четверг, 7 февраля 2013 г.

задачи на химическое равновесие алгоритм решения

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Обратите внимание: в справочнике при 298 К

MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadkebcaWGYqaabeaaaaa@395F@

Отсюда следует ответ:

MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqyWdi3aaSbaaSqaaiaadAdbaeqaaaaa@3898@

Tпл = 505 K; M = 0.119 кг/моль;

MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdq0aaSbaaSqaaiaad+dbcaWG7qaabeaakmaanaaabaGaamisaaaacqGH9aqpcaaI3aGaaiilaiaaikdacaaMe8UaaeOoeiaabsbbcaqG2qGaae4laiaabYdbcaqG+qGaae4oeiaabYebaaa@449B@

=7,2 кДж/моль

      

MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaamiuaaqaaiaadsgacaWGubaaaiabg2da9maalaaabaGaaGymaaqaaiaaicdacaGGUaGaaGimaiaaicdacaaIZaGaaGOmaaaacaqGXqGaaeimeiaabcebcaqGVaGaaeOgeiabg2da9iaaiodacaGGSaGaaGymaiaaikdacqGHflY1caaIXaGaaGimamaaCaaaleqabaGaaG4naaaakiaab+bbcaqGWqGaae4laiaabQbbcaGG7aaaaa@4F40@

бар/К=3,12⋅

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Решение. Для решения необходимо использовать уравнение Клаузиуса Клапейрона:

Молекулярная масса олова 119.

ж = 6,988 г/см3.

тв, учитывая, что Qпл = 7,2 кДж/моль и r

Найти плотность твердого олова r

 (1/2-00). Зависимость температуры плавления Sn от давления (в бар) передается выражением: t(`С) = 231,8 + 0,0032(P 1).

Химическое равновесие в гетерогенных системах (решение задач)

Химическое равновесие в гетерогенных системах (решение задач)

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