 торр (количество введенной воды составит 6,8δ + 1 моль). При дальнейшем добавлении воды происходит превращение моногидрата в тригидрат. Сосуществование этих фаз в равновесии требует давления паров воды 30 торр. Полное превращение моногидрата в тригидрат будет достигнуто при n_H2O = 6,8^.δ + 3 моль.

При достижении давления 47 торр в равновесии находится реакция (1). Равновесное давление 47 торр поддерживается в диапазоне введенной воды от 10,7^.δ + 3 моль до 10,7^.δ + 5 моль. Дальше давление паров воды опять будет линейно повышаться с увеличением количества введенной воды.

107. (2/1-07).Зависимости Δ r G 0 (T) MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiaacIcacaWGubGaaiykaaaa@3C6F@ реакций окисления ряда металлов, графита и СО (диаграммы Эллингхэма) приведены на рисунке. Определите: 1) при какой T и какие металлы могут самопроизвольно восста-навливаться из соответ-ствующих оксидов;
2) при какой T и какие металлы можно восста-новить монооксидом углерода;
3) при какой T и какие металлы можно восстано-вить графитом?
Парциальные давления всех газообразных веществ считать равными 1 атм.

Решение. Условие самопроизвольного процесса Δ r G <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaaaaOGaeyipaWJaaGimaaaa@3B41@ , при парциальных давлениях 1 атм это условие переходит в Δ r G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabgYda8iaaicdaaaa@3BFB@ .

Самопроизвольно восстанавливаются:
Ag при Т > 490 К по реакции Ag₂O = 2Ag + 0,5O₂, − Δ r1 G 0 (T>490)<0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaeuiLdq0aaSbaaSqaaiaadkhacaaIXaaabeaakiaadEeadaqhaaWcbaaabaGaaGimaaaakiaacIcacaWGubGaeyOpa4JaaGinaiaaiMdacaaIWaGaaiykaiabgYda8iaaicdaaaa@4318@ и Cu при Т > 1720 К по реакции CuO = Cu + 0,5O₂, − Δ r2 G 0 (T>1720)<0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaeuiLdq0aaSbaaSqaaiaadkhacaaIYaaabeaakiaadEeadaqhaaWcbaaabaGaaGimaaaakiaacIcacaWGubGaeyOpa4JaaGymaiaaiEdacaaIYaGaaGimaiaacMcacqGH8aapcaaIWaaaaa@43D0@ . Cd и Ca не восстанавливаются.

Монооксидом углерода восстанавливаются Ag, Cu и Сd при любой температуре: Ag₂O +CO = 2Ag + CO₂, Δ r G 0 = Δ r4 G 0 − Δ r1 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGinaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaigdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4807@ ,

CuO + CO = Cu + CO₂, Δ r G 0 = Δ r4 G 0 − Δ r2 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGinaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaikdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4808@ ,

Не восстанавливается кальций.

Графитом восстанавливаются:

Ag, при любой T: Ag₂O + C = 2Ag + CO, Δ r G 0 = Δ r5 G 0 − Δ r1 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGynaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaigdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4808@ ,

Cu при Т > 280 К: CuO + C = Cu + CO, Δ r G 0 = Δ r5 G 0 − Δ r2 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGynaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaikdaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@4809@ ,

Cd при Т > 850 К: CdO + C = Cd + CO, Δ r G 0 = Δ r5 G 0 − Δ r3 G 0 <0 MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaadkhaaeqaaOGaam4ramaaCaaaleqabaGaaGimaaaakiabg2da9iabfs5aenaaBaaaleaacaWGYbGaaGynaaqabaGccaWGhbWaa0baaSqaaaqaaiaaicdaaaGccqGHsislcqqHuoardaWgaaWcbaGaamOCaiaaiodaaeqaaOGaam4ramaaDaaaleaaaeaacaaIWaaaaOGaeyipaWJaaGimaaaa@480A@ .

Не восстанавливается кальций.