IRN-2 C-0 C-1

IRN-2 C-0 C
Ehsan Naderi Donig

IChO
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‫‪G0-1‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-0 G-1‬‬‫ﺍﻟﻤﭙﯿﺎﺩ ﺑﯿﻦﺍﻟﻤﻠﻠﯽ ﺷﯿﻤﯽ ‪ 2021‬ﮊﺍﭘﻦ‬‫ﺩﻭﺭﻩ ﭘﻨﺠﺎﻩﻭﺳﻮﻡ‪ IChO2021 ،‬ﮊﺍﭘﻦ‬‫‪ 25‬ﺟﻮﻻﯼ ﺗﺎ ‪ 2‬ﺁﮔﻮﺳﺖ ‪2021‬‬‫‪https://www.icho2021.org‬‬‫‪IChO‬‬‫‪G0-2‬‬‫‪IRN-2 C-0 G-2‬‬‫)‪persian (Iran‬‬‫ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ ﻋﻤﻮﻣﯽ‬‫• ﺑﺮﺍﯼ ﻧﻮﺷﺘﻦ ﭘﺎﺳﺦﻫﺎ ﻓﻘﻂ ﻣﺠﺎﺯﯾﺪ ﺍﺯ ﺧﻮﺩﮐﺎﺭ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫• ﻣﺎﺷﯿﻦﺣﺴﺎﺏ ﺷﻤﺎ ﺑﺎﯾﺪ ﺍﺯ ﻧﻮﻉ ﻏﯿﺮ ﻗﺎﺑﻞ ﺑﺮﻧﺎﻣﻪﺭﯾﺰﯼ ﺑﺎﺷﺪ‪.‬‬‫• ﺍﯾﻦ ﺁﺯﻣﻮﻥ ﺷﺎﻣﻞ ‪ ۹‬ﻣﺴﺎﻟﻪ ﺍﺳﺖ‪.‬‬‫• ﻣﯽﺗﻮﺍﻧﯿﺪ ﻣﺴﺎﺋﻞ ﺭﺍ ﺑﺎ ﻫﺮ ﺗﺮﺗﯿﺒﯽ ﺣﻞ ﮐﻨﯿﺪ‪.‬‬‫• ﺑﺮﺍﯼ ﺣﻞ ﻣﺴﺎﺋﻞ ‪ ۵‬ﺳﺎﻋﺖ ﺯﻣﺎﻥ ﺩﺍﺭﯾﺪ‪.‬‬‫• ﭘﺲ ﺍﺯ ﺩﺍﺩﻩ ﺷﺪﻥ ﺩﺳﺘﻮﺭ ﺷﺮﻭﻉ ﻣﯽﺗﻮﺍﻧﯿﺪ ﺷﺮﻭﻉ ﺑﻪ ﮐﺎﺭ ﮐﻨﯿﺪ‪.‬‬‫• ﺗﻤﺎﻡ ﻧﺘﺎﯾﺞ ﺑﺎﯾﺪ ﺑﺎ ﺧﻮﺩﮐﺎﺭ ﺩﺭ ﮐﺎﺩﺭﻫﺎﯼ ﻣﺸﺨﺺﺷﺪﻩ ﺩﺭ ﺑﺮﮔﻪﻫﺎﯼ ﭘﺎﺳﺦﻧﺎﻣﻪ ﻧﻮﺷﺘﻪ ﺷﻮﺩ‪ .‬ﻣﯽﺗﻮﺍﻧﯿﺪ ﺍﺯ ﭘﺸﺖ ﺑﺮﮔﻪﻫﺎﯼ ﺳﺆﺍﻝ ﺑﻪﻋﻨﻮﺍﻥ‬‫ﭼﺮﮐﻨﻮﯾﺲ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﺑﻪ ﺟﻮﺍﺏﻫﺎﯼ ﻧﻮﺷﺘﻪﺷﺪﻩ ﺩﺭ ﺧﺎﺭﺝ ﺍﺯ ﮐﺎﺩﺭﻫﺎﯼ ﻣﺸﺨﺺ‪ ،‬ﻧﻤﺮﻩﺍﯼ ﺗﻌﻠﻖ ﻧﻤﯽﮔﯿﺮﺩ‪.‬‬‫ﭘﺎﺳﺦ ﺩﺭﺳﺖ ﺑﺎﯾﺪ‬‫• ﻫﺮ ﺟﺎ ﮐﻪ ﻻﺯﻡ ﺍﺳﺖ ﻣﺤﺎﺳﺒﺎﺕ ﻣﺮﺑﻮﻁ ﺑﻪ ﭘﺎﺳﺦﻫﺎ ﺭﺍ ﻫﻢ ﺩﺭ ﮐﺎﺩﺭﻫﺎ ﺑﻨﻮﯾﺴﯿﺪ‪ .‬ﺑﺮﺍﯼ ﮔﺮﻓﺘﻦ ﻧﻤﺮﻩ ﮐﺎﻣﻞ ﻋﻼﻭﻩ ﺑﺮ ﻧﻮﺷﺘﻦ‬‫ِ‬‫ﻧﺸﺎﻥ ﺩﻫﯿﺪ ﮐﻪ ﺭﻭﯼ ﺳﺆﺍﻝ ﮐﺎﺭ ﮐﺮﺩﻩﺍﯾﺪ‪.‬‬‫ﺩﺳﺘﻮﺭ ﺗﻮﻗﻒ‪ ،‬ﺍﺧﻄﺎﺭ ﺧﻮﺍﻫﺪ ﺩﺍﺩ‪.‬‬‫• ﻣﺮﺍﻗﺐ ﺁﺯﻣﻮﻥ ‪ ۳۰‬ﺩﻗﯿﻘﻪ ﻗﺒﻞ ﺍﺯ‬‫ِ‬‫• ﻭﻗﺘﯽ ﺩﺳﺘﻮﺭ ﺗﻮﻗﻒ ﺩﺍﺩﻩ ﺷﺪﻩ ﺑﺎﯾﺪ ﮐﺎﺭ ﺭﺍ ﻣﺘﻮﻗﻒ ﮐﻨﯿﺪ‪ .‬ﺍﮔﺮ ﺍﺯ ﻧﻮﺷﺘﻦ ﺩﺳﺖ ﺑﺮﻧﺪﺍﺭﯾﺪ ﺑﺮﺍﯼ ﺁﺯﻣﻮﻥ ﺷﻤﺎ ﻧﻤﺮﻩ ﺻﻔﺮ ﻣﻨﻈﻮﺭ ﺧﻮﺍﻫﺪ ﺷﺪ‪.‬‬‫• ﻧﺴﺨﻪ ﺭﺳﻤﯽ ﺩﻓﺘﺮﭼﻪ ﺁﺯﻣﻮﻥ ﺑﻪ ﺯﺑﺎﻥ ﺍﻧﮕﻠﯿﺲ ﺑﻨﺎ ﺑﻪ ﺩﺭﺧﻮﺍﺳﺖ ﺷﻤﺎ ﻭ ﻓﻘﻂ ﺑﺮﺍﯼ ﺭﻓﻊ ﺍﺑﻬﺎﻡ ﺩﺭ ﺍﺧﺘﯿﺎﺭﺗﺎﻥ ﻗﺮﺍﺭ ﺧﻮﺍﻫﺪ ﮔﺮﻓﺖ‪.‬‬‫• ﺷﻤﺎ ﻣﺠﺎﺯ ﻧﯿﺴﺘﯿﺪ ﺑﺪﻭﻥ ﮔﺮﻓﺘﻦ ﺍﺟﺎﺯﻩ ﺻﻨﺪﻟﯽ ﺧﻮﺩ ﺭﺍ ﺗﺮﮎ ﮐﻨﯿﺪ‪ .‬ﺍﮔﺮ ﺑﻪ ﮐﻤﮏ ﻧﯿﺎﺯ ﺩﺍﺷﺘﯿﺪ )ﻣﺸﮑﻞ ﻣﺎﺷﯿﻦ ﺣﺴﺎﺏ‪ ،‬ﻧﯿﺎﺯ ﺑﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺳﺮﻭﯾﺲ‬‫ﺑﻬﺪﺍﺷﺘﯽ‪ ،(... ،‬ﺩﺳﺖ ﺧﻮﺩ ﺭﺍ ﺑﺎﻻ ﺑﺒﺮﯾﺪ ﻭ ﻣﻨﺘﻈﺮ ﻣﺮﺍﻗﺐ ﺑﻤﺎﻧﯿﺪ‪.‬‬‫ﻣﻮﻓﻖ ﺑﺎﺷﯿﺪ!‬‫ﻣﺴﺎﺋﻞ ﻭ ﺑﺎﺭﻡﻫﺎ‬‫ﺷﻤﺎﺭﻩ ﻣﺴﺄﻟﻪ‬
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‫ﺟﻤﻊ ﮐﻞ‬‫ﻋﻨﻮﺍﻥ‬‫ﻧﻤﺮﻩ ﮐﻞ ﺳﺆﺍﻝ‬‫ﺩﺭﺻﺪ ﺩﺭ ﻧﻤﺮﻩ ﮐﻞ‬‫‪35‬‬‫‪11‬‬‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺳﻄﺢ ﯾﮏ ﻓﻠﺰ‬‫‪24‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪ -‬ﺑﯿﺮ؟‬‫‪22‬‬‫ﮐﭙﺴﻮﻝ ﺯﻣﺎﻥ ﺍﯾﺰﻭﺗﻮﭘﯽ‬‫ﺷﯿﻤﯽ ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ ﺭﻭﯼ‬‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﺮﻣﻮﺯ‬‫ﺷﯿﻤﯽ ﺣﺎﻟﺖ ﺟﺎﻣﺪ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ‬‫ﺑﺎﺯﯼ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻭﺋﯿﺪﯼ‬‫ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺁﻟﯽ ﺩﯾﻨﺎﻣﯿﮏ ﻭ ﮐﺎﯾﺮﺍﻟﯿﻪ ﺁﻥﻫﺎ‬‫ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﮐﭙﺴﻮﻝ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﻭ ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﺩﻭﺳﺖ ﻧﺪﺍﺭﺩ‬‫‪32‬‬‫‪60‬‬‫‪45‬‬‫‪36‬‬‫‪26‬‬‫‪23‬‬‫‪11‬‬‫‪8‬‬‫‪11‬‬‫‪12‬‬‫‪13‬‬‫‪13‬‬‫‪11‬‬‫‪10‬‬‫‪100‬‬

‫‪G0-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-0 G-3‬‬‫ﺛﺎﺑﺖﻫﺎﯼ ﻓﯿﺰﯾﮑﯽ ﻭ ﻣﻌﺎﺩﻟﻪﻫﺎ‬‫ﺙﺍﺏﺕﻩﺍ‬‫ﺳﺮﻋﺖ ﻧﻮﺭ ﺩﺭ ﺧﻸ‬‫‪𝑐 = 2.99792458 × 108 m s−1‬‬‫ﺑﺎﺭ ﺑﻨﯿﺎﺩﯼ‬‫‪𝑒 = 1.602176634 × 10−19 C‬‬‫ﺛﺎﺑﺖ ﭘﻼﻧﮏ‬‫ﺟﺮﻡ ﺍﻟﮑﺘﺮﻭﻥ‬‫ﺛﺎﺑﺖ ﺍﻟﮑﺘﺮﯾﮏ )ﮔﺬﺭﺩﻫﯽ ﺧﻸ(‬‫ﺛﺎﺑﺖ ﺁﻭﻭﮔﺎﺩﺭﻭ‬‫ﺛﺎﺑﺖ ﺑﻮﻟﺘﺴﻤﺎﻥ‬‫ﺛﺎﺑﺖ ﻓﺎﺭﺍﺩﯼ‬‫ﺛﺎﺑﺖ ﮔﺎﺯﻫﺎ‬‫ﻭﺍﺣﺪ ﺟﺮﻡ ﺍﺗﻤﯽ‬‫ﻓﺸﺎﺭ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ‬‫ﻓﺸﺎﺭ ﺍﺗﻤﺴﻔﺮﯼ‬‫ﺻﻔﺮ ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ‬‫ﺁﻧﮕﺴﺘﺮﻭﻡ‬‫ﭘﯿﮑﻮﻣﺘﺮ‬‫ﺍﻟﮑﺘﺮﻭﻥ ﻭﻟﺖ‬‫ﺟﺰﺀ ﺩﺭ ﻣﯿﻠﯿﻮﻥ‬‫ﺟﺰﺀ ﺩﺭ ﺑﯿﻠﯿﻮﻥ‬‫ﺟﺰﺀ ﺩﺭ ﺗﺮﯾﻠﯿﻮﻥ‬‫ﻋﺪﺩ ﭘﯽ‬‫ﭘﺎﯾﻪ ﻟﮕﺎﺭﯾﺘﻢ ﻃﺒﯿﻌﯽ )ﻋﺪﺩ ﺍﻭﯾﻠﺮ(‬‫‪ℎ = 6.62607015 × 10−34 J s‬‬‫‪𝑚e = 9.10938370 × 10−31 kg‬‬‫‪𝜀0 = 8.85418781 × 10−12 F m−1‬‬‫‪mol‬‬‫‪−1‬‬‫‪−1‬‬‫‪−1‬‬‫‪JK‬‬‫‪−23‬‬‫‪𝑁A = 6.02214076 × 10‬‬‫‪𝑘B = 1.380649 × 10‬‬‫‪𝐹 = 𝑁A × 𝑒 = 9.64853321233100184 × 10 C mol‬‬‫‪4‬‬‫‪mol‬‬‫‪−1‬‬‫‪23‬‬‫‪mol−1‬‬‫‪−1‬‬‫‪−1‬‬‫‪𝑅 = 𝑁A × 𝑘B = 8.31446261815324 J K‬‬‫‪L atm K‬‬‫‪−2‬‬‫‪= 8.2057366081 × 10‬‬‫‪𝑢 = 1 Da = 1.66053907 × 10−27 kg‬‬‫‪𝑝 = 1 bar = 10 Pa‬‬‫‪5‬‬‫‪𝑝atm = 1.01325 × 10 Pa‬‬‫‪5‬‬‫‪0 C = 273.15 K‬‬
‫∘‬

‫‪m‬‬
‫‪m‬‬

‫‪J‬‬

‫‪−10‬‬‫‪−12‬‬‫‪−19‬‬‫‪−6‬‬‫‪−9‬‬‫‪1 Å = 10‬‬‫‪1 pm = 10‬‬‫‪1 eV = 1.602176634 × 10‬‬‫‪1 ppm = 10‬‬‫‪1 ppb = 10‬‬‫‪−12‬‬‫‪1 ppt = 10‬‬‫‪𝜋 = 3.141592653589793‬‬‫‪𝑒 = 2.718281828459045‬‬

‫‪IChO‬‬‫‪G0-4‬‬‫‪IRN-2 C-0 G-4‬‬‫)‪persian (Iran‬‬‫ﻡﻉﺍﺩﻝﻩﻩﺍ‬‫ﻗﺎﻧﻮﻥ ﮔﺎﺯ ﺍﯾﺪﻩﺍﻝ‬‫ﻗﺎﻧﻮﻥ ﮐﻮﻟﻦ‬‫ﻗﺎﻧﻮﻥ ﺍﻭﻝ ﺗﺮﻣﻮﺩﯾﻨﺎﻣﯿﮏ‬‫ﺁﻧﺘﺎﻟﭙﯽ ‪H‬‬‫ﺁﻧﺘﺮﻭﭘﯽ ﺑﺮ ﻣﺒﻨﺎﯼ ﺍﺻﻞ ﺑﻮﻟﺘﺴﻤﺎﻥ‬‫‪S‬‬‫ﺗﻐﯿﯿﺮ ﺁﻧﺘﺮﻭﭘﯽ 𝑆‪Δ‬‬‫ﺍﻧﺮﮊﯼ ﺁﺯﺍﺩ ﮔﯿﺒﺲ ‪G‬‬
‫ﺧﺎﺭﺝ ﻗﺴﻤﺖ ﻭﺍﮐﻨﺶ ‪Q‬‬

‫𝑇𝑅𝑛 = 𝑉 𝑃‬‫ﮐﻪ ‪ P‬ﻓﺸﺎﺭ‪ V ،‬ﺣﺠﻢ‪ n ،‬ﻣﻘﺪﺍﺭ ﻣﺎﺩﻩ ﻭ ‪ T‬ﺩﻣﺎﯼ ﻣﻄﻠﻖ ﮔﺎﺯ ﺍﯾﺪﻩﺍﻝ ﺍﺳﺖ‪.‬‬‫𝑞 𝑞‬‫‪𝐹 = 𝑘e 1 2 2‬‬‫𝑟‬‫ﮐﻪ ‪ F‬ﻧﯿﺮﻭﯼ ﺍﻟﮑﺘﺮﻭﺍﺳﺘﺎﺗﯿﮑﯽ‪ 𝑘e (≃ 9.0 × 109 N m2 C−2 ) ،‬ﺛﺎﺑﺖ ﮐﻮﻟﻦ‪ 𝑞1 ،‬ﻭ ‪ 𝑞2‬ﻣﻘﺎﺩﯾﺮ ﺑﺎﺭﻫﺎ ﻭ ‪r‬‬‫ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺑﺎﺭﻫﺎﺳﺖ‪.‬‬‫𝑤 ‪Δ𝑈 = 𝑞 +‬‬‫ﮐﻪ 𝑈‪ Δ‬ﺗﻐﯿﯿﺮ ﺍﻧﺮﮊﯼ ﺩﺭﻭﻧﯽ‪ 𝑞 ،‬ﮔﺮﻣﺎﯼ ﺩﺍﺩﻩﺷﺪﻩ ﻭ 𝑤 ﮐﺎﺭ ﺍﻧﺠﺎﻡﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫𝑉𝑃 ‪𝐻 = 𝑈 +‬‬‫𝑊 ‪𝑆 = 𝑘B ln‬‬‫ﮐﻪ ‪ W‬ﺗﻌﺪﺍﺩ ﺭﯾﺰﺣﺎﻟﺖﻫﺎﺳﺖ‪.‬‬‫𝑞‬‫‪Δ𝑆 = rev‬‬‫𝑇‬‫ﮐﻪ ‪ 𝑞rev‬ﮔﺮﻣﺎﯼ ﻣﺮﺑﻮﻁ ﺑﻪ ﻓﺮﺍﯾﻨﺪ ﺑﺮﮔﺸﺖﭘﺬﯾﺮ ﺍﺳﺖ‪.‬‬‫𝑆𝑇 ‪𝐺 = 𝐻 −‬‬‫∘ 𝐸 𝐹𝑧‪Δr 𝐺∘ = −𝑅𝑇 ln 𝐾 = −‬‬‫ﮐﻪ ‪ K‬ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ‪ 𝑧 ،‬ﺗﻌﺪﺍﺩ ﺍﻟﮑﺘﺮﻭﻥﻫﺎ ﻭ ∘ 𝐸 ﭘﺘﺎﻧﺴﯿﻞ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﺍﻟﮑﺘﺮﻭﺩ ﺍﺳﺖ‪.‬‬‫𝑄 ‪Δr 𝐺 = Δr 𝐺∘ + 𝑅𝑇 ln‬‬‫ﻭﺍﮐﻨﺶ‬‫ﺑﺮﺍﯼ‬‫ِ‬‫‪𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D‬‬‫𝑐‬‫𝑑‬‫]‪[C] [D‬‬
‫=𝑄‬
‫𝑎‬
‫𝑏‬
‫]‪[A] [B‬‬‫ﮐﻪ ]‪ [A‬ﻏﻠﻈﺖ ‪ A‬ﺍﺳﺖ‪.‬‬‫‪IChO‬‬‫‪G0-5‬‬‫‪IRN-2 C-0 G-5‬‬‫)‪persian (Iran‬‬‫ﺗﻐﯿﯿﺮ ﮔﺮﻣﺎ 𝑞‪Δ‬‬‫ﻣﻌﺎﺩﻟﻪ ﻧﺮﻧﺴﺖ ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶﻫﺎﯼ‬‫ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ‬‫ﻣﻌﺎﺩﻟﻪ ﺁﺭﻧﯿﻮﺱ‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪ -‬ﺑﯿﺮ‬‫ﻣﻌﺎﺩﻟﻪ ﻫﻨﺪﺭﺳﻮﻥ‪ -‬ﻫﺴﻞﺑﺎﺥ‬‫ﺍﻧﺮﮊﯼ ﯾﮏ ﻓﻮﺗﻮﻥ‬‫𝑇‪Δ𝑞 = 𝑛𝑐m Δ‬‬‫ﻣﻮﻟﯽ ﻣﺴﺘﻘﻞ ﺍﺯ ﺩﻣﺎﺳﺖ‪.‬‬‫ﮔﺮﻣﺎﯾﯽ‬‫ﮐﻪ ‪ 𝑐m‬ﻇﺮﻓﯿﺖ‬
‫ِ‬
‫𝐶‬
‫𝑇𝑅‬
‫‪ln ox‬‬‫‪𝐸 = 𝐸∘ +‬‬‫𝐹𝑧‬‫‪𝐶red‬‬‫ﮐﻪ ‪ 𝐶ox‬ﻏﻠﻈﺖ ﻣﺎﺩﻩﯼ ﺍﮐﺴﯿﺪﺷﺪﻩ ﻭ ‪ 𝐶red‬ﻏﻠﻈﺖ ﻣﺎﺩﻩﯼ ﺍﺣﯿﺎﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫𝐸‬‫) 𝑎 ‪𝑘 = 𝐴 exp (−‬‬‫𝑇𝑅‬‫ﮐﻪ ‪ k‬ﺛﺎﺑﺖ ﺳﺮﻋﺖ‪ A ،‬ﺿﺮﯾﺐ ﭘﯿﺶﻧﻤﺎﯾﯽ ﻭ 𝑎𝐸 ﺍﻧﺮﮊﯼ ﻓﻌﺎﻟﺴﺎﺯﯼ ﺍﺳﺖ‪.‬‬‫𝑥𝑒 = )𝑥(‪exp‬‬‫𝑐𝑙𝜀 = 𝐴‬‫ﮐﻪ ‪ A‬ﻣﯿﺰﺍﻥ ﺟﺬﺏ‪ 𝜀 ،‬ﺿﺮﯾﺐ ﺟﺬﺏ ﻣﻮﻟﯽ‪ l ،‬ﻃﻮﻝ ﻣﺴﯿﺮ ﻧﻮﺭ ﻭ ‪ c‬ﻏﻠﻈﺖ ﻣﺤﻠﻮﻝ ﺍﺳﺖ‪.‬‬‫ﺗﻌﺎﺩﻝ‬‫ﺑﺮﺍﯼ ﯾﮏ‬‫ِ‬‫‪HA ⇌ H+ + A−‬‬‫ﮐﻪ ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ ﺑﺮﺍﺑﺮ ‪ 𝐾a‬ﺑﺎﺷﺪ‪،‬‬‫] ‪[A−‬‬‫)‬‫( ‪pH = p𝐾a + log‬‬‫]‪[HA‬‬‫𝑐‬‫‪𝐸 = ℎ𝜈 = ℎ‬‬‫𝜆‬‫ﮐﻪ 𝜈 ﻓﺮﮐﺎﻧﺲ ﻭ 𝜆 ﻃﻮﻝ ﻣﻮﺝ ﻧﻮﺭ ﺍﺳﺖ‪.‬‬‫ﻣﺠﻤﻮﻉ ﯾﮏ ﺳﺮﯼ ﻫﻨﺪﺳﯽ‬‫ﻭﻗﺘﯽ ‪،𝑥 ≠ 1‬‬‫ﻣﻌﺎﺩﻟﻪ ﺗﻘﺮﯾﺒﯽ ﮐﻪ ﻣﯽﺗﻮﺍﻥ ﺑﺮﺍﯼ‬‫ﺣﻞ ﻣﺴﺎﺋﻞ ﺑﻪ ﮐﺎﺭ ﺑﺮﺩ‬‫ﻭﻗﺘﯽ ﮐﻪ ‪،𝑥 ≪ 1‬‬‫‪1‬‬‫𝑥‪≃1+‬‬‫𝑥‪1−‬‬‫‪1 − 𝑥𝑛+1‬‬‫𝑛‬‫= 𝑖𝑥 ‪1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 = ∑𝑖=0‬‬‫𝑥‪1−‬‬IChO
IRN-2 C-0 G-6

G0-6
persian (Iran)‫ﺟﺪﻭﻝ ﺗﻨﺎﻭﺑﯽ‬IChO
IRN-2 C-0 G-7

G0-7
persian (Iran)

NMR ‫ﺟﺎﺑﻪﺟﺎﯾﯽﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ﭘﺮﻭﺗﻮﻥ‬‫‪G0-8‬‬‫)‪persian (Iran‬‬‫ً‬‫ﺗﻘﺮﯾﺒﺎ 𝑚𝑝𝑝 ‪+0.4‬‬‫𝛿‪ Δ‬ﺑﺮﺍﯼ ﯾﮏ ﺍﺳﺘﺨﻼﻑ ﮔﺮﻭﻩ ﺁﻟﮑﯿﻞ‪:‬‬‫‪IChO‬‬‫‪IRN-2 C-0 G-8‬‬

IRN-2 C-1 C-1

IRN-2 C-1 C
Ehsan Naderi Donig

IChO
Problem 1
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q1-1‬‬‫‪IChO‬‬‫‪IRN-2 C-1 Q-1‬‬‫)‪persian (Iran‬‬‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺳﻄﺢ ﯾﮏ ﻓﻠﺰ‬‫‪11% of the total‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪A.2‬‬‫‪B.1‬‬‫‪B.2‬‬‫‪B.3‬‬‫‪B.4‬‬

‫‪6‬‬

‫‪4‬‬

‫‪5‬‬

‫‪3‬‬

‫‪3‬‬

‫‪3‬‬

‫‪Total‬‬‫‪24‬‬‫‪Score‬‬‫ﺍﻧﺘﻈﺎﺭ ﻣﯽﺭﻭﺩ ﮐﻪ ﺩﺭ ﺁﯾﻨﺪﻩ ﺑﺘﻮﺍﻥ ﻫﯿﺪﺭﻭﮊﻥ ﺭﺍ ﺑﻪﻋﻨﻮﺍﻥ ﯾﮏ ﻣﻨﺒﻊ ﺍﻧﺮﮊﯼ ﺍﺳﺘﻔﺎﺩﻩ ﮐﺮﺩ ﮐﻪ ﺑﻪ ﺳﻮﺧﺖﻫﺎﯼ ﻓﺴﯿﻠﯽ ﻭﺍﺑﺴﺘﮕﯽ ﻧﺪﺍﺭﺩ‪ .‬ﺩﺭ ﺍﯾﻦ ﻣﺴﺄﻟﻪ ﻓﺮﺍﯾﻨﺪ‬‫ﺫﺧﯿﺮﻩﺳﺎﺯﯼ ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﯾﮏ ﻓﻠﺰ ﺭﺍ ﺑﺮﺭﺳﯽ ﻣﯽﮐﻨﯿﻢ ﮐﻪ ﺑﻪ ﺗﮑﻨﻮﻟﻮﮊﯼ ﺩﺧﯿﺮﻩﺳﺎﺯﯼ ﺍﻧﺮﮊﯼ ﻭ ﺣﻤﻞ ﻭ ﻧﻘﻞ ﺑﺮ ﭘﺎﯾﻪ ﻫﯿﺪﺭﻭﮊﻥ ﻣﺮﺑﻮﻁ ﺍﺳﺖ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﺳﻄﺢ ﻓﻠﺰﺍﺕ ﺑﻪ ﺩﺭﻭﻥ ﺁﻧﻬﺎ ﺟﺬﺏ ﻣﯽﺷﻮﺩ‪ ،‬ﺩﺭ ﺍﺑﺘﺪﺍ ﻓﺮﺍﯾﻨﺪ ﺟﺬﺏ ﺳﻄﺤﯽ ﻫﯿﺪﺭﻭﮊﻥ ﺭﻭﯼ ﺳﻄﺢ ﻓﻠﺰ‪H2 (g) → ،‬‬‫ﺍﺯ ﺁﻧﺠﺎ ﮐﻪ ﻫﯿﺪﺭﻭﮊﻥ ﺍﺯ ﻃﺮﯾﻖ‬‫ِ‬‫)‪ ،2H(ad‬ﺭﺍ ﺑﺮﺭﺳﯽ ﻣﯽﮐﻨﯿﻢ‪ .‬ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ ﺣﺎﻟﺖﻫﺎﯼ ﮔﺎﺯﯼ ﻭ ﺟﺬﺏﺷﺪﻩﯼ ﻫﯿﺪﺭﻭﮊﻥ ﺑﻪﺗﺮﺗﯿﺐ ﺑﺎ )‪ (g‬ﻭ )‪ (ad‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﻣﻮﻟﮑﻮﻝﻫﺎﯼ‬‫ﺟﺬﺏ ﺳﻄﺢ ﻣﯽﺷﻮﻧﺪ )ﺷﮑﻞ ‪ .(1‬ﺩﺭ ﺍﯾﻨﺠﺎ ﺍﻧﺮﮊﯼ ﭘﺘﺎﻧﺴﯿﻞ‬‫ﺻﻮﺭﺕ ﺍﺗﻢﻫﺎﯼ ‪H‬‬‫ﻫﯿﺪﺭﻭﮊﻥ ) ‪ (H2‬ﮐﻪ ﺑﻪ ﺳﻄﺢ ﻓﻠﺰ )‪ (M‬ﻣﯽﺭﺳﻨﺪ‪ ،‬ﺭﻭﯼ ﺳﻄﺢ ﺗﻔﮑﯿﮏ ﻭ ﺑﻪ‬‫ِ‬‫ِ‬‫ﺳﻄﺤﯽ ﻓﻠﺰ )‪ .(z‬ﻓﺮﺽ ﻣﯽﺷﻮﺩ ﻣﺤﻮﺭﯼ ﮐﻪ ﺩﻭ ﺍﺗﻢ ‪ H‬ﺭﺍ ﺑﻪ ﻫﻢ‬‫‪ H2‬ﺑﺎ ﺩﻭ ﻣﺘﻐﯿﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪ :‬ﻓﺎﺻﻠﻪ ﺑﯿﻦﺍﺗﻤﯽ )‪ (d‬ﻭ ﺍﺭﺗﻔﺎﻉ ﻧﺴﺒﺖ ﺑﻪ ﺍﺗﻢ‬‫ِ‬‫ﻋﻤﻮﺩﯼ ﻧﻘﻄﻪﭼﯿﻦ ﺩﺭ ﺷﮑﻞ ‪ 1‬ﻗﺮﺍﺭ ﺩﺍﺭﺩ‪ .‬ﺷﮑﻞ ‪ 2‬ﻧﻤﻮﺩﺍﺭ ﮐﺎﻧﺘﻮﺭﯼ ﺍﻧﺮﮊﯼ‬‫ﻣﺘﺼﻞ ﻣﯽﮐﻨﺪ ﺑﺎ ﺳﻄﺢ ﻓﻠﺰ ﻣﻮﺍﺯﯼ ﺍﺳﺖ ﻭ ﻣﺮﮐﺰ ﮔﺮﺍﻧﺶ ﻫﻤﯿﺸﻪ ﺭﻭﯼ ﺧﻂ‬‫ِ‬‫ﻣﻮﻝ ‪ H2‬ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪ .‬ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺧﻂﻫﺎﯼ‬‫ﭘﺘﺎﻧﺴﯿﻞ ﺭﺍ ﺑﺮﺍﯼ ﺗﻔﮑﯿﮏ ﺭﻭﯼ ﺳﻄﺢ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ .‬ﻣﻘﺎﺩﯾﺮ ﻋﺪﺩﯼ ﺍﻧﺮﮊﯼ ﭘﺘﺎﻧﺴﯿﻞ ﺭﺍ ﺑﺮﺣﺴﺐ ‪ kJ‬ﺑﺮ‬‫ِ‬‫ﺗﻮﭘﺮ ‪ ،20 kJ mol−1‬ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺧﻂﻫﺎﯼ ﺑﺮﯾﺪﻩ ‪ 100 kJ mol−1‬ﻭ ﻓﺎﺻﻠﻪ ﺑﯿﻦ ﺧﻂﻫﺎﯼ ﺗﻮﭘﺮ ﻭ ﺑﺮﯾﺪﻩ ‪ 80 kJ mol−1‬ﺍﺳﺖ‪ .‬ﺍﺯ ﺍﻧﺮﮊﯼ ﻧﻘﻄﻪ ﺻﻔﺮ‬‫ﺍﺭﺗﻌﺎﺷﯽ ﺻﺮﻓﻨﻈﺮ ﻣﯽﺷﻮﺩ‪.‬‬‫‪Q1-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-1 Q-2‬‬‫ﺷﮑﻞ ‪ .1‬ﺗﻌﺮﯾﻒ ﻣﺘﻐﯿﺮﻫﺎ‪ .‬ﺷﮑﻞ ﺑﺎ ﻣﻘﯿﺎﺱﻫﺎﯼ ﻭﺍﻗﻌﯽ ﺭﺳﻢ ﻧﺸﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺷﮑﻞ ‪2‬‬

‫‪Q1-3‬‬‫‪IChO‬‬‫‪IRN-2 C-1 Q-3‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫ﺑﺮﺍﯼ ﻫﺮ ﯾﮏ ﺍﺯ ﻣﻮﺍﺭﺩ )‪ (i‬ﺗﺎ )‪ (iii‬ﺯﯾﺮ‪ ،‬ﻧﺰﺩﯾﮏﺗﺮﯾﻦ ﻣﻘﺪﺍﺭ ﺭﺍ ﺍﺯ ﺑﯿﻦ ﻣﻘﺎﺩﯾﺮ ‪ A‬ﺗﺎ ‪ G‬ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫)‪ (i‬ﻓﺎﺻﻠﻪ ﺑﯿﻦﺍﺗﻤﯽ ﺑﺮﺍﯼ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ‪ H2‬ﮔﺎﺯﯼ ﺷﮑﻞ‬‫)‪ (ii‬ﻓﺎﺻﻠﻪ ﺑﯿﻦﺍﺗﻤﯽ ﺑﯿﻦ ﺍﺗﻢﻫﺎﯼ ﻓﻠﺰ ) ‪ 𝑑M‬ﺩﺭ ﺷﮑﻞ ‪(1‬‬‫)‪ (iii‬ﻓﺎﺻﻠﻪ ﺍﺗﻢﻫﺎﯼ ‪ H‬ﺟﺬﺏﺷﺪﻩ ﺍﺯ ﺳﻄﺢ ) ‪ ℎad‬ﺩﺭ ﺷﮑﻞ ‪(1‬‬‫‪D. 0.15 nm‬‬‫‪A.2‬‬‫‪C. 0.11 nm‬‬‫‪G. 0.27 nm‬‬‫‪B. 0.07 nm‬‬‫‪F. 0.23 nm‬‬‫‪A. 0.03 nm‬‬‫‪E. 0.19 nm‬‬‫ﺑﺮﺍﯼ ﻫﺮ ﯾﮏ ﺍﺯ ﻣﻮﺍﺭﺩ )‪ (i‬ﻭ )‪ (ii‬ﺯﯾﺮ‪ ،‬ﻧﺰﺩﯾﮏﺗﺮﯾﻦ ﻣﻘﺪﺍﺭ ﺭﺍ ﺍﺯ ﺑﯿﻦ ﻣﻘﺎﺩﯾﺮ ‪ A‬ﺗﺎ ‪ H‬ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫)‪ (i‬ﺍﻧﺮﮊﯼ ﻣﻮﺭﺩ ﻧﯿﺎﺯ ﺑﺮﺍﯼ ﺗﻔﮑﯿﮏ ‪ H2‬ﮔﺎﺯﯼ ﺑﻪ ‪ H‬ﮔﺎﺯﯼ‬‫)‪H2 (g) → 2H(g‬‬‫)‪ (ii‬ﺍﻧﺮﮊﯼ ﺁﺯﺍﺩﺷﺪﻩ ﻃﯽ ﺟﺬﺏ ﺳﻄﺤﯽ ‪ H2‬ﮔﺎﺯﯼ‬‫)‪H2 (g) → 2H(ad‬‬‫‪D. 100 kJ mol−1‬‬‫‪H. 400 kJ mol−1‬‬‫‪C. 60 kJ mol−1‬‬‫‪G. 300 kJ mol−1‬‬‫‪6pt‬‬‫‪B. 40 kJ mol−1‬‬‫‪F. 200 kJ mol−1‬‬‫‪A. 20 kJ mol−1‬‬‫‪E. 150 kJ mol−1‬‬‫‪4pt‬‬

‫‪Q1-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-1 Q-4‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﻫﻤﺎﻧﻄﻮﺭ ﮐﻪ ﺩﺭ ﻭﺍﮐﻨﺶﻫﺎﯼ )‪ (1a‬ﻭ )‪ (1b‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺍﺗﻢﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻧﯽ ﮐﻪ ﺟﺬﺏ ﺳﻄﺢ ﺷﺪﻩﺍﻧﺪ‪ ،‬ﺩﺭ ﻣﺮﺣﻠﻪ ﺑﻌﺪﯼ ﯾﺎ ﺑﻪ ﺩﺭﻭﻥ ﻓﻠﺰ ﺟﺬﺏ‬‫ﻣﯽﺷﻮﻧﺪ ﯾﺎ ﺩﻭﺑﺎﺭﻩ ﺑﺎ ﻫﻢ ﺗﺮﮐﯿﺐ ﻣﯽﺷﻮﻧﺪ ﻭ ﺑﺎ ﻭﺍﺟﺬﺏ‪ ،‬ﻭﺍﺭﺩ ﻓﺎﺯ ﮔﺎﺯﯼ ﻣﯽﺷﻮﻧﺪ‪ .‬ﺩﺭ ﺍﯾﻦ ﻭﺍﮐﻨﺶ‪ H(ab) ،‬ﺍﺗﻢ ﻫﯿﺪﺭﻭﮊﻧﯽ ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ ﮐﻪ ﺑﻪ‬‫ﺩﺭﻭﻥ ﻓﻠﺰ ﺟﺬﺏ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺳﺮﻋﺖ ﻭﺍﮐﻨﺶﻫﺎﯼ ﺟﺬﺏ ﺳﻄﺤﯽ )‪ ، (adsorption‬ﻭﺍﺟﺬﺏ )‪ (desorption‬ﻭ ﺟﺬﺏ )‪ (absorption‬ﺑﻪﺗﺮﺗﯿﺐ ﺑﺎ ] ‪ 𝑟2 [s−1 ] ،𝑟1 [s−1‬ﻭ ] ‪𝑟3 [s−1‬‬‫ﺑﺮﺍﺑﺮ ﺍﺳﺖ‪ .‬ﺍﯾﻦ ﺳﺮﻋﺖﻫﺎ ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﭼﻨﯿﻦ ﻧﻮﺷﺖ‪:‬‬‫ﺩﺭ ﺍﯾﻦ ﺭﻭﺍﺑﻂ ] ‪ 𝑘2 [s−1 ] ، 𝑘1 [s−1 Pa−1‬ﻭ ] ‪ 𝑘3 [s−1‬ﺛﺎﺑﺖﻫﺎﯼ ﺳﺮﻋﺖ ﻭﺍﮐﻨﺶ ﻫﺴﺘﻨﺪ ﻭ ‪ 𝑃H2‬ﻓﺸﺎﺭ ‪ H2‬ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ 𝜃 .‬ﮐﺴﺮﯼ ﺍﺯ ﺟﺎﯾﮕﺎﻩﻫﺎﯼ‬‫ﺩﺭ ﺩﺳﺘﺮﺱ ﺭﻭﯼ ﺳﻄﺢ ﺍﺳﺖ ﮐﻪ ﺑﺎ ﺍﺗﻢﻫﺎﯼ ‪ H‬ﺍﺷﻐﺎﻝ ﺷﺪﻩﺍﻧﺪ )‪ .(0 ≤ 𝜃 ≤ 1‬ﻓﺮﺽ ﻣﯽﺷﻮﺩ ﮐﻪ ﺟﺬﺏ ﺳﻄﺤﯽ )‪ (adsorption‬ﻭ ﻭﺍﺟﺬﺏ ﺩﺭ‬‫ﻣﻘﺎﯾﺴﻪ ﺑﺎ ﺟﺬﺏ )‪ (absorption‬ﺳﺮﯾﻊﺍﻧﺪ ) ‪ (𝑟1 , 𝑟2 ≫ 𝑟3‬ﻭ 𝜃 ﺛﺎﺑﺖ ﻣﯽﻣﺎﻧﺪ‪.‬‬‫‪B.1‬‬‫‪ 𝑟3‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﭼﻨﯿﻦ ﻧﻮﺷﺖ‪:‬‬‫𝐶 ﺭﺍ ﺑﺮﺣﺴﺐ ‪ 𝑘1‬ﻭ ‪ 𝑘2‬ﺑﻪﺩﺳﺖ ﺁﻭﺭﯾﺪ‪.‬‬‫‪5pt‬‬

‫‪Q1-5‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-1 Q-5‬‬‫ﻧﻤﻮﻧﻪﺍﯼ ﺍﺯ ﻓﻠﺰ ﺑﺎ ﻣﺴﺎﺣﺖ ﺳﻄﺢ ‪ 𝑆 = 1.0×10−3 m2‬ﺩﺭ ﻇﺮﻓﯽ ﺑﻪ ﺣﺠﻢ ﯾﮏ ﻟﯿﺘﺮ ) ‪ (1.0×10−3 m3‬ﺷﺎﻣﻞ ‪ H2‬ﺑﺎ ﻓﺸﺎﺭ ‪𝑃H2 = 1.0×102 Pa‬‬‫ﺳﻄﺤﯽ ﺍﺗﻢ ﻫﯿﺪﺭﻭﮊﻥ ﺭﻭﯼ ﺳﻄﺢ ﻓﻠﺰ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪ .𝑁 = 1.3 × 1018 m−2‬ﺩﻣﺎﯼ ﺳﻄﺢ ﺩﺭ‬‫ﻗﺮﺍﺭ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺩﺍﻧﺴﯿﺘﻪﯼ ﺟﺎﯾﮕﺎﻩﻫﺎﯼ ﺟﺬﺏ‬‫ِ‬‫‪−4‬‬‫‪−1‬‬‫ﺛﺎﺑﺖ ‪ 𝑣 = 4.0 × 10 Pa s‬ﮐﺎﻫﺶ ﻣﯽﯾﺎﺑﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ‬‫‪ 𝑇 = 400 K‬ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺑﺎ ﭘﯿﺸﺮﻓﺖ ﻭﺍﮐﻨﺶ )‪ 𝑃H2 ،(1‬ﺑﺎ ﺳﺮﻋﺖ‬‫ِ‬‫‪ H2‬ﮔﺎﺯﯼ ﺍﯾﺪﻩﺍﻝ ﺍﺳﺖ ﻭ ﻣﯽﺗﻮﺍﻥ ﺍﺯ ﺣﺠﻢ ﻧﻤﻮﻧﻪ ﻓﻠﺰ ﺻﺮﻓﻨﻈﺮ ﮐﺮﺩ‪.‬‬‫‪B.2‬‬‫ﻣﻘﺪﺍﺭ ﺍﺗﻢﻫﺎﯼ ‪ H‬ﺟﺬﺏﺷﺪﻩ ﺭﺍ ﺑﺮﺣﺴﺐ ﻣﻮﻝ‪ ،‬ﺑﻪﺍﺯﺍﯼ ﻭﺍﺣﺪ ﻣﺴﺎﺣﺖ ﺳﻄﺢ ﻭ ﺑﻪﺍﺯﺍﯼ ﻭﺍﺣﺪ ﺯﻣﺎﻥ‪،‬‬‫] ‪ ،𝐴 [mol s−1 m−2‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬‫‪B.3‬‬‫ﺩﺭ ﺩﻣﺎﯼ ‪ 𝐶 ، 𝑇 = 400 K‬ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪ . 1.0 × 102 Pa−1‬ﻣﻘﺪﺍﺭ ‪ 𝑘3‬ﺭﺍ ﺩﺭ ‪ 400‬ﮐﻠﻮﯾﻦ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﺍﮔﺮ‬‫ﺟﻮﺍﺏ ﻗﺴﻤﺖ ‪ B.2‬ﺭﺍ ﺑﻪﺩﺳﺖ ﻧﯿﺎﻭﺭﺩﻩﺍﯾﺪ‪ ،‬ﺍﺯ ﻣﻘﺪﺍﺭ ‪ 𝐴 = 3.6 × 10−7 mol s−1 m−2‬ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫ِ‬‫‪3pt‬‬‫‪B.4‬‬‫ﺩﺭ ﯾﮏ ﺩﻣﺎﯼ ‪ T‬ﻣﺘﻔﺎﻭﺕ‪ ،‬ﻣﻘﺎﺩﯾﺮ ‪ 𝐶 = 2.5 × 103 Pa−1‬ﻭ ‪ 𝑘3 = 4.8 × 10−2 s−1‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺍﺯ ﺑﯿﻦ‬‫ﻣﻨﺤﻨﯽﻫﺎﯼ )‪ (a‬ﺗﺎ )‪ (h‬ﺩﺭ ﺩﻭ ﻧﻤﻮﺩﺍﺭ ﺯﯾﺮ‪ ،‬ﻣﻨﺤﻨﯽ ﺩﺭﺳﺖ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ ﮐﻪ ‪ 𝑟3‬ﺭﺍ ﺑﻪﺻﻮﺭﺕ ﺗﺎﺑﻌﯽ ﺍﺯ ‪ 𝑃H2‬ﺩﺭ ﺍﯾﻦ‬‫ﺩﻣﺎ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪.‬‬‫‪3pt‬‬

IChO

A1-1

IRN-2 C-1 A-1

persian (Iran)‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺳﻄﺢ ﯾﮏ ﻓﻠﺰ‬A ‫ﻗﺴﻤﺖ‬(6 pt) A.1
(iii)

(ii)

(i)

(4 pt) A.2
(ii)

(i)

IChO

A1-2

IRN-2 C-1 A-2

persian (Iran)B ‫ﻗﺴﻤﺖ‬(5 pt) B.1𝐶=(3 pt) B.2

𝐴=

mol s−1 m−2

IChO

A1-3

IRN-2 C-1 A-3

persian (Iran)(3 pt) B.3

𝑘3 =

s−1
(3 pt) B.4

IRN-2 C-2 C-1

IRN-2 C-2 C
Ehsan Naderi Donig

IChO
Problem 2
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q2-1‬‬‫‪IChO‬‬‫‪IRN-2 C-2 Q-1‬‬‫)‪persian (Iran‬‬‫ﮐﭙﺴﻮﻝ ﺯﻣﺎﻥ ﺍﯾﺰﻭﺗﻮﭘﯽ‬‫‪11% of the total‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪8‬‬‫‪A.2‬‬‫‪8‬‬‫‪A.3‬‬‫‪10‬‬‫‪A.4‬‬‫‪9‬‬‫‪Total‬‬‫‪35‬‬‫‪Score‬‬‫ﻣﻮﻟﮑﻮﻝﻫﺎﯾﯽ ﻣﺎﻧﻨﺪ ‪ CH4‬ﻭ ‪ CH3 D‬ﺭﺍ ﮐﻪ ﻓﻘﻂ ﺩﺭ ﺗﺮﮐﯿﺐ ﺍﯾﺰﻭﺗﻮﭘﯽ ﺑﺎ ﻫﻢ ﺗﻔﺎﻭﺕ ﺩﺍﺭﻧﺪ‪ ،‬ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒ ﻣﯽﻧﺎﻣﻨﺪ‪ .‬ﺧﻮﺍﺹ ﺷﯿﻤﯿﺎﯾﯽ ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒﻫﺎ‬‫ﯾﮑﺴﺎﻥ ﺩﺭ ﻧﻈﺮﮔﺮﻓﺘﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺑﺎ ﺍﯾﻦ ﺣﺎﻝ‪ ،‬ﺩﺭ ﻃﺒﯿﻌﺖ ﺗﻔﺎﻭﺕﻫﺎﯼ ﺟﺰﯾﯽ ﺩﺭ ﺧﻮﺍﺹ ﺷﯿﻤﯿﺎﯾﯽ ﺁﻧﻬﺎ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪.‬‬‫ﻓﺮﺽ ﮐﻨﯿﺪ ﺗﻤﺎﻡ ﻣﻮﺍﺩﯼ ﮐﻪ ﺩﺭ ﺍﯾﻦ ﻣﺴﺎﻟﻪ ﻣﻄﺮﺡ ﺷﺪﻩﺍﻧﺪ‪ ،‬ﺩﺭ ﻓﺎﺯ ﮔﺎﺯﯼ ﻫﺴﺘﻨﺪ‪.‬‬‫ﺗﻌﺎﺩﻝ ﺯﯾﺮ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﯽﮔﯿﺮﯾﻢ‪:‬‬‫ﺁﻧﺘﺮﻭﭘﯽ )‪ (S‬ﺑﺎ ﺍﻓﺰﺍﯾﺶ ﺗﻌﺪﺍﺩ ﺣﺎﻟﺖﻫﺎﯼ ﻣﯿﮑﺮﻭﺳﮑﻮﭘﯽ ﻣﻤﮑﻦ ﺑﺮﺍﯼ ﺳﯿﺴﺘﻢ )‪ (W‬ﺍﻓﺰﺍﯾﺶ ﻣﯽﯾﺎﺑﺪ‪:‬‬‫ﺑﺮﺍﯼ ‪ 12 C16 O2‬ﻭ ‪ 𝑊 = 1 ،12 C18 O2‬ﺍﺳﺖ‪ .‬ﺩﺭ ﻣﻘﺎﺑﻞ‪ ،‬ﺑﺮﺍﯼ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ‪ 𝑊 = 2 ،12 C16 O18 O‬ﺍﺳﺖ‪ ،‬ﺯﯾﺮﺍ ﺍﺗﻢﻫﺎﯼ ﺍﮐﺴﯿﮋﻥ ﺩﺭ ﺍﯾﻦ ﻣﻮﻟﮑﻮﻝ‬‫ﻣﻮﻟﮑﻮﻝ ‪ 12 C16 O18 O‬ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪.𝑊 = 22 = 4 ،‬‬‫ﺗﻤﯿﺰﭘﺬﯾﺮﻧﺪ‪ .‬ﺍﺯ ﺁﻧﺠﺎ ﮐﻪ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﺗﻌﺎﺩﻝ ﻧﺸﺎﻥ ﺩﺍﺩﻩﺷﺪﻩ ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ ،(1‬ﺩﻭ‬‫ِ‬‫‪Q2-2‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪IRN-2 C-2 Q-2‬‬‫ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ )𝐻‪ (Δ‬ﺑﺮﺍﯼ ﻣﻌﺎﺩﻟﻪ )‪ (3‬ﺑﺪﻭﻥ ﺗﻮﺟﻪ ﺑﻪ ﺩﻣﺎ ﻣﺜﺒﺖ ﺍﺳﺖ‪.‬‬‫‪8pt‬‬‫ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ )‪ (K‬ﺭﺍ ﺑﺮﺍﯼ ﻣﻌﺎﺩﻟﻪ )‪ (3‬ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﺑﺴﯿﺎﺭ ﭘﺎﯾﯿﻦ )ﺩﺭ ﺣﺪ ‪ (𝑇 → 0‬ﻭ ﺩﻣﺎﻫﺎﯼ ﺑﺴﯿﺎﺭ ﺑﺎﻻ )ﺩﺭ ﺣﺪ‬‫∞‪ (𝑇 → +‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﻭﺍﮐﻨﺶ ﺩﺭ ﺍﯾﻦ ﺩﻣﺎﻫﺎ ﺗﻐﯿﯿﺮ ﻧﻤﯽﮐﻨﺪ ﻭ 𝐻‪ Δ‬ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﺑﺎﻻ ﺑﻪ ﻣﻘﺪﺍﺭﯼ‬‫ﺛﺎﺑﺖ ﻫﻤﮕﺮﺍ ﻣﯽﺷﻮﺩ‪.‬‬‫ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ ﺯﯾﺮ‪ ،‬ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ )𝐻‪ (Δ‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﺎ ﺍﺭﺗﻌﺎﺷﺎﺕ ﻣﻮﻟﮑﻮﻟﯽ ﺗﻮﺻﯿﻒ ﮐﺮﺩ‪.‬‬‫ﺍﺭﺗﻌﺎﺷﯽ ] ‪ 𝜈 [s−1‬ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ‪:‬‬‫ﺩﺭ ﺩﻣﺎﯼ ﺻﻔﺮ ﮐﻠﻮﯾﻦ‪ ،‬ﺍﻧﺮﮊﯼ ﺍﺭﺗﻌﺎﺷﯽ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﻭﺍﺗﻤﯽ ﺑﺎ ﻓﺮﮐﺎﻧﺲ‬‫ِ‬‫ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ ‪ k‬ﺛﺎﺑﺖ ﻧﯿﺮﻭ ﻭ 𝜇 ﺟﺮﻡ ﮐﺎﻫﺶﯾﺎﻓﺘﻪ ﺍﺳﺖ‪ .‬ﺟﺮﻡ ﮐﺎﻫﺶﯾﺎﻓﺘﻪ ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﺮ ﺍﺳﺎﺱ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﺑﺮﺣﺴﺐ ﺟﺮﻡ ﺍﺗﻢﻫﺎ ) ‪ 𝑚1‬ﻭ ‪ (𝑚2‬ﺩﺭ‬‫ﻣﻮﻟﮑﻮﻝ ﺩﻭﺍﺗﻤﯽ ﺑﯿﺎﻥ ﮐﺮﺩ‪:‬‬‫‪A.2‬‬‫ﺍﺭﺗﻌﺎﺵ ‪ H2‬ﺑﺮﺣﺴﺐ ﻋﺪﺩ ﻣﻮﺟﯽ ﺩﺭ ‪ 4161.0 cm−1‬ﺭﺥ ﻣﯽﺩﻫﺪ‪ .‬ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ ﺯﯾﺮ‪ Δ𝐻 ،‬ﺭﺍ ﺩﺭ ﺩﻣﺎﯼ ﺻﻔﺮ‬‫ﮐﻠﻮﯾﻦ ﺑﺮﺣﺴﺐ ﻭﺍﺣﺪ ‪ J mol−1‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫ﻓﺮﺽ ﮐﻨﯿﺪ‪:‬‬‫• ﻓﻘﻂ ﺍﻧﺮﮊﯼ ﺍﺭﺗﻌﺎﺷﯽ ﺩﺭ 𝐻‪ Δ‬ﺳﻬﻢ ﺩﺍﺭﺩ‪.‬‬‫• ﻣﻘﺎﺩﯾﺮ ‪ k‬ﺑﺮﺍﯼ ‪ HD ،H2‬ﻭ ‪ D2‬ﯾﮑﺴﺎﻥ ﺍﺳﺖ‪.‬‬‫• ﺟﺮﻡ ‪ H‬ﺑﺮﺍﺑﺮ ‪ 1‬ﺩﺍﻟﺘﻮﻥ )‪ (Da‬ﻭ ﺟﺮﻡ ‪ D‬ﺑﺮﺍﺑﺮ ‪ 2‬ﺩﺍﻟﺘﻮﻥ ﺍﺳﺖ‪.‬‬‫‪8pt‬‬

‫‪Q2-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-2 Q-3‬‬‫ﻣﻮﻟﯽ ‪ D2‬ﺗﻌﺮﯾﻒ ﻣﯽﺷﻮﺩ‪.‬‬‫ﻧﺴﺒﺖ ﻣﻮﻟﯽ ‪ HD ،H2‬ﻭ ‪ D2‬ﺩﺭ ﺳﯿﺴﺘﻤﯽ ﮐﻪ ﺩﺭ ﺗﻌﺎﺩﻝ ﺑﺎﺷﺪ‪ ،‬ﺑﻪ ﺩﻣﺎ ﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ‪ .‬ﺩﺭ ﺍﯾﻨﺠﺎ ‪ ΔD2‬ﺑﻪﺻﻮﺭﺕ ﺗﻐﯿﯿﺮ ﻧﺴﺒﺖ‬‫ِ‬‫] ‪[D2‬‬‫] ‪[D‬‬‫ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ‪ 𝑅D2 ،‬ﻧﺸﺎﻥﺩﻫﻨﺪﻩﯼ ‪ 2‬ﺩﺭ ﻧﻤﻮﻧﻪ ﺍﺳﺖ ﻭ ‪ 𝑅D∗ 2‬ﺑﻪ‬‫] ‪[H2‬‬‫] ‪[H2‬‬‫∞‪ ،𝑇 → +‬ﺗﻮﺯﯾﻊ ﺍﯾﺰﻭﺗﻮﭖﻫﺎ ﺗﺼﺎﺩﻓﯽ ﻣﯽﺷﻮﺩ‪.‬‬‫‪A.3‬‬‫ﺩﺭ ∞‪ 𝑇 → +‬ﺍﺷﺎﺭﻩ ﻣﯽﮐﻨﺪ‪ .‬ﺩﺭ ﺍﯾﻨﺠﺎ ﻻﺯﻡ ﺑﻪ ﺫﮐﺮ ﺍﺳﺖ ﮐﻪ ﺩﺭ ﺣﺪ‬‫ﻃﺒﯿﻌﯽ ‪ ، D‬ﺩﺭ ﺣﺎﻟﺘﯽ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ ﮐﻪ ﺗﺒﺎﺩﻝ ﺍﯾﺰﻭﺗﻮﭘﯽ ﺩﺭ ﺗﻌﺎﺩﻝ ﺍﺳﺖ ﻭ ﺩﺭ ﺩﻣﺎﯼ‬‫‪ ΔD2‬ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻓﺮﺍﻭﺍﻧﯽ‬‫ِ‬‫ﺗﻌﺎﺩﻝ‪ ،‬ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ )‪ (K‬ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ (4‬ﺑﺎ ‪ 0.300‬ﺑﺮﺍﺑﺮ ﺍﺳﺖ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ ‪ D‬ﻭ ‪ H‬ﺑﻪﺗﺮﺗﯿﺐ‬‫ﺑﺮﺍﺑﺮ ‪ 1.5576 × 10–4‬ﻭ ‪ 1 − 1.5576 × 10–4‬ﺍﺳﺖ‪.‬‬‫‪10pt‬‬‫‪Q2-4‬‬‫‪IChO‬‬‫‪IRN-2 C-2 Q-4‬‬‫)‪persian (Iran‬‬‫ﺩﺭ ﺣﺎﻟﺖ ﮐﻠﯽ‪ ،‬ﺑﺎ ﮐﺎﻫﺶ ﺩﻣﺎ ﻧﺴﺒﺖ ﻣﻮﻟﯽ ﺑﺮﺍﯼ ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒﻫﺎﯼ ﺩﻭﺍﺳﺘﺨﻼﻓﯽ )ﮐﻪ ﺩﻭ ﺍﺗﻢ ﺍﯾﺰﻭﺗﻮﭖ ﺳﻨﮕﯿﻦ ﺩﺭ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﺍﺭﻧﺪ(‪ ،‬ﺍﻓﺰﺍﯾﺶ ﻣﯽﯾﺎﺑﺪ‪.‬‬‫ﻧﺴﺒﺖ ﻣﻮﻟﯽ ﺩﻭ ﻣﻮﻟﮑﻮﻝ ‪ CO2‬ﺑﺎ ﻭﺯﻥﻫﺎﯼ ﻣﻮﻟﮑﻮﻟﯽ ‪ 44‬ﻭ ‪ 47‬ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﯽﮔﯿﺮﯾﻢ‪ .‬ﺍﯾﻦ ﺩﻭ ﻣﻮﻟﮑﻮﻝ ﺭﺍ ﺩﺭ ﺍﺩﺍﻣﻪ ﺑﺎ ]‪ CO2 [44‬ﻭ ]‪ CO2 [47‬ﻧﺸﺎﻥ‬‫ﻣﯽﺩﻫﯿﻢ‪ .‬ﮐﻤﯿﺖ ‪ Δ47‬ﺑﻪﺻﻮﺭﺕ ﺯﯾﺮ ﺗﻌﺮﯾﻒ ﻣﯽﺷﻮﺩ‪:‬‬‫]]‪[CO2 [47‬‬‫]]‪[CO2 [47‬‬‫∗‬‫‪ 𝑅47‬ﺑﻪ‬‫ﺩﺭ ﻧﻤﻮﻧﻪ ﺍﺳﺖ ﻭ‬‫‪ 𝑅47‬ﻧﺸﺎﻥﺩﻫﻨﺪﻩﯼ‬‫]]‪[CO2 [44‬‬‫]]‪[CO2 [44‬‬‫ﺩﺭ ﺟﺪﻭﻝ ﺯﯾﺮ ﺁﻣﺪﻩ ﺍﺳﺖ‪ .‬ﺍﺯ ﺍﯾﺰﻭﺗﻮﭖﻫﺎﯾﯽ ﮐﻪ ﺩﺭ ﺍﯾﻨﺠﺎ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻧﺸﺪﻩﺍﻧﺪ‪ ،‬ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪.‬‬‫ﺩﺭ ∞‪ 𝑇 → +‬ﺍﺷﺎﺭﻩ ﻣﯽﮐﻨﺪ‪ .‬ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ ﺍﺗﻢﻫﺎﯼ ﮐﺮﺑﻦ ﻭ ﺍﮐﺴﯿﮋﻥ‬‫‪C‬‬‫ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ‬‫ﻓﺮﺍﻭﺍﻧﯽ ﻃﺒﯿﻌﯽ‬‫‪O‬‬‫‪12‬‬‫‪C‬‬‫‪0.988888‬‬‫‪16‬‬‫‪0.997621‬‬‫‪O‬‬‫‪13‬‬‫‪0.011112‬‬‫‪17‬‬‫‪0.0003790‬‬‫‪O‬‬‫‪18‬‬‫‪0.0020000‬‬‫ﻭﺍﺑﺴﺘﮕﯽ ﺩﻣﺎﯾﯽ ‪ Δ47‬ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﺗﻌﯿﯿﻦ ﻣﯽﺷﻮﺩ ﮐﻪ ‪ T‬ﺩﻣﺎﯼ ﻣﻄﻠﻖ ﺑﺮﺣﺴﺐ ﻭﺍﺣﺪ ﮐﻠﻮﯾﻦ ﺍﺳﺖ‪:‬‬‫‪A.4‬‬‫ﺑﺮﺍﯼ ﯾﮏ ﭘﻼﻧﮑﺘﻮﻥ ﻓﺴﯿﻞﺷﺪﻩ ﮐﻪ ﺍﺯ ﺑﺴﺘﺮ ﺩﺭﯾﺎﯼ ﻗﻄﺐ ﺟﻨﻮﺏ ﺑﻪﺩﺳﺖ ﺁﻣﺪﻩ ‪ 𝑅47‬ﺑﺮﺍﺑﺮ ﺑﺎ ‪4.50865 × 10–5‬‬‫ﺍﺳﺖ‪ .‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﯾﻦ ‪ ،𝑅47‬ﺩﻣﺎ ﺭﺍ ﺗﺨﻤﯿﻦ ﺑﺰﻧﯿﺪ‪ .‬ﺍﯾﻦ ﺩﻣﺎ ﺭﺍ ﺑﺮﺍﺑﺮ ﺑﺎ ﺩﻣﺎﯼ ﻫﻮﺍ ﺩﺭ ﺩﻭﺭﺍﻥ ﺯﻧﺪﮔﯽ ﺁﻥ ﭘﻼﻧﮑﺘﻮﻥ‬‫ﺍﯾﺰﻭﺗﻮﭘﻮﻟﻮﮒ ]‪ CO2 [47‬ﺭﺍ ﺩﺭ ﻣﺤﺎﺳﺒﻪ ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪.‬‬‫ﺩﺭ ﻧﻈﺮ ﻣﯽﮔﯿﺮﻧﺪ‪ .‬ﻓﻘﻂ ﻣﺘﺪﺍﻭﻝﺗﺮﯾﻦ‬‫ِ‬‫‪9pt‬‬

IChO

A2-1

IRN-2 C-2 A-1

persian (Iran)‫ﮐﭙﺴﻮﻝ ﺯﻣﺎﻥ ﺍﯾﺰﻭﺗﻮﭘﯽ‬(8 pt) A.1

𝑇 → +∞ ∶ 𝐾 =,𝑇 → 0 ∶ 𝐾 =

IChO

A2-2

IRN-2 C-2 A-2

persian (Iran)(8 pt) A.2

Δ𝐻 =J mol−1

IChO

A2-3

IRN-2 C-2 A-3

persian (Iran)(10 pt) A.3

ΔD2 =

IChO
IRN-2 C-2 A-4

A2-4
persian (Iran)(9 pt) A.4

𝑇 =K

IRN-2 C-3 C-1

IRN-2 C-3 C
Ehsan Naderi Donig

IChO
Problem 3
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q3-1‬‬‫‪IChO‬‬‫‪IRN-2 C-3 Q-1‬‬‫)‪persian (Iran‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪ -‬ﺑﯿﺮ؟‬‫‪Question‬‬‫‪Points‬‬‫‪8% of the total‬‬‫‪A.1‬‬‫‪10‬‬‫‪B.1‬‬‫‪6‬‬‫‪B.2‬‬‫‪6‬‬‫‪Total‬‬‫‪22‬‬‫‪Score‬‬‫ﺩﺭ ﺍﯾﻦ ﻣﺴﺄﻟﻪ‪ ،‬ﺍﺯ ﺟﺬﺏﻫﺎﯼ ﺳﻞ ﻭ ﺣﻼﻝ ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪ .‬ﺩﻣﺎﯼ ﺗﻤﺎﻡ ﻣﺤﻠﻮﻝﻫﺎ ﻭ ﮔﺎﺯﻫﺎ ﺩﺭ ‪ 25‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﻣﺤﻠﻮﻝ ‪ X‬ﺑﻪﺗﺮﺗﯿﺐ ﺑﺮﺍﺑﺮﻧﺪ ﺑﺎ ‪،1.00 × 10−2 mol L−1‬‬‫ﺁﺑﯽ ‪ X‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ HA‬ﻭ ‪ NaA‬ﺗﻬﯿﻪ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﻏﻠﻈﺖﻫﺎﯼ] ‪ [HA] ،[A−‬ﻭ ] ‪ [H+‬ﺩﺭ‬‫ﻣﺤﻠﻮﻝ ِ‬‫ِ‬‫‪ 1.00 × 10−3 mol L−1‬ﻭ ‪ .1.00 × 10−4 mol L−1‬ﺍﯾﻦ ﻏﻠﻈﺖﻫﺎ ﺍﺯ ﻃﺮﯾﻖ ﺗﻌﺎﺩﻝ ﺍﺳﯿﺪ‪-‬ﺑﺎﺯ ﺯﯾﺮ ﺑﺎ ﻫﻢ ﺍﺭﺗﺒﺎﻁ ﺩﺍﺭﻧﺪ‪:‬‬‫] ‪[A− ][H+‬‬‫]‪[HA‬‬‫)‪(1‬‬‫=𝐾‬‫‪HA ⇌ A− + H+‬‬‫ﻃﻮﻝ ﻣﺴﯿﺮ ﻧﻮﺭ ﺩﺭ ﻗﺴﻤﺖ ‪ A‬ﺍﯾﻦ ﻣﺴﺄﻟﻪ ﺑﺮﺍﺑﺮ 𝑙 ﺍﺳﺖ‪ .‬ﺍﺯ ﺗﻐﯿﯿﺮ ﺩﺍﻧﺴﯿﺘﻪ ﺑﺮ ﺍﺛﺮ ﺭﻗﯿﻖ ﺷﺪﻥ ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﻫﯿﭻ ﻭﺍﮐﻨﺶ ﺷﯿﻤﯿﺎﯾﯽ ﺩﯾﮕﺮﯼ‬‫ﺑﻪ ﺟﺰ ﻣﻌﺎﺩﻟﻪ )‪ (1‬ﺭﺥ ﻧﻤﯽﺩﻫﺪ‪.‬‬‫‪A.1‬‬‫ﺟﺬﺏ ﻣﺤﻠﻮﻝ ‪ X‬ﺩﺭ ﻃﻮﻝ ﻣﻮﺝ ‪ 𝜆1‬ﺑﺮﺍﺑﺮ ‪ 𝐴1‬ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﻣﯽﺷﻮﺩ‪ .‬ﺳﭙﺲ ﻣﺤﻠﻮﻝ ‪ X‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻫﯿﺪﺭﻭﮐﻠﺮﯾﮏ‬‫ﺑﺮﺍﺑﺮ ﺣﺠﻢ ﺍﻭﻟﯿﻪﺍﺵ ﺭﻗﯿﻖ ﻣﯽﺷﻮﺩ‪ .‬ﭘﺲ ﺍﺯ ﺭﻗﯿﻖﺳﺎﺯﯼ‪ ،‬ﺟﺬﺏ ﺩﺭ ‪ 𝜆1‬ﻫﻨﻮﺯ ﺑﺮﺍﺑﺮ‬‫ﺍﺳﯿﺪ ﺑﺎ ‪ pH = 2.500‬ﺗﺎ ﺩﻭ‬‫ِ‬‫ﺟﺬﺏ ‪ HA‬ﻭ ‪ A−‬ﺭﺍ ﺩﺭ ‪ 𝜆1‬ﻧﺸﺎﻥ‬‫ﺿﺮﺍﯾﺐ‬‫ﺗﺮﺗﯿﺐ‬
‫ﺑﻪ‬
‫𝜀‬
‫ﻭ‬
‫𝜀‬
‫)‬
‫ﮐﻨﯿﺪ‬‫ﺗﻌﯿﯿﻦ‬
‫ﺭﺍ‬
‫𝜀‬
‫‪−‬‬
‫ِ‬
‫‪A‬‬
‫‪HA‬‬‫‪ 𝐴1‬ﺍﺳﺖ‪ .‬ﻧﺴﺒﺖ ‪HA /𝜀A−‬‬‫ﻣﯽﺩﻫﻨﺪ(‪.‬‬‫‪10pt‬‬‫‪Q3-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-3 Q-2‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺗﻌﺎﺩﻝ ﺯﯾﺮ ﺭﺍ ﺩﺭ ﻓﺎﺯ ﮔﺎﺯﯼ ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪.‬‬‫‪D ⇌ 2M‬‬‫(‪)2‬‬‫ﺧﺎﻟﺺ ‪ D‬ﺑﺎ ﻓﺸﺎﺭ 𝑃 ﭘﺮ ﻣﯽﺷﻮﺩ‪ .‬ﺩﺭ ﺍﯾﻦ ﻣﺤﻔﻈﻪ ﺩﯾﻮﺍﺭ ﻣﺘﺤﺮﮎ ﺷﻔﺎﻓﯽ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﮐﻪ ﻣﺴﺎﺣﺖ ﺳﻄﺢ‬‫ﻣﺤﻔﻈﻪﺍﯼ ﺑﻪ ﺷﮑﻞ ﻣﮑﻌﺐ ﻣﺴﺘﻄﯿﻞ ﺑﺎ ﮔﺎﺯ‬‫ِ‬‫ﻣﻘﻄﻊ ﺁﻥ 𝑆 ﺍﺳﺖ )ﺷﮑﻞ ﺯﯾﺮ ﺭﺍ ﺑﺒﯿﻨﯿﺪ(‪ .‬ﺩﺭ ﺣﺎﻟﯽ ﮐﻪ ﻓﺸﺎﺭ ﮐﻞ ﺩﺭ ‪ P‬ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺗﻌﺎﺩﻝ ﺑﺮﻗﺮﺍﺭ ﻣﯽﺷﻮﺩ‪ .‬ﺟﺬﺏ ﮔﺎﺯ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ‬‫𝑙) 𝑉‪ 𝐴 = 𝜀(𝑛/‬ﮐﻪ 𝜀‪ 𝑉 ،𝑛 ،‬ﻭ 𝑙 ﺑﻪﺗﺮﺗﯿﺐ ﺿﺮﯾﺐ ﺟﺬﺏ‪ ،‬ﻣﻘﺪﺍﺭ ﮔﺎﺯ ﺑﺮﺣﺴﺐ ﻣﻮﻝ‪ ،‬ﺣﺠﻢ ﮔﺎﺯ ﻭ ﻃﻮﻝ ﻣﺴﯿﺮ ﻧﻮﺭ ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ‬‫ﺗﻤﺎﻡ ﺍﺟﺰﺍﯼ ﻣﺨﻠﻮﻁ ﮔﺎﺯﯼ ﻣﺎﻧﻨﺪ ﮔﺎﺯ ﺍﯾﺪﻩﺍﻝ ﺭﻓﺘﺎﺭ ﻣﯽﮐﻨﻨﺪ‪.‬‬‫ﺩﺭ ﺻﻮﺭﺕ ﻟﺰﻭﻡ ﺍﺯ ﺗﻌﺮﯾﻒﻫﺎﯼ ﺯﯾﺮ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫ﻓﺸﺎﺭ ﺟﺰﯾﯽ‬‫ﻣﻘﺪﺍﺭ ﺑﺮ ﺣﺴﺐ ﻣﻮﻝ‬
‫ﺣﺠﻢ‬

‫‪D‬‬

‫ﺣﺎﻟﺖ ﺍﻭﻟﯿﻪ‬‫ﺑﻌﺪ ﺍﺯ ﺗﻌﺎﺩﻝ‬

‫‪M‬‬

‫‪D‬‬

‫𝑃‬

‫‪0‬‬

‫‪𝑝D‬‬‫‪𝑝M‬‬‫‪𝑛0‬‬‫‪0‬‬‫‪𝑛D‬‬‫‪𝑛M‬‬‫‪𝑉0‬‬‫‪M‬‬‫𝑉‬‫‪B.1‬‬‫ﺟﺬﺏ ﮔﺎﺯ ﺩﺭ ‪ 𝜆B1‬ﮐﻪ ﺍﺯ ﺟﻬﺖ 𝑥 ) 𝑥𝑙 = 𝑙( ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻫﻢ ﺩﺭ ﺣﺎﻟﺖ ﺍﻭﻟﯿﻪ ﻭ ﻫﻢ ﭘﺲ ﺍﺯ ﺗﻌﺎﺩﻝ‪ ،‬ﺑﺮﺍﺑﺮ‬‫ﺟﺬﺏ ‪ D‬ﻭ ‪ M‬ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪.‬‬‫‪ 𝐴B1‬ﺍﺳﺖ‪ .‬ﻧﺴﺒﺖ ‪ 𝜀D /𝜀M‬ﺭﺍ ﺩﺭ ‪ 𝜆B1‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪ 𝜀D .‬ﻭ ‪ 𝜀M‬ﺑﻪﺗﺮﺗﯿﺐ ﺿﺮﺍﯾﺐ‬‫ِ‬‫‪6pt‬‬‫‪B.2‬‬‫ﺟﺬﺏ ﮔﺎﺯ ﺩﺭ ‪ 𝜆B2‬ﮐﻪ ﺍﺯ ﺟﻬﺖ 𝑦 ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻫﻢ ﺩﺭ ﺣﺎﻟﺖ ﺍﻭﻟﯿﻪ ) ‪ (𝑙 = 𝑙𝑦0‬ﻭ ﻫﻢ ﭘﺲ ﺍﺯ ﺗﻌﺎﺩﻝ‬‫) 𝑦𝑙 = 𝑙(‪ ،‬ﺑﺮﺍﺑﺮ ‪ 𝐴B2‬ﺍﺳﺖ‪ .‬ﻧﺴﺒﺖ ‪ 𝜀D /𝜀M‬ﺭﺍ ﺩﺭ ‪ 𝜆B2‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫‪6pt‬‬

‫‪A3-1‬‬‫)‪persian (Iran‬‬‫ﻗﺎﻧﻮﻥ ﻻﻣﺒﺮﺕ‪-‬ﺑﯿﺮ؟‬‫ﻗﺴﻤﺖ ‪A‬‬‫‪(10 pt) A.1‬‬‫)ﺍﺩﺍﻣﻪ ﺩﺭ ﺻﻔﺤﻪ ﺑﻌﺪ(‬‫‪IChO‬‬‫‪IRN-2 C-3 A-1‬‬

IChO
IRN-2 C-3 A-2

A3-2
persian (Iran)(cont.) A.1

𝜀HA /𝜀A− =IChO
IRN-2 C-3 A-3

A3-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(6 pt) B.1

𝜀D /𝜀M =IChO
IRN-2 C-3 A-4

A3-4
persian (Iran)(6 pt) B.2

𝜀D /𝜀M =IRN-2 C-4 C-1

IRN-2 C-4 C
Ehsan Naderi Donig

IChO
Problem 4
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q4-1‬‬‫‪IChO‬‬‫‪IRN-2 C-4 Q-1‬‬‫)‪persian (Iran‬‬‫ﺷﯿﻤﯽ ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ ﺭﻭﯼ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪5‬‬‫‪11% of the total‬‬‫‪B.1‬‬‫‪4‬‬‫‪B.2‬‬‫‪3‬‬‫‪B.3‬‬‫‪5‬‬‫‪B.4‬‬‫‪9‬‬‫‪Total‬‬‫‪32‬‬‫‪Score‬‬‫ﺯﺩﺍﯾﯽ ﺁﺏ‪ ،‬ﺭﻭﯼ )‪ (Zn‬ﻣﻮﺟﻮﺩ ﺩﺭ ﭘﺴﺎﺏﻫﺎﯼ ﺻﻨﻌﺘﯽ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ‬‫ﺍﺯ ﻣﺪﺕﻫﺎ ﻗﺒﻞ ﺍﺯ ﺭﻭﯼ ﺩﺭ ﺁﻟﯿﺎﮊﻫﺎﯼ ﺑﺮﻧﺞ ﻭ ﺍﺳﺘﯿﻞ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺑﻪ ﻣﻨﻈﻮﺭ ﺳﻢ‬‫ِ‬‫ﺍﺯ ﺭﻭﺵ ﺗﻪﻧﺸﯿﻨﯽ ﺟﺪﺍ ﻣﯽﺷﻮﺩ‪ .‬ﺭﺳﻮﺑﺎﺗﯽ ﮐﻪ ﺑﻪ ﺩﺳﺖ ﻣﯽﺁﯾﺪ ﺍﺣﯿﺎ ﻣﯽﺷﻮﺩ ﺗﺎ ﺭﻭﯼ ﻓﻠﺰﯼ ﺑﺎﺯﯾﺎﺑﯽ ﻭ ﺩﻭﺑﺎﺭﻩ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﺍﻧﺤﻼﻝ ﺭﻭﯼ ﻫﯿﺪﺭﻭﮐﺴﯿﺪ )‪ Zn(OH)2 (s‬ﺩﺭ ‪ 25 ∘ C‬ﻭ ﺛﺎﺑﺖﻫﺎﯼ ﺗﻌﺎﺩﻝ ﻣﺮﺑﻮﻃﻪ ﺩﺭ ﻣﻌﺎﺩﻻﺕ ‪ 1‬ﺗﺎ ‪ 4‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺗﻌﺎﺩﻝ‬‫ِ‬‫ﺣﻼﻟﯿﺖ )𝑆( ﺭﻭﯼ )ﻏﻠﻈﺖ ﺭﻭﯼ ﺩﺭ ﯾﮏ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺍﺷﺒﺎﻉﺷﺪﻩ( ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ (5‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪Q4-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-4 Q-2‬‬‫‪A.1‬‬‫ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﺗﻌﺎﺩﻝﻫﺎﯼ ﻣﻌﺎﺩﻻﺕ )‪ (1‬ﺗﺎ )‪ (4‬ﺑﺮﻗﺮﺍﺭ ﻣﯽﺷﻮﺩ‪ ،‬ﻣﺤﺪﻭﺩﻩﺍﯼ ﺍﺯ ‪ pH‬ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ ﮐﻪ ﺩﺭ ﺁﻥ‪ ،‬ﺩﺭ‬‫‪2+‬‬‫‪ ،[Zn(OH)2−‬ﺑﺰﺭﮒﺗﺮﯾﻦ ﻣﻘﺪﺍﺭ ﻣﺮﺑﻮﻁ ﺑﻪ‬‫ﺑﯿﻦ ﻏﻠﻈﺖﻫﺎﯼ‬‫ﺗﻌﺎﺩﻟﯽ ])‪ [Zn(OH)2 (aq)] ،[Zn (aq‬ﻭ ])‪4 (aq‬‬‫ِ‬‫])‪ [Zn(OH)2 (aq‬ﺑﺎﺷﺪ‪.‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﻣﺤﻠﻮﻝ ﺯﯾﺮ ﺻﺎﻓﯽ‬‫ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺍﺷﺒﺎﻉﺷﺪﻩﺍﯼ ﺍﺯ )‪ Zn(OH)2 (s‬ﺑﺎ ‪ pH = 7.00‬ﺗﻬﯿﻪ ﻭ ﺻﺎﻑ ﻣﯽﺷﻮﺩ‪ .‬ﺑﻪ‬‫ِ‬‫‪ NaOH‬ﺍﺿﺎﻓﻪ ﻣﯽﺷﻮﺩ ﺗﺎ ‪ pH‬ﺁﻥ ﺗﺎ ‪ 12.00‬ﺍﻓﺰﺍﯾﺶ ﯾﺎﺑﺪ‪ .‬ﺩﺭﺻﺪ ﻣﻮﻟﯽﺍﯼ ﺍﺯ ﺭﻭﯼ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ ﮐﻪ ﺩﺭ ﺍﺛﺮ‬‫ﺍﻓﺰﺍﯾﺶ ‪ pH‬ﺍﺯ ‪ 7.00‬ﺗﺎ ‪ 12.00‬ﺭﺳﻮﺏ ﻣﯽﮐﻨﺪ‪ .‬ﺍﺯ ﺗﻐﯿﯿﺮﺍﺕ ﺣﺠﻢ ﻭ ﺩﻣﺎ ﺻﺮﻓﻨﻈﺮ ﮐﻨﯿﺪ‪.‬‬‫ِ‬‫‪5pt‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺩﺭ ﻣﺮﺣﻠﻪ ﺑﻌﺪﯼ‪ ،‬ﺭﻭﯼ ﻫﯿﺪﺭﻭﮐﺴﯿﺪِ ﺑﺎﺯﯾﺎﺑﯽﺷﺪﻩ ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ ﺗﺎ ﺑﺮ ﺍﺳﺎﺱ ﻭﺍﮐﻨﺶ ﺯﯾﺮ‪ ،‬ﺭﻭﯼ ﺍﮐﺴﯿﺪ ﺑﻪ ﺩﺳﺖ ﺁﯾﺪ‪:‬‬‫ﺭﻭﯼ ﻓﻠﺰﯼ ﺍﺣﯿﺎ ﻣﯽﺷﻮﺩ‪:‬‬‫ﺳﭙﺲ ﺭﻭﯼ ﺍﮐﺴﯿﺪ ﺩﺭ ﻭﺍﮐﻨﺶ ﺑﺎ ﻫﯿﺪﺭﻭﮊﻥ ﺑﻪ ِ‬‫‪B.1‬‬‫ﺟﺰﯾﯽ ﺑﺨﺎﺭ‬‫ﺷﺮﺍﯾﻂ ﻓﺸﺎﺭ ﺛﺎﺑﺖ ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ‪ 1‬ﺑﺎﺭ )‪ (bar‬ﭘﯿﺶ ﺑﺮﻭﺩ‪ ،‬ﻻﺯﻡ ﺍﺳﺖ ﻓﺸﺎﺭ‬‫ﺑﺮﺍﯼ ﺁﻧﮑﻪ ﻭﺍﮐﻨﺶ )‪ (7‬ﺩﺭ‬‫ِ‬‫ِ‬‫ﺁﺑﯽ ﮐﻪ ﺗﻮﻟﯿﺪ ﻣﯽﺷﻮﺩ‪ ،‬ﮐﻢ ﺷﻮﺩ‪ .‬ﺣﺪ ﺑﺎﻻﯾﯽ ﻓﺸﺎﺭ ﺟﺰﯾﯽ ﺑﺨﺎﺭ ﺁﺏ ﺭﺍ ﺑﺮﺍﯼ ﺁﻧﮑﻪ ﻭﺍﮐﻨﺶ )‪ (7‬ﺑﺘﻮﺍﻧﺪ ﺩﺭ ﺩﻣﺎﯼ ‪300‬‬‫ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﭘﯿﺶ ﺑﺮﻭﺩ‪ ،‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﺩﺭ ﺍﯾﻨﺠﺎ ﺍﻧﺮﮊﯼﻫﺎﯼ ﺗﺸﮑﯿﻞ ﮔﯿﺒﺲ ﺑﺮﺍﯼ ﺭﻭﯼ ﺍﮐﺴﯿﺪ ﻭ ﺑﺨﺎﺭ ﺁﺏ ﺩﺭ‬‫ﺩﻣﺎﯼ ‪ 300‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﻭ ﻓﺸﺎﺭ ‪ 1‬ﺑﺎﺭ ﺑﺮﺍﯼ ﺗﻤﺎﻡ ﮔﻮﻧﻪﻫﺎﯼ ﮔﺎﺯﯼ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ‪:‬‬‫‪Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1‬‬‫ﻭ‬‫‪Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1‬‬‫‪4pt‬‬‫ﺭﻭﯼ ﻓﻠﺰﯼ ﺑﺮﺍﯼ ﺳﺎﺧﺘﻦ ﺍﻟﮑﺘﺮﻭﺩ ﻣﻨﻔﯽ )ﺁﻧﺪ( ﺩﺭ ﺑﺎﺗﺮﯼﻫﺎﯼ ﻓﻠﺰ‪-‬ﻫﻮﺍ ﺍﺳﺘﻔﺎﺩﻩ ﻣﯽﺷﻮﺩ‪ .‬ﺍﯾﻦ ﺍﻟﮑﺘﺮﻭﺩ ﺍﺯ ‪ Zn‬ﻭ ‪ ZnO‬ﺗﺸﮑﯿﻞ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﺍﺯ ﻭﺍﮐﻨﺶ‬‫ﺍﺯ ِ‬‫ﺍﮐﺴﺎﯾﺶ‪ -‬ﮐﺎﻫﺶ ﺯﯾﺮ ﺑﺮﺍﯼ ﺗﻮﻟﯿﺪ ﺟﺮﯾﺎﻥ ﺍﻟﮑﺘﺮﯾﮑﯽ ﺍﺳﺘﻔﺎﺩﻩ ﻣﯽﮐﻨﺪ‪ .‬ﻧﯿﺮﻭﯼ ﻣﺤﺮﮐﻪ ﺍﻟﮑﺘﺮﯾﮑﯽ )‪ (e.m.f.‬ﯾﺎ ∘ 𝐸 ﺍﯾﻦ ﻭﺍﮐﻨﺶ ﺩﺭ ﺩﻣﺎﯼ ‪ 25‬ﺩﺭﺟﻪ‬‫ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﻭ ﻓﺸﺎﺭ ‪ 1‬ﺑﺎﺭ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪B.2‬‬‫ﯾﮏ ﺑﺎﺗﺮﯼ ﺭﻭﯼ‪-‬ﻫﻮﺍ ﺑﺎ ﺟﺮﯾﺎﻥ ‪ 20‬ﻣﯿﻠﯽﺁﻣﭙﺮ ﺩﺭ ﻣﺪﺕ ‪ 24‬ﺳﺎﻋﺖ ﺗﺨﻠﯿﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺗﻐﯿﯿﺮ ﺟﺮﻡ ﺍﻟﮑﺘﺮﻭﺩ ﻣﻨﻔﯽ )ﺁﻧﺪ(‬‫ﺍﯾﻦ ﺑﺎﺗﺮﯼ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬

‫‪Q4-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-4 Q-3‬‬‫ﮐﻮﻩ ﻓﻮﺟﯽ‬‫‪B.3‬‬‫ﺍﺛﺮ ﺗﻐﯿﯿﺮﺍﺕ ﻣﺤﯿﻄﯽ ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪ e.m.f. .‬ﺭﺍ ﺩﺭ ﻗﻠﻪ ﮐﻮﻩ ﻓﻮﺟﯽ‬‫ﺗﻐﯿﯿﺮ ‪ e.m.f.‬ﯾﮏ ﺑﺎﺗﺮﯼ ﺭﻭﯼ‪-‬ﻫﻮﺍ ﺭﺍ ﺩﺭ ِ‬‫ﺍﺭﺗﻔﺎﻉ‬‫ﺩﺭ‬‫ﺍﺗﻤﺴﻔﺮﯼ‬‫ﻓﺸﺎﺭ‬‫ﺑﺎﺷﺪ‪.‬‬‫ﻣﺘﺮ‬‫‪3776‬‬‫ﺍﺭﺗﻔﺎﻉ‬‫ﻭ‬‫ﻓﻮﺭﯾﻪ(‬
‫ﻣﺎﻩ‬
‫)ﺩﺭ‬
‫ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ ،‬ﺍﮔﺮ ﺩﺭ ﺍﯾﻦ ﻗﻠﻪ ﺩﻣﺎ ‪−38 ∘ C‬‬‫ِ‬‫]‪ ℎ [m‬ﻭ ﺩﻣﺎﯼ ]‪ 𝑇 [∘ C‬ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪:‬‬‫‪5pt‬‬‫ﻧﺴﺒﺖ ﻣﻮﻟﯽ ﺍﮐﺴﯿﮋﻥ ﺩﺭ ﺍﺗﻤﺴﻔﺮ ‪ 21%‬ﺍﺳﺖ‪ .‬ﺗﻐﯿﯿﺮ ﺍﻧﺮﮊﯼ ﮔﯿﺒﺲ ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ )‪ (8‬ﺩﺭ ﺩﻣﺎﯼ ‪ −38 ∘ C‬ﻭ ﻓﺸﺎﺭ ‪1‬‬‫ﺑﺎﺭ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ‪Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1‬‬‫‪B.4‬‬‫ﺗﻐﯿﯿﺮ ﺍﻧﺮﮊﯼ ﮔﯿﺒﺲ ﺭﺍ ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ )‪ (6‬ﺩﺭ ‪ 25 ∘ C‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﭘﺘﺎﻧﺴﯿﻞﻫﺎﯼ ﮐﺎﻫﺶ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ‬‫)‪ 𝐸 ∘ (Zn2+ /Zn‬ﻭ )‪ 𝐸 ∘ (O2 /H2 O‬ﺩﺭ ‪ 25 ∘ C‬ﻭ 𝑟𝑎𝑏 ‪ 1‬ﺑﻪﺗﺮﺗﯿﺐ ﺩﺭ ﻣﻌﺎﺩﻟﻪﻫﺎﯼ )‪ (10‬ﻭ )‪ (11‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪9pt‬‬IChO
IRN-2 C-4 A-1

A4-1
persian (Iran)‫ ﮐﺎﻫﺶ ﺭﻭﯼ‬-‫ﺷﯿﻤﯽ ﺍﮐﺴﺎﯾﺶ‬A ‫ﻗﺴﻤﺖ‬(6 pt) A.1< pH <IChO
IRN-2 C-4 A-2

A4-2
persian (Iran)(5 pt) A.2

%

IChO
IRN-2 C-4 A-3

A4-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(4 pt) B.1

𝑝H2 O=bar(3 pt) B.2

g

IChO
IRN-2 C-4 A-4

A4-4
persian (Iran)(5 pt) B.3

V

IChO

A4-5

IRN-2 C-4 A-5

persian (Iran)(9 pt) B.4

Δ𝐺∘ =J mol−1

IRN-2 C-5 C-1

IRN-2 C-5 C
Ehsan Naderi Donig

IChO
Problem 5
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q5-1‬‬‫‪IChO‬‬‫‪IRN-2 C-5 Q-1‬‬‫)‪persian (Iran‬‬‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﺮﻣﻮﺯ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪9‬‬‫‪A.2‬‬‫‪7‬‬‫‪12% of the total‬‬‫‪A.3‬‬‫‪6‬‬‫‪A.4‬‬‫‪10‬‬‫‪B.1‬‬‫‪5‬‬‫‪B.2‬‬‫‪15‬‬‫‪B.3‬‬‫‪8‬‬‫‪Total‬‬‫‪60‬‬‫‪Score‬‬‫ﺍﮔﺮﭼﻪ ﺳﯿﻠﯿﺴﯿﻢ ﻣﺎﻧﻨﺪ ﮐﺮﺑﻦ ﺩﺭ ﮔﺮﻭﻩ ‪ 14‬ﺍﺳﺖ‪ ،‬ﺧﻮﺍﺻﺶ ﺑﻪ ﻃﻮﺭ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪﺍﯼ ﻓﺮﻕ ﺩﺍﺭﺩ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬

‫ﺑﺮﺧﻼﻑ ﭘﯿﻮﻧﺪ ﺳﻪ ﮔﺎﻧﻪ ﮐﺮﺑﻦ‪-‬ﮐﺮﺑﻦ‪ ،‬ﭘﯿﻮﻧﺪ ﺳﻪ ﮔﺎﻧﻪ ﺳﯿﻠﯿﺴﯿﻢ ‪-‬ﺳﯿﻠﯿﺴﯿﻢ ﺩﺭ ﺗﺮﮐﯿﺒﯽ ﺑﺎ ﻓﺮﻣﻮﻝ ‪ :R) R1 –Si ≡ Si–R1‬ﺍﺳﺘﺨﻼﻑ ﺁﻟﯽ( ﮐﺎﻣﻼ ﻓﻌﺎﻝ‬‫ﺍﺳﺖ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ‪ ،‬ﺑﺎ ﺍﺗﯿﻠﻦ ﻭﺍﮐﻨﺶ ﻣﯽﺩﻫﺪ ﻭ ﯾﮏ ﻣﺤﺼﻮﻝ ﺣﻠﻘﻮﯼ ﺗﺸﮑﯿﻞ ﻣﯽﺩﻫﺪ ﮐﻪ ﯾﮏ ﺣﻠﻘﻪ ﭼﻬﺎﺭﻋﻀﻮﯼ ﺩﺍﺭﺩ‪.‬‬‫ﻭﻗﺘﯽ ‪ R1 –Si ≡ Si–R1‬ﺑﺎ ﯾﮏ ﺁﻟﮑﯿﻦ ) ‪ (R2 –C ≡ C–R2‬ﻭﺍﮐﻨﺶ ﻣﯽﺩﻫﺪ‪ ،‬ﺗﺮﮐﯿﺐ ‪ A‬ﮐﻪ ﺣﻠﻘﻪ ﭼﻬﺎﺭ ﻋﻀﻮﯼ ﺩﺍﺭﺩ ﯾﻪ ﻋﻨﻮﺍﻥ ﯾﮏ ﺣﺪﻭﺍﺳﻂ ﺍﻭﻟﯿﻪ‬‫ﺑﯿﺸﺘﺮ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﯾﮕﺮ ‪ R2 –C ≡ C–R2‬ﺑﺎ ﺗﺮﮐﯿﺐ ‪ A‬ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ‪ B‬ﻭ ‪ C‬ﺭﺍ ﻣﯽﺩﻫﺪ ﮐﻪ ﻫﺮ ﺩﻭﯼ ﺁﻥﻫﺎ ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﻣﺰﺩﻭﺝ‬‫ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪ .‬ﻭﺍﮐﻨﺶ‬‫ِ‬‫ﺣﻠﻘﻮﯼ ﺷﺒﯿﻪ ﺑﻨﺰﻥ ﺩﺍﺭﻧﺪ ﻭ‘‪ ’disilabenzenes‬ﻧﺎﻣﯿﺪﻩ ﻣﯽﺷﻮﻧﺪ‪ .‬ﻫﺮ ﺩﻭﯼ ﺁﻧﻬﺎ ﺷﺎﻣﻞ ﯾﮏ ﺣﻠﻘﻪ ﺷﺶ ﻋﻀﻮﯼ ﻫﺴﺘﻨﺪ ﻭ ﻣﯽﺗﻮﺍﻥ ﻓﺮﻣﻮﻝ ﺁﻧﻬﺎ ﺭﺍ ﺑﻪ‬‫ﺻﻮﺭﺕ ‪ (R1 –Si)2 (R2 –C)4‬ﻧﺸﺎﻥ ﺩﺍﺩ‪.‬‬‫‪Q5-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-5 Q-2‬‬‫ﻃﯿﻒ‪ 13 C NMR‬ﺑﺮﺍﯼ ﺍﺳﮑﻠﺖ ﻫﺎﯼ ﺳﺎﺧﺘﺎﺭﯼ ‪ Si2 C4‬ﺑﺎ ﺣﻠﻘﻪ ﺷﺶ ﻋﻀﻮﯼ ﺩﺭ ‪ B‬ﻭ ‪ ، C‬ﺩﻭ ﺳﯿﮕﻨﺎﻝ ﺑﺮﺍﯼ ‪ B‬ﻭ ﯾﮏ ﺳﯿﮕﻨﺎﻝ ﺑﺮﺍﯼ ‪ C‬ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪.‬‬‫‪A.1‬‬‫ﻓﺮﻣﻮﻝ ﻫﺎﯼ ﺳﺎﺧﺘﺎﺭﯼ ‪ B ، A‬ﻭ ‪ C‬ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ Si ،R2 ،R1‬ﻭ ‪ C‬ﺩﺭ ﯾﮑﯽ ﺍﺯ ﻓﺮﻡﻫﺎﯼ ﺭﺯﻭﻧﺎﻧﺴﯽ ﻣﺤﺘﻤﻞ‬‫ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ ﺩﺭ ﺑﺮﺧﯽ ﺍﺯ ﻭﺍﮐﻨﺸﻬﺎﯼ ﻫﯿﺪﺭﻭﮊﻥﺩﺍﺭ ﺷﺪﻥ ﺳﯿﺴﺘﻢ ﻫﺎﯼ ﻏﯿﺮ ﺍﺷﺒﺎﻉ ﮐﻪ ﺩﺭ ﭘﺎﯾﯿﻦ )ﺷﮑﻞ‬‫‪ (1‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﺍﻧﺮﮊﯼ ﭘﺎﯾﺪﺍﺭﯼ ﺁﺭﻭﻣﺎﺗﯿﮏ )‪ (ASE‬ﺭﺍ ﺑﺮﺍﯼ ﺑﻨﺰﻥ ﻭ ﺗﺮﮐﯿﺐ ‪ (R1 = R2 = H ) C‬ﺑﻪ ﺻﻮﺭﺕ‬‫ﻣﻘﺎﺩﯾﺮ ﻣﺜﺒﺖ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪7pt‬‬‫ﺷﮑﻞ ‪1‬‬

‫‪Q5-3‬‬‫‪IChO‬‬‫‪IRN-2 C-5 Q-3‬‬‫)‪persian (Iran‬‬‫ﻭﻗﺘﯽ ﻣﺤﻠﻮﻝ ‪ C‬ﺩﺭ ﺯﺍﯾﻠﻦ ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪ ،‬ﺩﭼﺎﺭ ﺍﯾﺰﻭﻣﺮﯼ ﻣﯽﺷﻮﺩ ﻭ ﯾﮏ ﻣﺨﻠﻮﻁ ﺗﻌﺎﺩﻟﯽ ﺍﺯ ﺗﺮﮐﯿﺒﺎﺕ ‪ D‬ﻭ ‪ E‬ﺑﺪﺳﺖ ﻣﯽﺁﯾﺪ‪ .‬ﻧﺴﺒﺖ ﻣﻮﻟﯽ ‪ D‬ﺑﻪ‬‫‪ E‬ﺩﺭ ﺩﻣﺎﯼ ‪ 50.0‬ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ ﻣﻌﺎﺩﻝ ‪ 1‬ﺑﻪ ‪ 40.0‬ﻭ ﻧﺴﺒﺖ ﻣﻮﻟﯽ ‪ D‬ﺑﻪ ‪ E‬ﺩﺭ ﺩﻣﺎﯼ ‪ 120.0‬ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ ﻣﻌﺎﺩﻝ ‪ 1‬ﺑﻪ ‪ 20.0‬ﺍﺳﺖ‪.‬‬‫‪A.3‬‬‫‪6pt‬‬‫𝐻‪ Δ‬ﺭﺍ ﺑﺮﺍﯼ ﺗﺒﺪﯾﻞ ‪ D‬ﺑﻪ ‪ E‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ 𝐻‪ Δ‬ﺑﻪ ﺩﻣﺎ ﺑﺴﺘﮕﯽ ﻧﺪﺍﺭﺩ‪.‬‬‫ﺍﯾﺰﻭﻣﺮﯼ ﺷﺪﻥ ‪ C‬ﺑﻪ ‪ D‬ﻭ ﺑﻪ ‪ E‬ﺑﺎ ﺗﺒﺪﯾﻞ ﭘﯿﻮﻧﺪﻫﺎﯼ 𝜋 ﺑﻪ ﭘﯿﻮﻧﺪﻫﺎﯼ 𝜎 ﻭ ﺑﺪﻭﻥ ﺷﮑﺴﺘﻪ ﺷﺪﻥ ﭘﯿﻮﻧﺪﻫﺎﯼ 𝜎 ﺍﻧﺠﺎﻡ ﻣﯽﺷﻮﺩ‪ .‬ﺁﻧﺎﻟﯿﺰ ‪ 13 C NMR‬ﺑﺮﺍﯼ‬‫ﺍﺳﮑﻠﺖ ﺳﺎﺧﺘﺎﺭﯼ ‪ Si2 C4‬ﺍﺯ ‪ D‬ﯾﮏ ﺳﯿﮕﻨﺎﻝ ﻭ ﺑﺮﺍﯼ ﺍﺳﮑﻠﺖ ﺳﺎﺧﺘﺎﺭﯼ ‪ Si2 C4‬ﺍﺯ ‪ E‬ﺩﻭ ﺳﯿﮕﻨﺎﻝ ﻣﯽﺩﻫﺪ‪ .‬ﺩﺭ ﺳﺎﺧﺘﺎﺭ ‪ D‬ﺣﻠﻘﻪ ﺳﻪ ﻋﻀﻮﯼ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ‬‫ﺩﺭ ﺣﺎﻟﯿﮑﻪ ‪ E‬ﺩﻭ ﺣﻠﻘﻪ ﺳﻪ ﻋﻀﻮﯼ ﺩﺍﺭﺩ ﮐﻪ ﺩﺭ ﯾﮏ ﺿﻠﻊ ﻣﺸﺘﺮﮎ ﻫﺴﺘﻨﺪ‪.‬‬‫‪A.4‬‬‫ﻓﺮﻣﻮﻝ ﺳﺎﺧﺘﺎﺭﯼ ‪ D‬ﻭ ‪ E‬ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ Si ، R2 ، R1‬ﻭ ‪ C‬ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪10pt‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﯽﺗﻮﺍﻧﺪ ﺑﺎ ﻋﻨﺎﺻﺮ ﺍﻟﮑﺘﺮﻭﻧﮕﺎﺗﯿﻮ ﻣﺎﻧﻨﺪ ﻓﻠﻮﺋﻮﺭ ﺗﺮﮐﯿﺒﺎﺗﯽ ﺑﺎ ﮐﺌﻮﺭﺩﯾﻨﺎﻧﺴﯿﻮﻥ ﺑﺎﻻ )ﺑﺎ ﺑﯿﺸﺘﺮ ﺍﺯ ﭼﻬﺎﺭ ﺍﺳﺘﺨﻼﻑ( ﺗﺸﮑﯿﻞ ﺩﻫﺪ‪ .‬ﺷﺒﯿﻪ‬‫ﻓﻠﻮﺋﻮﺭﯾﺪﻫﺎﯼ ﻓﻠﺰﯼ ﮐﻪ ﺍﻏﻠﺐ ﺑﻪ ﻋﻨﻮﺍﻥ ﻭﺍﮐﻨﺸﮕﺮﻫﺎﯼ ﻓﻠﻮﺋﻮﺭﺩﺍﺭ ﮐﺮﺩﻥ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﯽ ﮔﯿﺮﻧﺪ ‪ ،‬ﻓﻠﻮﺋﻮﺭﯾﺪﻫﺎﯼ ﺳﯿﻠﯿﺴﯿﻮﻣﯽ ﺑﺎ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻥ‬‫ﺑﺎﻻ ﻧﯿﺰ ﺑﻪ ﻋﻨﻮﺍﻥ ﻭﺍﮐﻨﺸﮕﺮﻫﺎﯼ ﻓﻠﻮﺋﻮﺭﺩﺍﺭ ﮐﺮﺩﻥ ﻋﻤﻞ ﻣﯽﮐﻨﻨﺪ‪.‬‬‫ﻭﺍﮐﻨﺶ ﻓﻠﻮﺋﻮﺭﺩﺍﺭﮐﺮﺩﻥ ‪ CCl4‬ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ‪ Na2 SiF6‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺍﻧﺠﺎﻡ ﺷﺪ‪.‬‬‫• ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﮐﺮﺩﻥ ﻣﺤﻠﻮﻝ ‪Na2 SiF6‬‬‫· ﺗﻬﯿﻪ‬‫ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪ : F‬ﻣﻘﺪﺍﺭ ‪ 0.855‬ﮔﺮﻡ ﺍﺯ ‪ 188.053 ) Na2 SiF6‬ﮔﺮﻡ ﺑﺮ ﻣﻮﻝ( ﺩﺭ ﺁﺏ ﺣﻞ ﺷﺪ )ﺣﺠﻢ ﮐﻞ ‪ 200 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ(‪.‬‬‫ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪ : G‬ﻣﻘﺪﺍﺭ ‪ 6.86‬ﮔﺮﻡ ﺍﺯ ‪ 568.424) Ce2 (SO4 )3‬ﮔﺮﻡ ﺑﺮ ﻣﻮﻝ( ﺩﺭ ﺁﺏ ﺣﻞ ﺷﺪ )ﺣﺠﻢ ﮐﻞ ‪ 200 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ(‪.‬‬‫· ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ‬‫ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﺭﺳﻮﺑﯽ ‪ 50.0‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ F‬ﺍﺯ ﻃﺮﯾﻖ ﻗﻄﺮﻩ ﻗﻄﺮﻩ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ﻣﺤﻠﻮﻝ ‪ ، G‬ﺩﺭ ﺣﻀﻮﺭ ﺷﻨﺎﺳﺎﮔﺮ ‪ orange xylenol‬ﺍﻧﺠﺎﻡ‬‫ﻣﯽﺷﻮﺩ ﮐﻪ ﺑﺎ ‪ Ce3+‬ﮐﺌﻮﺭﺩﯾﻨﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺑﻌﺪ ﺍﺯ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ‪ 18.8‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﻣﺤﻠﻮﻝ ‪ ، G‬ﺭﻧﮓ ﻣﺤﻠﻮﻝ ﺍﺯ ﺯﺭﺩ ﺑﻪ ﺳﺮﺧﺎﺑﯽ ﺗﻐﯿﯿﺮ ﻣﯽﮐﻨﺪ‪ .‬ﺭﺳﻮﺏ‬‫ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﯾﮏ ﺗﺮﮐﯿﺐ ﺩﻭﺗﺎﯾﯽ ﻭ ﺷﺎﻣﻞ ‪ Ce3+‬ﺍﺳﺖ ﻭ ﺗﻨﻬﺎ ﺗﺮﮐﯿﺐ ﺳﯿﻠﯿﺴﯿﻮﻣﯽ ﺣﺎﺻﻞ ‪ Si(OH)4 ،‬ﺍﺳﺖ‪.‬‬‫‪B.1‬‬‫‪5pt‬‬‫ﺑﺮﺍﯼ ﻭﺍﮐﻨﺶ ‪ Na2 SiF6‬ﺑﺎ ‪ Ce2 (SO4 )3‬ﻣﻌﺎﺩﻟﻪ ﻣﻮﺍﺯﻧﻪ ﺷﺪﻩ ﺑﻨﻮﯾﺴﯿﺪ‪.‬‬‫• ﻭﺍﮐﻨﺶ ‪ CCl4‬ﺑﺎ ‪Na2 SiF6‬‬‫)ﺍﺯ ﺩﺳﺖ ﺭﻓﺘﻦ ﻣﻮﺍﺩ ﻣﺜﻼ ﺩﺭ ﺍﺛﺮ ﺗﺒﺨﯿﺮ ﺷﺪﻥ ﺩﺭ ﺣﯿﻦ ﺍﻧﺠﺎﻡ ﺩﺳﺘﻮﺭﺍﻟﻌﻤﻞ ﺯﯾﺮ‪ ،‬ﻗﺎﺑﻞ ﺻﺮﻑ ﻧﻈﺮ ﺍﺳﺖ‪(.‬‬‫ﻣﻘﺪﺍﺭ 𝑥 ﮔﺮﻡ ﺍﺯ ‪ Na2 SiF6‬ﺑﻪ ‪ 500.0‬ﮔﺮﻡ ‪ CCl4‬ﺍﺿﺎﻓﻪ ﺷﺪﻩ ﻭ ﺩﺭ ﯾﮏ ﻇﺮﻑ ﻭﺍﮐﻨﺶ ﺩﺭ ﺑﺴﺘﻪ ﻭ ﻣﻘﺎﻭﻡ ﺩﺭ ﺑﺮﺍﺑﺮ ﻓﺸﺎﺭ ﺗﺎ ‪ 300‬ﺩﺭﺟﻪ ﺳﻠﺴﯿﻮﺱ‬‫ﺣﺮﺍﺭﺕ ﺩﺍﺩﻩ ﺷﺪ‪ .‬ﺳﭙﺲ ‪ Na2 SiF6‬ﻭﺍﮐﻨﺶ ﻧﺪﺍﺩﻩ ﻭ ‪ NaCl‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﺑﻪ ﮐﻤﮏ ﺻﺎﻑ ﮐﺮﺩﻥ ﺟﺪﺍ ﺷﺪﻧﺪ‪ .‬ﺯﯾﺮ ﺻﺎﻓﯽ ﺑﺎ ‪ CCl4‬ﺗﺎ ﺣﺠﻢ ‪ 1.00‬ﻟﯿﺘﺮ‬‫ﺭﻗﯿﻖ ﺷﺪ )ﻣﺤﻠﻮﻝ ‪ .(H‬ﻃﯿﻒ ﻫﺎﯼ‪ 29 Si NMR‬ﻭ ‪ 19 F NMR‬ﻣﺤﻠﻮﻝ ‪ SiF4 ، H‬ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﺗﻨﻬﺎ ﺗﺮﮐﯿﺐ ﺳﯿﻠﯿﺴﯿﻮﻣﯽ ﻧﺸﺎﻥ ﺩﺍﺩ‪ .‬ﺩﺭ ﻃﯿﻒ ‪NMR‬‬‫‪19‬‬‫‪ 19 F‬ﻋﻼﻭﻩ ﺑﺮ ‪ SiF4‬ﺳﯿﮕﻨﺎﻝ ﻫﺎﯼ ﻣﺮﺑﻮﻁ ﺑﻪ ‪ CF3 Cl ، CF2 Cl2 ، CFCl3‬ﻭ ‪ CF4‬ﻣﺸﺎﻫﺪﻩ ﺷﺪﻧﺪ )ﺟﺪﻭﻝ ‪ (1‬ﻧﺴﺒﺖ ﻫﺎﯼ ﺍﻧﺘﮕﺮﺍﻟﯽ ﻃﯿﻒ ‪F NMR‬‬‫ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺗﻌﺪﺍﺩ ﻫﺴﺘﻪ ﻫﺎﯼ ﻓﻠﻮﺋﻮﺭ ﺍﺳﺖ‪.‬‬‫‪F NMR data‬‬‫‪19‬‬‫‪Integration ratio‬‬‫‪CFCl3‬‬‫ﺟﺪﻭﻝ ‪1‬‬‫‪45.0‬‬‫‪ SiF4‬ﻣﻄﺎﺑﻖ ﻣﻌﺎﺩﻟﻪ )‪ (8‬ﻫﯿﺪﺭﻭﻟﯿﺰ ﺷﺪﻩ ﻭ ‪ H2 SiF6‬ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪.‬‬‫‪CF2 Cl2‬‬‫‪65.0‬‬‫‪CF3 Cl‬‬‫‪18.0‬‬‫‪CF4‬‬‫‪2.0‬‬

‫‪Q5-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-5 Q-4‬‬‫ﻣﻘﺪﺍﺭ ‪ 10‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ H‬ﺑﻪ ﻣﻘﺪﺍﺭ ﺍﺿﺎﻓﯽ ﺁﺏ ﺍﻓﺰﻭﺩﻩ ﺷﺪ ﮐﻪ ﻣﻨﺠﺮ ﺑﻪ ﻫﯿﺪﺭﻭﻟﯿﺰ ﮐﺎﻣﻞ ‪ SiF4‬ﮔﺮﺩﯾﺪ‪ H2 SiF6 .‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﺩﺭ ﺍﺛﺮ ﻫﯿﺪﺭﻭﻟﯿﺰ‬‫ﺩﺭ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪ ،‬ﭘﺲ ﺍﺯ ﺟﺪﺍﺳﺎﺯﯼ ﺧﻨﺜﯽ ﺷﺪﻩ ﻭ ﮐﺎﻣﻼ ﺑﻪ ‪ Na2 SiF6‬ﺗﺒﺪﯾﻞ ﺷﺪ )ﻣﺤﻠﻮﻝ ﺁﺑﯽ ‪(J‬‬‫ﺭﺳﻮﺏ ﻭﺍﮐﻨﺶ ﻧﺪﺍﺩﻩ ﺍﺯ ‪ Na2 SiF6‬ﻭ ﻫﻤﭽﻨﯿﻦ ‪ NaCl‬ﮐﻪ ﺍﺯ ﻃﺮﯾﻖ ﺻﺎﻑ ﮐﺮﺩﻥ ﺩﺭ ﻣﺮﺣﻠﻪ ﺍﺑﺘﺪﺍﯾﯽ ﺟﺪﺍ ﺷﺪﻩ ﺑﻮﺩﻧﺪ )ﺩﺭ ﻣﺘﻦ ﺯﯾﺮ ﺁﻥ ﺧﻂ ﮐﺸﯿﺪﻩ ﺷﺪﻩ‬‫ﺍﺳﺖ(‪ ،‬ﺑﻪ ﻃﻮﺭ ﮐﺎﻣﻞ ﺩﺭ ﺁﺏ ﺣﻞ ﺷﺪﻧﺪ ﺗﺎ ﯾﮏ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺑﺪﺳﺖ ﺁﯾﺪ )‪ 10.0‬ﻟﯿﺘﺮ ﻣﺤﻠﻮﻝ‪( K‬‬‫ﺳﭙﺲ ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﻫﺎﯼ ﺭﺳﻮﺑﯽ ﺩﯾﮕﺮﯼ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ G‬ﺍﻧﺠﺎﻡ ﺷﺪ ﻭ ﻧﻘﺎﻁ ﭘﺎﯾﺎﻧﯽ ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﺑﺎ ‪ G‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺑﻮﺩ‪:‬‬‫ﺑﺮﺍﯼ ﻣﺤﻠﻮﻝ ‪) J‬ﻣﻘﺪﺍﺭ ﮐﺎﻣﻞ ﺁﻥ( ﺑﺮﺍﺑﺮ ﺷﺪ ﺑﺎ‪ 61.6 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ‬‫ﺑﺮﺍﯼ ‪ 100‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ ﺍﺯ ﻣﺤﻠﻮﻝ ‪ K‬ﺑﺮﺍﺑﺮ ﺷﺪ ﺑﺎ‪ 44.4 :‬ﻣﯿﻠﯽ ﻟﯿﺘﺮ‬‫ﻻﺯﻡ ﺑﻪ ﺫﮐﺮ ﺍﺳﺖ ‪ ،‬ﺣﻀﻮﺭ ﻫﻤﺰﻣﺎﻥ ‪ NaCl‬ﯾﺎ ‪ SiO2‬ﺗﺎﺛﯿﺮﯼ ﺑﺮ ﺗﯿﺘﺮﺍﺳﯿﻮﻥ ﺭﺳﻮﺑﯽ ﻧﺪﺍﺷﺖ‪.‬‬‫‪B.2‬‬‫ﺟﺮﻡ ‪ NaCl‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ ﺩﺭ ﻇﺮﻑ ﻭﺍﮐﻨﺶ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ )ﺍﻃﻼﻋﺎﺗﯽ ﮐﻪ ﺯﯾﺮ ﺁﻥ ﺧﻂ ﮐﺸﯿﺪﻩ ﺷﺪﻩ ﺑﻮﺩ(‪ .‬ﺟﺮﻡ 𝑥‬‫ﺑﺮ ﺣﺴﺐ ﮔﺮﻡ ﻣﺮﺑﻮﻁ ﺑﻪ ‪ Na2 SiF6‬ﮐﻪ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺎﺩﻩ ﺍﻭﻟﯿﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪15pt‬‬‫‪B.3‬‬‫ﺟﺮﻡ ‪ CF3 Cl‬ﺗﻮﻟﯿﺪ ﺷﺪﻩ‬‫‪8pt‬‬‫‪ 77.8‬ﺩﺭﺻﺪ ﺍﺯ ‪ CCl4‬ﮐﻪ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺎﺩﻩ ﺍﻭﻟﯿﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ ﻭﺍﮐﻨﺶ ﻧﺪﺍﺩ‪.‬‬‫ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬

IChO

A5-1

IRN-2 C-5 A-1

persian (Iran)‫ﺳﯿﻠﯿﺴﯿﻢ ﻣﺮﻣﻮﺯ‬A ‫ﻗﺴﻤﺖ‬(9 pt) A.1
pt) 3) Cpt) 3) Bpt) 3) A(7 pt) A.2

C6 H6 ∶

kJ mol–1 , C ∶kJ mol–1

IChO
IRN-2 C-5 A-2

A5-2
persian (Iran)(6 pt) A.3

kJ mol–1

Δ𝐻 =(10 pt) A.4
pt) 5) Ept) 5) DIChO
IRN-2 C-5 A-3

A5-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(5 pt) B.1(15 pt) B.2( ‫)ﺍﺩﺍﻣﻪ ﺩﺭ ﺻﻔﺤﻪ‬IChO
IRN-2 C-5 A-4

A5-4
persian (Iran)(cont.) B.2

NaCl ∶

g, Na2 SiF6 ∶

g

IChO
IRN-2 C-5 A-5

A5-5
persian (Iran)(8 pt) B.3

CF3 Cl ∶g

IRN-2 C-6 C-1

IRN-2 C-6 C
Ehsan Naderi Donig

IChO
Problem 6
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q6-1‬‬‫‪IChO‬‬‫‪IRN-2 C-6 Q-1‬‬‫)‪persian (Iran‬‬‫ﺷﯿﻤﯽ ﺣﺎﻟﺖ ﺟﺎﻣﺪ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ‬‫‪13% of the total‬‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪6‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪3‬‬‫‪B.1‬‬‫‪6‬‬‫‪B.2‬‬‫‪4‬‬‫‪B.3‬‬‫‪4‬‬‫‪B.4‬‬‫‪4‬‬‫‪C.1‬‬‫‪5‬‬‫‪C.2‬‬‫‪5‬‬‫‪C.3‬‬‫‪5‬‬‫‪Total‬‬‫‪45‬‬‫‪Score‬‬‫ﺁﺗﺸﻔﺸﺎﻧﯽ ﺩﺭ ﺟﺰﯾﺮﻩ ﺳﺎﮐﻮﺭﺍﺟﯿﻤﺎ‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﺳﯿﻠﯿﮑﺎﺗﯽ ﻣﻮﺟﻮﺩ ﺩﺭ ﻣﻮﺍﺩ ﻣﺬﺍﺏ ﺁﺗﺸﻔﺸﺎﻧﯽ‬‫ﮊﺍﭘﻦ ﯾﮑﯽ ﺍﺯ ﮐﺸﻮﺭﻫﺎﯾﯽ ﺍﺳﺖ ﮐﻪ ﺑﺎﻻﺗﺮﯾﻦ ﺗﻌﺪﺍﺩ ﺁﺗﺸﻔﺸﺎﻥﻫﺎ ﺭﺍ ﺩﺭ ﺟﻬﺎﻥ ﺩﺍﺭﺩ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﮐﺎﻧﯽﻫﺎﯼ‬‫ِ‬‫)ﻣﺎﮔﻤﺎ( ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﯽﺷﻮﻧﺪ‪ ،‬ﺑﺨﺸﯽ ﺍﺯ ﯾﻮﻥﻫﺎﯼ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ ) ‪ (Mn+‬ﺩﺭ ﻣﺎﮔﻤﺎ ﻭﺍﺭﺩِ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﻣﯽﺷﻮﻧﺪ‪ .‬ﯾﻮﻥ ‪ Mn+‬ﻣﻮﺭﺩ ﻣﻄﺎﻟﻌﻪ ﺩﺭ ﺍﯾﻦ‬‫ﻣﺴﺎﻟﻪ ﺑﺎ ﯾﻮﻥﻫﺎﯼ ﺍﮐﺴﯿﺪ ) ‪ (O2−‬ﮐﺌﻮﺭﺩﯾﻨﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺳﺎﺧﺘﺎﺭ ﻫﻨﺪﺳﯽ ﺍﻃﺮﺍﻑ ﯾﻮﻥ ﻓﻠﺰ ﺩﺭ ﻣﺎﮔﻤﺎ ﺑﻪﺻﻮﺭﺕ ﭼﻬﺎﺭ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﻭ ﭼﻬﺎﺭﻭﺟﻬﯽ ) ‪(𝑇d‬‬‫ﻭ ﺩﺭ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﺑﻪﺻﻮﺭﺕ ﺷﺶ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﻭ ﻫﺸﺖﻭﺟﻬﯽ ) ‪ (𝑂h‬ﺍﺳﺖ ﮐﻪ ﻫﺮ ﺩﻭ ﺣﺎﻟﺖ ﺁﺭﺍﯾﺶ ﺍﻟﮑﺘﺮﻭﻧﯽ ﭘﺮﺍﺳﭙﯿﻦ )‪(high-spin‬‬‫ﺩﺍﺭﻧﺪ‪ .‬ﺿﺮﯾﺐ ﺗﻮﺯﯾﻊ ‪ Mn+‬ﺑﯿﻦ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﻭ ﻣﺎﮔﻤﺎ )𝐷( ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﭼﻨﯿﻦ ﻧﻮﺷﺖ‪:‬‬‫ﺩﺭ ﺍﯾﻦ ﺭﺍﺑﻄﻪ ‪ [M]s‬ﻭ ‪ [M]l‬ﺑﻪﺗﺮﺗﯿﺐ ﻏﻠﻈﺖ ‪ Mn+‬ﺭﺍ ﺩﺭ ﮐﺎﻧﯽﻫﺎﯼ ﺳﯿﻠﯿﮑﺎﺗﯽ ﻭ ﻣﺎﮔﻤﺎ ﻧﺸﺎﻥ ﻣﯽﺩﻫﻨﺪ‪ .‬ﺩﺭ ﺟﺪﻭﻝ ﺯﯾﺮ ﻣﻘﺎﺩﯾﺮ 𝐷 ﺑﺮﺍﯼ ‪ Cr2+‬ﻭ ‪Mn2+‬‬‫ﺑﻪﻋﻨﻮﺍﻥ ﻧﻤﻮﻧﻪ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪Q6-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-6 Q-2‬‬‫ﻣﯿﺪﺍﻥ ‪ 𝑂h‬ﺑﺎﺷﺪ‪ .‬ﻫﻤﭽﻨﯿﻦ ‪ ΔT‬ﻭ ‪CFSET‬‬‫ﻓﺮﺽ ﮐﻨﯿﺪ ‪ ΔO‬ﻓﺎﺻﻠﻪﯼ ﺍﻧﺮﮊﯼ ﺍﻭﺭﺑﯿﺘﺎﻝﻫﺎﯼ ‪ d‬ﯾﻮﻥ ‪ Mn+‬ﻭ ‪ CFSEO‬ﺍﻧﺮﮊﯼ ﭘﺎﯾﺪﺍﺭﯼ ﻣﯿﺪﺍﻥ ﺑﻠﻮﺭ ﺩﺭ ﯾﮏ‬‫ِ‬‫ﻣﯿﺪﺍﻥ ‪ 𝑇d‬ﺩﺭ ﻧﻈﺮ ﺑﮕﯿﺮﯾﺪ‪.‬‬‫ﺭﺍ ﻫﻤﯿﻦ ﮐﻤﯿﺖﻫﺎ ﺩﺭ ﯾﮏ‬‫ِ‬‫‪A.1‬‬‫‪ |CFSEO −CFSET | =ΔCFSE‬ﺭﺍ ﺑﺮﺣﺴﺐ ‪ ΔO‬ﺑﺮﺍﯼ ﯾﻮﻥﻫﺎﯼ ‪ Mn2+ ،Cr2+‬ﻭ ‪ Co2+‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪ .‬ﻓﺮﺽ‬‫ﮐﻨﯿﺪ ‪ΔT = 4/9ΔO‬‬‫‪6pt‬‬‫‪A.2‬‬‫ﺑﺎ ﺭﺳﻢ 𝐷‪ ln‬ﺩﺭ ﺑﺮﺍﺑﺮ ‪ ΔCFSE/ΔO‬ﺩﺭ ﺳﯿﺴﺘﻢ ﻣﺨﺘﺼﺎﺕ ﺩﮐﺎﺭﺗﯽ ﺯﯾﺮ‪ ،‬ﺭﺍﺑﻄﻪﺍﯼ ﺧﻄﯽ ﻣﺸﺎﻫﺪﻩ ﻣﯽﺷﻮﺩ‪.‬‬‫ﻣﻘﺪﺍﺭ 𝐷 ﺭﺍ ﺑﺮﺍﯼ ‪ Co2+‬ﺗﺨﻤﯿﻦ ﺑﺰﻧﯿﺪ‪.‬‬‫‪3pt‬‬‫ﺳﺎﺧﺘﺎﺭ ﻧﻤﮏ ﻃﻌﺎﻡ )‪ (NaCl‬ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﯽﺷﻮﻧﺪ ﮐﻪ ﺩﺭ ﺁﻥ ‪ Mn+‬ﺳﺎﺧﺘﺎﺭ‬‫ﻓﻠﺰﯼ ‪) MO‬ﮐﻪ ‪ M‬ﻣﯽﺗﻮﺍﻧﺪ ‪ Mn ، V ، Ti ، Ca‬ﯾﺎ ‪ Co‬ﺑﺎﺷﺪ( ﺩﺭ‬‫ﺍﮐﺴﯿﺪﻫﺎﯼ‬
‫ِ‬
‫ِ‬
‫ً‬
‫ﻋﻤﺪﺗﺎ ﺑﺮﻫﻢﮐﻨﺶﻫﺎﯼ ﮐﻮﻟﻨﯽ ﺗﻌﯿﯿﻦ ﻣﯽﮐﻨﺪ‪.‬‬‫ﻫﻨﺪﺳﯽ ‪ 𝑂h‬ﺑﺎ ﺁﺭﺍﯾﺶ ﺍﻟﮑﺘﺮﻭﻧﯽ ﭘﺮﺍﺳﭙﯿﻦ )‪ (high-spin‬ﺑﻪ ﺧﻮﺩ ﻣﯽﮔﯿﺮﺩ‪ .‬ﺁﻧﺘﺎﻟﭙﯽ ﺷﺒﮑﻪ ﺍﯾﻦ ﺍﮐﺴﯿﺪﻫﺎ ﺭﺍ‬‫ﺍﯾﻦ ﺑﺮﻫﻢﮐﻨﺶﻫﺎ ﺑﺎ ﺷﻌﺎﻉ ﻭ ﺑﺎﺭ ﯾﻮﻥﻫﺎ ﺗﻌﯿﯿﻦ ﻣﯽﺷﻮﺩ‪ .‬ﻫﻤﭽﻨﯿﻦ ‪ CFSE‬ﻣﺮﺑﻮﻁ ﺑﻪ ‪ Mn+‬ﺩﺭ ﻣﯿﺪﺍﻥ ‪ 𝑂h‬ﻫﻢ ﺳﻬﻢﻫﺎﯾﯽ ﺩﺭ ﺁﻧﺘﺎﻟﭙﯽ ﺷﺒﮑﻪ ﺩﺍﺭﺩ‪.‬‬‫‪A.3‬‬‫ﺍﺯ ﺑﯿﻦ ﮔﺰﯾﻨﻪﻫﺎﯼ )‪ (a‬ﺗﺎ )‪ (f‬ﺩﺭ ﺟﺪﻭﻝ ﺯﯾﺮ‪ ،‬ﻣﺠﻤﻮﻋﻪ ﻣﻨﺎﺳﺐ ﺍﺯ ﺁﻧﺘﺎﻟﭙﯽﻫﺎﯼ ﺷﺒﮑﻪ ] ‪ [kJ mol−1‬ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬

‫‪Q6-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-6 Q-3‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫‪2+‬‬‫‪3+‬‬‫ﺍﯼ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﺷﮑﻞ ‪ 1‬ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﯽﺷﻮﺩ‪ .‬ﺩﺭ‬‫ﯾﮏ ﺍﮐﺴﯿﺪ ﻣﺨﻠﻮﻁ )‪ (A‬ﮐﻪ ﺷﺎﻣﻞ ﯾﻮﻥﻫﺎﯼ ‪ La‬ﻭ ‪ Cu‬ﺍﺳﺖ‪ ،‬ﺩﺭ ﺳﻠﻮﻝ ﻭﺍﺣﺪ ﭼﻬﺎﺭﮔﻮﺷﻪ ِ‬‫ﻭﺟﻬﯽ ] ‪ [CuO6‬ﻃﻮﻝ ﭘﯿﻮﻧﺪﻫﺎﯼ ‪ Cu-O‬ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ‪ (𝑙𝑧 ) z‬ﺑﻠﻨﺪﺗﺮ ﺍﺯ ﻃﻮﻝ ﭘﯿﻮﻧﺪﻫﺎﯼ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ‪ (𝑙𝑥 ) x‬ﺍﺳﺖ ﻭ ] ‪ [CuO6‬ﻧﺴﺒﺖ‬
‫ﻫﺸﺖ‬
‫ِ‬
‫ﭼﻨﺪﺣﺎﻟﺘﯽ ﺍﻭﺭﺑﯿﺘﺎﻝﻫﺎﯼ ‪ d𝑥2 −𝑦2 ) eg‬ﻭ ‪ (d𝑧2‬ﻣﯽﺷﻮﺩ‪.‬‬‫ﺭﻓﺘﻦ‬‫ﺑﯿﻦ‬‫ﺍﺯ‬‫ﺑﺎﻋﺚ‬‫ﺍﻧﺤﺮﺍﻑ‬‫ﺍﯾﻦ‬‫ﺷﻮﺩ‪.‬‬‫ﻣﯽ‬‫ﻣﻨﺤﺮﻑ‬‫ﻣﻨﻈﻢ‪،‬‬‫𝑂‬‫ﻫﻨﺪﺳﯽ‬‫ﺳﺎﺧﺘﺎﺭ‬
‫ﺑﻪ‬
‫‪h‬‬
‫ِ‬

‫ﺷﮑﻞ ‪1‬‬‫‪ A‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﺎ ﺗﺠﺰﯾﻪ ﺣﺮﺍﺭﺗﯽ )ﭘﯿﺮﻭﻟﯿﺰ( ﮐﻤﭙﻠﮑﺲ ‪ B‬ﺳﻨﺘﺰ ﮐﺮﺩ‪ .‬ﮐﻤﭙﻠﮑﺲ ‪ B‬ﺑﺎ ﻣﺨﻠﻮﻁ ﮐﺮﺩﻥ ﮐﻠﺮﯾﺪﻫﺎﯼ ﻓﻠﺰﯼ ﺩﺭ ﻣﺤﻠﻮﻝ ﺁﺑﯽ ﺁﻣﻮﻧﯿﺎﮎ ﺭﻗﯿﻖ )ﺷﺎﻣﻞ‬‫ﭘﯿﺮﻭﻟﯿﺰ ‪ B‬ﺩﺭ ﻫﻮﺍﯼ ﺧﺸﮏ‪ ،‬ﺗﺎ ﺩﻣﺎﯼ ‪ 200‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ ﮐﺎﻫﺶ‬‫ﺍﺳﮑﻮﺍﺭﯾﮏ ﺍﺳﯿﺪ ﺑﺎ ﻓﺮﻣﻮﻝ ‪ C4 H2 O4‬ﮐﻪ ﯾﮏ ﺩﯼﺍﺳﯿﺪ ﺍﺳﺖ( ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪ .‬ﺑﺎ‬‫ِ‬‫ﻭﺯﻥ ‪ 29.1‬ﺩﺭﺻﺪﯼ ﻣﺸﺎﻫﺪﻩ ﻣﯽﺷﻮﺩ ﮐﻪ ﻧﺎﺷﯽ ﺍﺯ ﺍﺯ ﺩﺳﺖ ﺭﻓﺘﻦ ﺁﺏ ﺑﻠﻮﺭﯼ ﺍﺳﺖ‪ .‬ﺑﻪ ﺩﻧﺒﺎﻝ ﺁﻥ‪ ،‬ﺗﺎ ﺩﻣﺎﯼ ‪ 700‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ‪ ،‬ﮐﺎﻫﺶ ﻭﺯﻥ ﺩﯾﮕﺮﯼ‬‫ِ‬‫ﻫﻢ ﺑﻪ ﺩﻟﯿﻞ ﺁﺯﺍﺩ ﺷﺪﻥ ‪ CO2‬ﺭﺥ ﻣﯽﺩﻫﺪ‪ .‬ﮐﺎﻫﺶ ﻭﺯﻥ ﮐﻠﯽ ﻃﯽ ﻓﺮﺍﯾﻨﺪ ﺗﺸﮑﯿﻞ ‪ A‬ﺍﺯ ‪ B‬ﺑﺮﺍﺑﺮ ‪ 63.6‬ﺩﺭﺻﺪ ﺍﺳﺖ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﻃﯽ ﻭﺍﮐﻨﺶ ﭘﯿﺮﻭﻟﯿﺰ‬‫ﻓﻘﻂ ﺁﺏ ﻭ ‪ CO2‬ﺁﺯﺍﺩ ﻣﯽﺷﻮﺩ‪.‬‬‫‪B.1‬‬‫ﻓﺮﻣﻮﻝﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ‪ A‬ﻭ ‪ B‬ﺭﺍ ﺑﻨﻮﯾﺴﯿﺪ‪.‬‬‫‪6pt‬‬‫‪B.2‬‬‫ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺷﮑﻞ ‪ 𝑙𝑥 ،1‬ﻭ 𝑧𝑙 ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪4pt‬‬‫‪B.3‬‬‫ﺑﺮﺍﯼ ‪ Cu2+‬ﺩﺭ ﻫﺸﺖﻭﺟﻬﯽ ﺍﻧﺤﺮﺍﻑﯾﺎﻓﺘﻪﯼ ] ‪ [CuO6‬ﺩﺭ ﺳﺎﺧﺘﺎﺭ ‪) A‬ﺷﮑﻞ ‪ ،(1‬ﻧﺎﻡ ﺍﻭﺭﺑﯿﺘﺎﻝﻫﺎﯼ ‪ eg‬ﺷﮑﺎﻓﺘﻪﺷﺪﻩ‬‫) ‪ d𝑥2 −𝑦2‬ﻭ ‪ (d𝑧2‬ﺭﺍ ﺩﺭ ﻗﺴﻤﺖﻫﺎﯼ )‪ (i‬ﻭ )‪ (ii‬ﭘﺎﺳﺦﻧﺎﻣﻪ ﺑﻨﻮﯾﺴﯿﺪ ﻭ ﺁﺭﺍﯾﺶ ﺍﻟﮑﺘﺮﻭﻧﯽ ﺭﺍ ﺩﺭ ﮐﺎﺩﺭ ﺧﻂﭼﯿﻦ‬‫ﭘﺎﺳﺦﻧﺎﻣﻪ ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪4pt‬‬

‫‪Q6-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-6 Q-4‬‬‫‪ A‬ﯾﮏ ﻋﺎﯾﻖ ﺍﺳﺖ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﯾﮏ ﯾﻮﻥ ‪ La3+‬ﺟﺎﯾﺶ ﺭﺍ ﺑﻪ ﯾﮏ ‪ Sr2+‬ﻣﯽﺩﻫﺪ‪ ،‬ﺩﺭ ﺷﺒﮑﻪ ﮐﺮﯾﺴﺘﺎﻟﯽ ﺣﻔﺮﻩﺍﯼ ﺍﯾﺠﺎﺩ ﻣﯽﺷﻮﺩ ﮐﻪ ﻣﯽﺗﻮﺍﻧﺪ ﺟﺮﯾﺎﻥ‬‫ﺍﻟﮑﺘﺮﯾﮑﯽ ﺭﺍ ﻫﺪﺍﯾﺖ ﮐﻨﺪ‪ .‬ﺩﺭ ﻧﺘﯿﺠﻪ‪ ،‬ﺳﺎﺧﺘﺎﺭ ‪ A‬ﺩﻭﭖﺷﺪﻩ ﺑﺎ ‪ Sr2+‬ﺩﺭ ﺩﻣﺎﯼ ﮐﻤﺘﺮ ﺍﺯ ‪ 38‬ﮐﻠﻮﯾﻦ‪ ،‬ﺧﺎﺻﯿﺖ ﺍﺑﺮﺭﺳﺎﻧﺎﯾﯽ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ‬‫ﻭﺍﮐﻨﺶ ﺟﺎﯾﮕﺰﯾﻨﯽ ﺑﺮﺍﯼ ‪ A‬ﺭﺥ ﻣﯽﺩﻫﺪ‪ ،‬ﺩﺭ ﻫﺮ ﻣﺘﺮ ﻣﮑﻌﺐ ﺗﻌﺪﺍﺩ ‪ 2.05 × 1027‬ﺣﻔﺮﻩ ﺍﯾﺠﺎﺩ ﻣﯽﺷﻮﺩ‪.‬‬‫‪B.4‬‬‫ﻗﺴﻤﺖ ‪C‬‬

‫ﺟﺎﯾﮕﺰﯾﻦ ‪ La3+‬ﺷﺪﻩﺍﻧﺪ ﺭﺍ ﺑﺮ ﺍﺳﺎﺱ ﻧﺴﺒﺖﻫﺎﯼ‬‫ﺩﺭﺻﺪ ﯾﻮﻥﻫﺎﯼ ‪ Sr2+‬ﮐﻪ‬‫ﻣﻮﻟﯽ ﻭﺍﮐﻨﺶ ﺟﺎﯾﮕﺰﯾﻨﯽ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫ِ‬‫ِ‬‫ﻭﺍﻻﻧﺲ ﺑﻘﯿﻪﯼ ﯾﻮﻥﻫﺎﯼ ﺗﺸﮑﯿﻞﺩﻫﻨﺪﻩﯼ ﮐﺮﯾﺴﺘﺎﻝ ﻭ ﺳﺎﺧﺘﺎﺭ ﮐﺮﯾﺴﺘﺎﻟﯽ ﺩﺭ ﺍﺛﺮ ﻭﺍﮐﻨﺶ ﺟﺎﯾﮕﺰﯾﻨﯽ‬‫ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ‬‫ِ‬‫ﺗﻐﯿﯿﺮ ﻧﻤﯽﮐﻨﺪ‪.‬‬‫‪−‬‬‫‪4pt‬‬‫‪2+‬‬‫‪ Cu2 (CH3 CO2 )4‬ﺍﺯ ﭼﻬﺎﺭ ﯾﻮﻥ ‪ CH3 CO2‬ﺗﺸﮑﯿﻞ ﺷﺪﻩ ﺍﺳﺖ ﮐﻪ ﺑﻪ ﺩﻭ ﯾﻮﻥ ‪ Cu‬ﮐﺌﻮﺭﺩﯾﻨﻪ ﺷﺪﻩﺍﻧﺪ )ﺷﮑﻞ ‪ .( 2A‬ﺗﺮﮐﯿﺐ ‪Cu2 (CH3 CO2 )4‬‬‫‪−‬‬
‫ﮐﺮﺑﻦ ﭼﻬﺎﺭ ﮔﺮﻭﻩ ‪ CH3 CO2‬ﻋﺒﻮﺭ ﻣﯽﮐﻨﻨﺪ ﻭ ﯾﮏ ﻣﺤﻮﺭ ﺍﺯ ﺩﻭ ﯾﻮﻥ‬‫ﺗﻘﺎﺭﻥ ﺳﺎﺧﺘﺎﺭﯼ ﺑﺎﻻﯾﯽ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ‪ .‬ﺩﺭ ﺍﯾﻦ ﺳﺎﺧﺘﺎﺭ ﺩﻭ ﻣﺤﻮﺭ ﺍﺯ ﺍﺗﻢﻫﺎﯼ‬
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‫‪2+‬‬‫‪ Cu‬ﻣﯽﮔﺬﺭﺩ‪ .‬ﺗﻤﺎﻡ ﺍﯾﻦ ﺷﺶ ﻣﺤﻮﺭ ﺩﺭ ﻭﺿﻌﯿﺖ ﻋﻤﻮﺩ ﺑﺮ ﻫﻢ ﺟﻬﺖﮔﯿﺮﯼ ﮐﺮﺩﻩﺍﻧﺪ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﺑﻪ ﺟﺎﯼ ‪ CH3 CO2‬ﺍﺯ ﯾﮏ ﻟﯿﮕﺎﻧﺪ ﺩﯼﮐﺮﺑﻮﮐﺴﯿﻼﺕ‬‫‪2+‬‬‫‪L1‬‬‫ﻣﺴﻄﺢ ‪) L1‬ﺷﮑﻞ ‪ ( 2B‬ﻭ ‪ Cu‬ﺗﺸﮑﯿﻞ‬‫ﮐﺮﺑﻮﮐﺴﯿﻼﺕ‬‫‪ Cu4 (L1‬ﺍﺯ ﺩﯼ‬‫ﻗﻔﺴﯽ ‪L1)4‬‬‫ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ ،‬ﯾﮏ »ﮐﻤﭙﻠﮑﺲ ﻗﻔﺴﯽ« ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ‪ .‬ﮐﻤﭙﻠﮑﺲ‬
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‫ﺷﺪﻥ ﺩﻭ ﮔﺮﻭﻩ ﮐﺮﺑﻮﮐﺴﯿﻼﺕ‪ ،‬ﮐﻪ ﺩﺭ ﺷﮑﻞ ‪ 2B‬ﺑﺎ ﺩﻭ ﻋﻼﻣﺖ ﭘﯿﮑﺎﻥ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩﺍﻧﺪ‪،‬‬‫ﺷﺪﻩ ﺍﺳﺖ )ﺷﮑﻞ ‪ .( 2C‬ﺯﺍﻭﯾﻪ ‪ θ‬ﺑﯿﻦ ﺟﻬﺖﻫﺎﯼ ﮐﺌﻮﺭﯾﻨﻪ‬‫ِ‬‫ﺳﺎﺧﺘﺎﺭ ﮐﻤﭙﻠﮑﺲ ﻗﻔﺴﯽ ﺭﺍ ﺗﻌﯿﯿﻦ ﻣﯽﮐﻨﺪ‪ .‬ﺑﺮﺍﯼ ‪ L1‬ﺯﺍﻭﯾﻪ ‪ θ‬ﺑﺮﺍﺑﺮ ﺑﺎ ﺻﻔﺮ ﺩﺭﺟﻪ ﺍﺳﺖ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﺍﺗﻢﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺷﮑﻞ ‪ 2‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ‬‫ﻧﺸﺪﻩﺍﻧﺪ‪.‬‬‫ﺷﮑﻞ ‪2‬‬

‫‪Q6-5‬‬‫)‪persian (Iran‬‬‫‪C.1‬‬‫‪IChO‬‬‫‪IRN-2 C-6 Q-5‬‬‫ﺷﯿﻤﯿﺎﯾﯽ‬‫ﻣﺴﻄﺢ ‪) L2‬ﺷﮑﻞ ﺯﯾﺮ( ﺯﺍﻭﯾﻪ ‪ θ‬ﺩﺭ ﻣﻘﺪﺍﺭ ‪ 90‬ﺩﺭﺟﻪ ﺗﺜﺒﯿﺖ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺍﮔﺮ ﺗﺮﮐﯿﺐ‬‫ﺑﺮﺍﯼ ﺩﯼﮐﺮﺑﻮﮐﺴﯿﻼﺕ‬‫ِ‬‫ِ‬‫‪2+‬‬‫‪L2‬‬‫‪ Cu𝑛 (L2‬ﺑﺎﺷﺪ‪ ،‬ﮐﻮﭼﮏﺗﺮﯾﻦ ﻣﻘﺎﺩﯾﺮ ﺻﺤﯿﺢ ‪ n‬ﻭ ‪m‬‬‫ﻗﻔﺴﯽ ﺗﺸﮑﯿﻞﺷﺪﻩ ﺍﺯ ‪ L2‬ﻭ ‪ Cu‬ﺑﻪﺻﻮﺭﺕ 𝑚)‪L2‬‬‫ﮐﻤﭙﻠﮑﺲ‬‫ِ‬‫‪−‬‬‫ﺭﺍ ﺑﻨﻮﯾﺴﯿﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ ﻓﻘﻂ ﮔﺮﻭﻩﻫﺎﯼ ‪ CO2‬ﺩﺭ ‪ L2‬ﺑﺎ ﯾﻮﻥﻫﺎﯼ‬‫ﻣﯽﺩﻫﻨﺪ‪.‬‬‫‪2+‬‬‫‪ Cu‬ﭘﯿﻮﻧﺪ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﺗﺸﮑﯿﻞ‬‫‪5pt‬‬

‫‪Q6-6‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-6 Q-6‬‬‫‪2+‬‬‫‪−‬‬‫ﮐﻤﭙﻠﮑﺴﯽ ﺍﺯ ﺭﻭﯼ ﺑﺎ ﻓﺮﻣﻮﻝ ‪ Zn4 O(CH3 CO2 )6‬ﺷﺎﻣﻞ ﭼﻬﺎﺭ ﯾﻮﻥ ‪ Zn‬ﭼﻬﺎﺭﻭﺟﻬﯽ‪ ،‬ﺷﺶ ‪ CH3 CO2‬ﻭ ﯾﮏ ‪ O2−‬ﺍﺳﺖ )ﺷﮑﻞ ‪ .( 3A‬ﺩﺭ‬
‫‪−‬‬
‫‪−‬‬
‫ﮐﺮﺑﻦ ‪ CH3 CO2‬ﻣﯽﮔﺬﺭﻧﺪ‪ ،‬ﺑﺮ ﻫﻢ ﻋﻤﻮﺩﻧﺪ‪ .‬ﻫﻨﮕﺎﻣﯽ ﮐﻪ ﺑﻪﺟﺎﯼ ‪CH3 CO2‬‬‫ﺍﯾﻦ ﮐﻤﭙﻠﮑﺲ ‪ O2−‬ﺩﺭ ﻣﺒﺪﺃ ﻗﺮﺍﺭ ﺩﺍﺭﺩ ﻭ ﺳﻪ ﻣﺤﻮﺭﯼ ﮐﻪ ﺍﺯ ﺍﺗﻢﻫﺎﯼ‬‫ِ‬‫‪2+‬‬‫ﺍﺯ ﭘﺎﺭﺍ‪-‬ﺑﻨﺰﻥﺩﯼﮐﺮﺑﻮﮐﺴﯿﻼﺕ )ﺷﮑﻞ ‪ 3B،‬ﺳﺎﺧﺘﺎﺭ ‪ ،L3‬ﺑﺎ ‪ (180° = θ‬ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ ،‬ﮐﻼﺳﺘﺮﻫﺎﯼ ‪ Zn‬ﺑﻪ ﯾﮑﺪﯾﮕﺮ ﻣﺘﺼﻞ ﻣﯽﺷﻮﻧﺪ ﻭ ﯾﮏ ﺟﺎﻣﺪ‬‫‪L3‬‬‫‪ [Zn4 O(L3‬ﺍﺳﺖ ﻭ ﯾﮏ‬‫ﮐﺮﯾﺴﺘﺎﻟﯽ )‪ (X‬ﺗﺸﮑﯿﻞ ﻣﯽﺷﻮﺩ ﮐﻪ ﻧﻮﻋﯽ »ﭘﻠﯿﻤﺮ ﮐﺌﻮﺭﺩﯾﻨﺎﺳﯿﻮﻧﯽ ﻣﺘﺨﻠﺨﻞ« ﺍﺳﺖ )ﺷﮑﻞ ‪ .( 3C‬ﺗﺮﮐﯿﺐ ‪ X‬ﺑﻪﺻﻮﺭﺕ 𝑛] ‪L3)3‬‬‫ﺳﺎﺧﺘﺎﺭ ﮐﺮﯾﺴﺘﺎﻟﯽ ﻣﮑﻌﺒﯽ ﻫﻤﺮﺍﻩ ﺑﺎ ﺣﻔﺮﻩﻫﺎﯾﯽ ﺩﺭ ﺍﻧﺪﺍﺯﻩﯼ ﻧﺎﻧﻮ ﺩﺍﺭﺩ‪ .‬ﺩﺭ ﺷﮑﻞ ‪ 3D‬ﯾﮑﯽ ﺍﺯ ﺍﯾﻦ ﺣﻔﺮﻩﻫﺎ ﺑﺎ ﯾﮏ ﮐﺮﻩ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﻫﻤﭽﻨﯿﻦ‬‫‪2+‬‬‫ﭼﻬﺎﺭﻭﺟﻬﯽ ‪ Zn‬ﺑﺎ ﺷﮑﻞﻫﺎﯼ ﭼﻨﺪﻭﺟﻬﯽ ﺑﻪ ﺭﻧﮓ ﺧﺎﮐﺴﺘﺮﯼ ﺗﯿﺮﻩ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩﺍﻧﺪ‪ .‬ﺗﻮﺟﻪ ﮐﻨﯿﺪ ﮐﻪ ﺍﺗﻢﻫﺎﯼ‬‫ﺩﺭ ﺷﮑﻞﻫﺎﯼ ‪ 3C‬ﻭ ‪ ، 3D‬ﮐﻼﺳﺘﺮﻫﺎﯼ‬‫ِ‬‫ﻫﯿﺪﺭﻭﮊﻥ ﺩﺭ ﺷﮑﻞ ‪ 3‬ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻧﺸﺪﻩﺍﻧﺪ‪.‬‬‫ﺷﮑﻞ ‪3‬‬‫‪C.2‬‬‫ﺗﺮﮐﯿﺐ ‪ X‬ﯾﮏ ﺳﻠﻮﻝ ﻭﺍﺣﺪ ﻣﮑﻌﺒﯽ ﺑﺎ ﻃﻮﻝ ﻟﺒﻪ ‪ a‬ﻭ ﭼﮕﺎﻟﯽ ‪ 0.592 g cm−3‬ﺩﺍﺭﺩ )ﺷﮑﻞ ‪ a .( 3C‬ﺭﺍ ﺑﺮﺣﺴﺐ‬‫‪ cm‬ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪5pt‬‬‫‪C.3‬‬‫ﺗﺮﮐﯿﺐ ‪ X‬ﺗﻌﺪﺍﺩ ﻗﺎﺑﻞ ﺗﻮﺟﻬﯽ ﺣﻔﺮﻩ ﺩﺍﺭﺩ ﻭ ﯾﮏ ﮔﺮﻡ ﺍﺯ ‪ X‬ﻣﯽﺗﻮﺍﻧﺪ ﺩﺭ ﻓﺸﺎﺭ ‪ 1‬ﺑﺎﺭ ﻭ ﺩﻣﺎﯼ ‪ 25‬ﺩﺭﺟﻪ ﺳﺎﻧﺘﯽﮔﺮﺍﺩ‪،‬‬‫ﻣﻘﺪﺍﺭ ‪ 3.0 × 102‬ﻣﯿﻠﯽﻟﯿﺘﺮ ﮔﺎﺯ ‪ CO2‬ﺭﺍ ﺩﺭ ﺣﻔﺮﻩﻫﺎﯾﺶ ﺟﺎﯼ ﺩﻫﺪ‪ .‬ﻣﯿﺎﻧﮕﯿﻦ ﺗﻌﺪﺍﺩ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ‪ CO2‬ﺩﺭ ﻫﺮ‬‫ﺣﻔﺮﻩ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﮐﻨﯿﺪ‪.‬‬‫‪5pt‬‬IChO
IRN-2 C-6 A-1

A6-1
persian (Iran)‫ﺷﯿﻤﯽ ﺣﺎﻟﺖ ﺟﺎﻣﺪ ﻓﻠﺰﺍﺕ ﻭﺍﺳﻄﻪ‬A ‫ﻗﺴﻤﺖ‬(6 pt) A.1

ΔO Co2+ ∶, ΔO Mn2+ ∶,ΔO Cr2+ ∶

IChO

A6-2

IRN-2 C-6 A-2

persian (Iran)(3 pt) A.2

𝐷∶(3 pt) A.3

IChO

A6-3

IRN-2 C-6 A-3

persian (Iran)B ‫ﻗﺴﻤﺖ‬(6 pt) B.1

B∶

,A ∶
(4 pt) B.2

𝑙𝑧 =nm ,𝑙𝑥 =

nm

IChO

A6-4

IRN-2 C-6 A-4

persian (Iran)(4 pt) B.3(ii) ∶,(i) ∶(4 pt) B.4

%

IChO

A6-5

IRN-2 C-6 A-5

persian (Iran)C ‫ﻗﺴﻤﺖ‬(5 pt) C.1𝑚=,𝑛 =(5 pt) C.2

𝑎=

cm

IChO
IRN-2 C-6 A-6

A6-6
persian (Iran)(5 pt) C.3

IRN-2 C-7 C-1

IRN-2 C-7 C
Ehsan Naderi Donig

IChO
Problem 7
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q7-1‬‬‫‪IChO‬‬‫‪IRN-2 C-7 Q-1‬‬‫)‪persian (Iran‬‬‫ﺑﺎﺯﯼ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻧﻮﺋﯿﺪﯼ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪5‬‬‫‪13% of the total‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪19‬‬‫‪B.1‬‬‫‪10‬‬‫‪Total‬‬‫‪36‬‬‫‪Score‬‬‫ﭘﺮﻭﻓﺴﻮﺭ ﻧﻮﺯﻭ )‪ (۱۹۹۶–۱۹۰۲‬ﺯﻣﯿﻨﻪ ﺗﺤﻘﯿﻘﺎﺗﯽ ﺗﺮﮐﯿﺒﺎﺕ ﺁﺭﻭﻣﺎﺗﯿﮏ ﻏﯿﺮ ﺑﻨﺰﻧﻮﺋﯿﺪﯼ ﺭﺍ ﺍﻓﺘﺘﺎﺡ ﮐﺮﺩ ﮐﻪ ﻫﻢ ﺍﮐﻨﻮﻥ ﺩﺭ ﻫﻤﻪ ﺟﺎ ﺩﺭ ﺷﯿﻤﯽ ﺁﻟﯽ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪.‬‬‫ﻗﺴﻤﺖ ‪A‬‬‫‪ Lineariifolianone‬ﯾﮏ ﻣﺤﺼﻮﻝ ﻃﺒﯿﻌﯽ ﺑﺎ ﺳﺎﺧﺘﺎﺭ ﻣﻨﺤﺼﺮ ﺑﻪ ﻓﺮﺩ ﺍﺳﺖ‪ ،‬ﮐﻪ ﺍﺯ ﮔﯿﺎﻩ ‪ linariifolia Inula‬ﮔﺮﻓﺘﻪ ﻣﯽﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺐ‬‫‪ ، (1) valencene‬ﻃﯽ ﯾﮏ ﻭﺍﮐﻨﺶ ﺗﮏ ﻣﺮﺣﻠﻪﺍﯼ ﺑﻪ ﺗﺮﮐﯿﺐ ‪ 2‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺐ ‪ 2‬ﻧﯿﺰ ﻃﯽ ﺳﻪ ﻣﺮﺣﻠﻪ ﻭﺍﮐﻨﺶ ﮐﻪ ﺑﺎ ﺗﺸﮑﯿﻞ ‪ 3‬ﻫﻤﺮﺍﻩ ﺍﺳﺖ‬‫ﺑﻪ ﮐﺘﻮﻥ ‪ 4‬ﺗﺒﺪﯾﻞ ﻣﯽ ﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺐ ‪ (5) Eremophilene‬ﺑﺎ ﺍﻧﺠﺎﻡ ﻫﻤﺎﻥ ﭼﻬﺎﺭ ﻣﺮﺣﻠﻪ ﺳﻨﺘﺰ‪ ،‬ﺑﻪ ‪ 6‬ﺗﺒﺪﯾﻞ ﻣﯽ ﺷﻮﺩ‪.‬‬‫ﮔﯿﺎﻩ ‪linariifolia Inula‬‬‫‪Q7-2‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪IRN-2 C-7 Q-2‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ‪ 2‬ﻭ ‪ 6‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ ﻭ ﻫﺮﺟﺎ ﻻﺯﻡ ﺍﺳﺖ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﻭﺍﺿﺢ ﻧﻤﺎﯾﺶ ﺩﻫﯿﺪ‪.‬‬‫‪5pt‬‬‫ﺳﭙﺲ ﮐﺘﻮﻥ ‪ 4‬ﺑﻪ ﺍﺳﺘﺮ ‪ 15‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪ .‬ﺗﻤﺎﻡ ﻣﺮﺍﮐﺰ ﻓﻀﺎﯾﯽ ‪ 7‬ﺩﺭ ﺗﺮﮐﯿﺐ ‪ 8‬ﮐﻪ ﻭﺯﻥ ﻣﻮﻟﮑﻮﻟﯽ ﺁﻥ ‪ 188‬ﺍﺳﺖ‪ ،‬ﺣﻔﻆ ﻣﯽﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺒﺎﺕ ‪ 9‬ﻭ ‪10‬‬‫ﺩﺍﺭﺍﯼ ﭘﻨﺞ ﻣﺮﮐﺰ ﻓﻀﺎﯾﯽ ﻫﺴﺘﻨﺪ ﻭ ﭘﯿﻮﻧﺪ ﺩﻭﮔﺎﻧﻪ ﮐﺮﺑﻦ‪-‬ﮐﺮﺑﻦ ﻧﺪﺍﺭﻧﺪ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﺍﺯ ‪ H2 18 O‬ﺑﻪ ﺟﺎﯼ ‪ H2 16 O‬ﺑﺮﺍﯼ ﺳﻨﺘﺰ ﻟﯿﻨﺮﯾﻔﻮﻟﯿﻨﻮﻥﻫﺎﯼ ‪ 13‬ﻭ‬‫‪ 14‬ﻋﻼﻣﺖﮔﺬﺍﺭﯼ ﺷﺪﻩ ﺑﺎ ‪ ، 18 O‬ﺑﻪ ﺗﺮﺗﯿﺐ ﺍﺯ ‪ 11‬ﻭ ‪ 12‬ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ‪ .‬ﺗﺮﮐﯿﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﺍﯾﺰﻭﺗﻮﭘﻮﻣﺮﻫﺎﯼ ﻋﻼﻣﺖﮔﺬﺍﺭﯼ ﺷﺪﻩ ﺑﺎ ‪ 18 O‬ﻫﺴﺘﻨﺪ‪ .‬ﺍﮔﺮ‬‫ﻋﻼﻣﺖﮔﺬﺍﺭﯼ ﺍﯾﺰﻭﺗﻮﭘﯽ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻧﮕﯿﺮﯾﻢ ﻫﺮ ﺩﻭ ﺗﺮﮐﯿﺐ ‪ 13‬ﻭ ‪ 14‬ﺗﺮﮐﯿﺐ ‪ 15‬ﺭﺍ ﺑﺎ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﯾﮑﺴﺎﻥ ﺑﺪﺳﺖ ﻣﯽﺩﻫﻨﺪ‪.‬‬IChO
IRN-2 C-7 Q-3

Q7-3
persian (Iran)‫‪Q7-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-7 Q-4‬‬‫‪A.2‬‬‫ﺳﺎﺧﺘﺎﺭ ﻣﻨﺎﺳﺒﯽ ﺑﺮﺍﯼ ‪ A‬ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﺗﺮﮐﯿﺒﺎﺕ ‪ 14–8‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ ﻭ ﻫﺮ ﺟﺎ ﻻﺯﻡ ﺍﺳﺖ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﻭﺍﺿﺢ ﻧﻤﺎﯾﺶ ﺩﻫﯿﺪ‪.‬‬‫ﺍﺗﻢﻫﺎﯼ ‪ 18 O‬ﺭﺍ ﺩﺭ ﺗﺮﮐﯿﺒﺎﺕ ‪ 13‬ﻭ ‪ 14‬ﻣﺎﻧﻨﺪ ﻣﺜﺎﻝ ﺯﯾﺮ ﻧﺸﺎﻥ ﺩﻫﯿﺪ‪.‬‬‫‪19pt‬‬‫‪Q7-5‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-7 Q-5‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺗﺮﮐﯿﺐ ‪ 19‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺳﻨﺘﺰ ﻣﯽﺷﻮﺩ‪ .‬ﺩﺭ ﺭﺍﺑﻄﻪ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻧﻮﺋﯿﺪﯼ‪ ،‬ﺍﺯ ﺗﺮﮐﯿﺐ ‪19‬ﻣﯽﺗﻮﺍﻥ ﺑﻪ ﻋﻨﻮﺍﻥ ﻓﻌﺎﻝﮐﻨﻨﺪﻩ ﺍﻟﮑﻞﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﮐﺮﺩ‪،‬‬‫ﻭ ﺍﺯ ﻫﻤﯿﻦ ﻃﺮﯾﻖ ﺗﺮﮐﯿﺐ ‪ 20‬ﺍﺯ ﻃﺮﯾﻖ ﺣﺪﻭﺍﺳﻂ ﺟﻔﺖ ﯾﻮﻧﯽ ‪ 21‬ﺑﻪ ‪ 22‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪ 21 .‬ﮐﻪ ﺗﺸﮑﯿﻞ ﺁﻥ ﺑﻪ ﮐﻤﮏ ‪ NMR‬ﺗﺎﯾﯿﺪ ﺷﺪﻩ ﺍﺳﺖ ﺑﺘﺪﺭﯾﺞ‬‫ﺗﺠﺰﯾﻪ ﺷﺪﻩ ﻭ ﺑﻪ ‪ 18‬ﻭ ‪ 22‬ﺗﺒﺪﯾﻞ ﻣﯽﺷﻮﺩ‪.‬‬‫‪B.1‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ‪ 19–17‬ﻭ ‪ 21‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﻧﻤﺎﯾﺶ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺿﺮﻭﺭﯼ ﻧﯿﺴﺖ‪.‬‬‫‪10pt‬‬IChO
IRN-2 C-7 A-1

A7-1
persian (Iran)‫ﺑﺎﺯﯼ ﺑﺎ ﺁﺭﻭﻣﺎﺗﯿﺴﯿﺘﻪ ﻏﯿﺮ ﺑﻨﺰﻭﺋﯿﺪﯼ‬A ‫ﻗﺴﻤﺖ‬(5 pt) A.1
pt) 3) 6pt) 2) 2(2 pt) A.2

IChO
IRN-2 C-7 A-2

A7-2
persian (Iran)(19 pt) A.3

IChO
IRN-2 C-7 A-3

A7-3
persian (Iran)B ‫ًﻗﺴﻤﺖ‬(10 pt) B.1
pt) 2) 18

pt) 2) 17pt) 3) 21pt) 3) 19

IRN-2 C-8 C-1

IRN-2 C-8 C
Ehsan Naderi Donig

IChO
Problem 8
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q8-1‬‬‫‪IChO‬‬‫‪IRN-2 C-8 Q-1‬‬‫)‪persian (Iran‬‬‫ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺁﻟﯽ ﺩﯾﻨﺎﻣﯿﮏ ﻭ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﺁﻥﻫﺎ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪9‬‬‫‪11% of the total‬‬‫‪A.2‬‬‫‪3‬‬‫‪A.3‬‬‫‪7‬‬‫‪B.1‬‬‫‪3‬‬‫‪B.2‬‬‫‪4‬‬‫‪Total‬‬‫‪26‬‬‫‪Score‬‬‫ﻗﺴﻤﺖ ‪A‬‬
‫ﻫﯿﺪﺭﻭﮐﺮﺑﻦﻫﺎﯼ ﺁﺭﻭﻣﺎﺗﯿﮏ ﭼﻨﺪﺣﻠﻘﻪﺍﯼ ﮐﻪ ﺣﻠﻘﻪﻫﺎﯼ ﺁﻥﻫﺎ ﺑﻪ ﻃﻮﺭ ﻣﺘﻮﺍﻟﯽ ﻭ ﺑﻪ ﺷﮑﻞ ﺍﺭﺗﻮ ﺑﻪ ﻫﻢ ﻣﺘﺼﻞ ﻣﯽﺷﻮﻧﺪ‪ [n] ،‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﻧﺎﻣﯿﺪﻩ ﻣﯽﺷﻮﻧﺪ‪.‬‬‫‪ (n‬ﺩﺭ ﺍﯾﻨﺠﺎ ﺗﻌﺪﺍﺩ ﺣﻠﻘﻪﻫﺎﯼ ﺷﺶ ﻋﻀﻮﯼ ﺭﺍ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ( ) ﺷﮑﻞ ﭘﺎﯾﯿﻦ ﺭﺍ ﺑﺒﯿﻨﯿﺪ(‪ .‬ﺩﺭ ﯾﮏ ﺭﻭﺵ‪ [4] ،‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ )‪ ([4]C‬ﺑﻪ ﮐﻤﮏ ﯾﮏ ﻭﺍﮐﻨﺶ‬‫ﻧﻮﺭﯼ ﮐﻪ ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﺑﺎ ﮐﺎﺭﺍﯾﯽ ﺑﺎﻻ ﺍﺯ ﻃﺮﯾﻖ ﺗﺸﮑﯿﻞ ﯾﮏ ﺣﺪﻭﺍﺳﻂ )‪ (Int‬ﺗﻬﯿﻪ ﻣﯽﺷﻮﺩ ﮐﻪ ﺍﯾﻦ ﺣﺪﻭﺍﺳﻂ ﺑﻪﺭﺍﺣﺘﯽ ﺑﻪ ﮐﻤﮏ ﯾﺪ‬‫ﺍﮐﺴﯿﺪ ﻣﯽﺷﻮﺩ‪.‬‬‫ﻭﺍﮐﻨﺶ ﻧﻮﺭﯼ ﻓﻮﻕ ﺑﻪ ﺷﯿﻮﻩﺍﯼ ﺷﺒﯿﻪ ﻭﺍﮐﻨﺶ ﺯﯾﺮ ﺍﻧﺠﺎﻡ ﻣﯽﺷﻮﺩ‪.‬‬‫‪Q8-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-8 Q-2‬‬‫ﺗﻮﺟﻪ ‪ :‬ﺩﺭ ﭘﺎﺳﺦﻫﺎﯼ ﺧﻮﺩ ﺩﺭ ﺗﻤﺎﻡ ﺑﺨﺶﻫﺎﯼ ﺳﻮﺍﻝ ‪ ، 8‬ﭘﯿﻮﻧﺪﻫﺎﯼ ﺳﺎﺩﻩ ﻭ ﺩﻭﮔﺎﻧﻪ ﻣﺘﻨﺎﻭﺏ ﺭﺍ ﺑﻪ ﺻﻮﺭﺗﯽ ﮐﻪ ﺑﺮﺍﯼ ﻣﺜﺎﻝﻫﺎﯼ ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ‬‫ﻧﻤﺎﯾﺶ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﺑﺮﺍﯼ ﺳﯿﺴﺘﻢﻫﺎﯼ ﻣﺰﺩﻭﺝ 𝜋 ﺍﺯ ﻋﻼﻣﺖ ﺩﺍﯾﺮﻩ ﺍﺳﺘﻔﺎﺩﻩ ﻧﮑﻨﯿﺪ‪.‬‬‫‪A.1‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ‪ A–C‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﺩﺭ ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﺭﺳﻢ ﺷﺪﻩ ﺍﯾﺰﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺑﺎﯾﺪ ﻗﺎﺑﻞ ﺗﺸﺨﯿﺺ ﺑﺎﺷﻨﺪ‪.‬‬‫‪9pt‬‬‫‪A.2‬‬‫ﺗﻼﺵ ﻫﺎ ﺑﺮﺍﯼ ﺳﻨﺘﺰ ]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺍﺯ ﻫﻤﺎﻥ ﻧﻤﮏ ﻓﺴﻔﻮﻧﯿﻮﻡ ﻭ ﻣﺎﺩﻩ ﺍﻭﻟﯿﻪ ﻣﻨﺎﺳﺐ ﻓﻘﻂ ﻣﻨﺠﺮ ﺑﻪ ﺗﺸﮑﯿﻞ ﻣﻘﺪﺍﺭ‬‫ﺑﺴﯿﺎﺭ ﮐﻤﯽ ﺍﺯ ]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺷﺪ ﻭ ﺩﺭ ﻋﻮﺽ ﻣﺤﺼﻮﻝ ‪ D‬ﮐﻪ ﻭﺯﻥ ﻣﻮﻟﮑﻮﻟﯽ ﺁﻥ ‪ 2‬ﺩﺍﻟﺘﻮﻥ )‪ (Da‬ﮐﻤﺘﺮ ﺍﺯ ]‪[5‬‬‫ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺍﺳﺖ ﺑﻪﺩﺳﺖ ﺁﻣﺪ‪ .‬ﺟﺎﺑﺠﺎﯾﯽﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺑﺮﺍﯼ ‪ D‬ﺩﺭ ﭘﺎﯾﯿﻦ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺳﺎﺧﺘﺎﺭ ‪ D‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪[D (δδ, ppm in CS2CS2, r.t.), 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H),‬‬‫])‪7.91 (2H‬‬‫‪3pt‬‬‫]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﻭ ]‪ [n‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦﻫﺎﯼ ﺑﺰﺭﮔﺘﺮ‪ ،‬ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﻣﺎﺭﭘﯿﭽﯽ ﺩﺍﺭﻧﺪ ﻭ ﺑﻪ ﻫﻢ ﺗﺒﺪﯾﻞ ﺷﺪﻥ ﺍﻧﺎﻧﺘﯿﻮﻣﺮﻫﺎ ﺩﺭ ﺍﯾﻦ ﻣﺎﺭﭘﯿﭻﻫﺎ ﺩﺭ ﺩﻣﺎﯼ ﺍﺗﺎﻕ ﺧﯿﻠﯽ‬‫ﺁﻫﺴﺘﻪ ﺍﺳﺖ‪ .‬ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ]‪ [n‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦﻫﺎ ﺑﻪ ﺻﻮﺭﺕ )‪ (M‬ﯾﺎ )‪ (P‬ﻣﺎﻧﻨﺪ ﺷﮑﻞ ﺯﯾﺮ ﻧﻤﺎﯾﺶ ﺩﺍﺩﻩ ﻣﯽﺷﻮﺩ‪.‬‬‫ﺍﻧﺎﻧﺘﯿﻮﻣﺮﻫﺎﯼ ]‪ [n‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦﻫﺎ ﺑﺎ ‪ n‬ﺑﺰﺭﮔﺘﺮ ﺍﺯ ‪ 4‬ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﻪ ﮐﻤﮏ ﮐﺮﻭﻣﺎﺗﻮﮔﺮﺍﻓﯽ ﺳﺘﻮﻧﯽ ﮐﺎﯾﺮﺍﻝ ﺍﺯ ﻫﻢ ﺟﺪﺍ ﮐﺮﺩ ﮐﻪ ﺑﻮﺳﯿﻠﻪ ﭘﺮﻭﻓﺴﻮﺭ ‪Yoshio‬‬‫‪ Okamoto‬ﺗﻮﺳﻌﻪ ﺩﺍﺩﻩ ﺷﺪ‪.‬‬‫‪Q8-3‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-8 Q-3‬‬‫ﻫﻠﯿﺴﯿﻦﻫﺎﯼ ﭼﻨﺪ ﺗﺎﯾﯽ ﻣﻮﻟﮑﻮﻝﻫﺎﯾﯽ ﺷﺎﻣﻞ ﺩﻭ ﯾﺎ ﭼﻨﺪ ﺳﺎﺧﺘﺎﺭ ﺷﺒﻪ ﻣﺎﺭﭘﯿﭻ ﻫﺴﺘﻨﺪ‪ .‬ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﻣﺎﺭﭘﯿﭽﯽ ‪ ،‬ﺑﺮﺍﯼ ﯾﮏ ﻫﻠﯿﺴﯿﻦ ﭼﻨﺪ‬‫ﺗﺎﯾﯽ ﭼﻨﺪﯾﻦ ﺍﯾﺰﻭﻣﺮ ﻓﻀﺎﯾﯽ ﻭﺟﻮﺩ ﺩﺍﺭﺩ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ ‪ ،‬ﺗﺮﮐﯿﺐ ‪ ، E‬ﺷﺎﻣﻞ ﺳﻪ ﺑﺨﺶ ﺷﺒﻪ ]‪ [5‬ﮐﺎﺭﺑﻮﻫﻠﯿﺴﯿﻦ ﺩﺭ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺍﺳﺖ‪ .‬ﯾﮑﯽ ﺍﺯ‬‫ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺁﻥ ﮐﻪ ﺑﻪ ﺻﻮﺭﺕ)‪ (P,P,P‬ﻧﺎﻡﮔﺬﺍﺭﯼ ﺷﺪﻩ ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﻤﺎﯾﺶ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪A.3‬‬‫ﺍﺯ ﺗﺮﯼﻣﺮﯾﺰﻩ ﺷﺪﻥ ‪-2،1‬ﺩﯼ ﺑﺮﻭﻣﻮ ﺑﻨﺰﻥ ﺩﺭ ﻣﺠﺎﻭﺭﺕ ﻧﯿﮑﻞ‪ ،‬ﺗﺮﯾﻔﻨﯿﻠﻦ ﺗﻮﻟﯿﺪ ﻣﯽﺷﻮﺩ‪ .‬ﻭﻗﺘﯽ ﻫﻤﯿﻦ ﻭﺍﮐﻨﺶ ﺭﻭﯼ‬‫ﯾﮏ ﺍﻧﺎﻧﺘﯿﻮﻣﺮ )‪ (P‬ﺍﺯ ﺗﺮﮐﯿﺐ‪ F‬ﺍﻧﺠﺎﻡ ﺷﻮﺩ ﻫﻠﯿﺴﯿﻦ ﭼﻨﺪ ﺗﺎﯾﯽ ‪ G‬ﺑﺎ ﻓﺮﻣﻮﻝ ﺑﺴﺘﻪ ‪ C66 H36‬ﺑﺪﺳﺖ ﻣﯽﺁﯾﺪ‪ .‬ﺑﺎ ﺩﺭ‬‫ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺍﯾﻨﮑﻪ ﺗﺒﺪﯾﻞ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺑﻪ ﻫﻤﺪﯾﮕﺮ ﺩﺭ ﺣﯿﻦ ﺍﻧﺠﺎﻡ ﻭﺍﮐﻨﺶ ﺍﺗﻔﺎﻕ ﻧﻤﯽﺍﻓﺘﺪ‪ ،‬ﺗﻤﺎﻡ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ‬‫ﻓﻀﺎﯾﯽ ﻣﺤﺘﻤﻞ ﺑﺮﺍﯼ ‪ G‬ﮐﻪ ﺩﺭ ﺍﯾﻦ ﻓﺮﺍﯾﻨﺪ ﺗﻮﻟﯿﺪ ﻣﯽﺷﻮﻧﺪ ﺭﺍ ﺑﺪﻭﻥ ﺗﮑﺮﺍﺭ ﺷﻨﺎﺳﺎﯾﯽ ﮐﻨﯿﺪ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺮﺟﻊ ‪ ،‬ﯾﮏ‬‫ﺍﯾﺰﻭﻣﺮ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﮐﺎﻣﻞ ﺭﺳﻢ ﮐﻨﯿﺪ ﻭ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﺭﺍ ﻣﺎﻧﻨﺪ ﻣﺜﺎﻝ ﺑﺎﻻ ﻭ ﺑﺎ ﺑﻪﮐﺎﺭﮔﯿﺮﯼ ﻋﻼﻣﺖﻫﺎﯼ ﻋﺪﺩﯼ‪ ،‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫ﺑﻘﯿﻪ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ﺑﺎﯾﺪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻫﻤﺎﻥ ﺍﻋﺪﺍﺩ ﮐﻪ ﺑﺮﺍﯼ ﻧﻤﺎﯾﺶ ﻣﻮﻗﻌﯿﺖﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ ﻭ ﻋﻼﻣﺖﻫﺎﯼ‬‫‪ M‬ﻭ ‪ P‬ﻟﯿﺴﺖ ﺷﻮﻧﺪ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ‪ ،‬ﺳﺎﯾﺮ ﺍﯾﺰﻭﻣﺮﻫﺎﯼ ﻓﻀﺎﯾﯽ ‪ E‬ﺑﺎﯾﺪ ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﻟﯿﺴﺖ ﺷﻮﻧﺪ‪:‬‬‫)‪(1, 2, 3) = (P, M, P), (P, M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and (M, P, M‬‬‫‪7pt‬‬

‫‪Q8-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-8 Q-4‬‬‫ﻗﺴﻤﺖ ‪B‬‬‫ﺳﻮﻣﺎﻧﻦ ﯾﮏ ﻫﯿﺪﺭﻭﮐﺮﺑﻦ ﮐﺎﺳﻪﺍﯼ ﺷﮑﻞ ﺍﺳﺖ ﮐﻪ ﺑﺮﺍﯼ ﺍﻭﻟﯿﻦ ﺑﺎﺭ ﺩﺭ ﺳﺎﻝ ‪ 2003‬ﺩﺭ ﮊﺍﭘﻦ ﮔﺰﺍﺭﺵ ﺷﺪ‪ .‬ﻧﺎﻡ ”‪ ”sumanene‬ﺑﺮﮔﺮﻓﺘﻪ ﺍﺯ ﯾﮏ ﮐﻠﻤﻪ‬‫ﺳﺎﻧﺴﮑﺮﯾﺖ‪-‬ﻫﻨﺪﯼ ”ﺳﻮﻣﺎﻥ” ﺍﺳﺖ ﮐﻪ ﺑﻪ ﻣﻌﻨﯽ ﺁﻓﺘﺎﺑﮕﺮﺩﺍﻥ ﺍﺳﺖ‪.‬‬‫ﺳﻨﺘﺰ ﺳﻮﻣﺎﻧﻦ ﺑﻪ ﮐﻤﮏ ﯾﮏ ﺗﻮﺍﻟﯽ ﺍﺯ ﻭﺍﮐﻨﺶﻫﺎ ﺍﻧﺠﺎﻡ ﺷﺪ ﮐﻪ ﺷﺎﻣﻞ ﯾﮏ ﻣﺮﺣﻠﻪ ﻭﺍﮐﻨﺶ ﻣﺘﺎﺗﺰﯾﺰ ﻫﻤﺮﺍﻩ ﺑﺎ ﺑﺎﺯ ﺷﺪﻥ ﺣﻠﻘﻪ ﻭ ﯾﮏ ﻣﺮﺣﻠﻪ ﻭﺍﮐﻨﺶ ﻣﺘﺎﺗﺰﯾﺰ‬‫ﻫﻤﺮﺍﻩ ﺑﺎ ﺑﺴﺘﻪ ﺷﺪﻥ ﺣﻠﻘﻪ ﺍﺳﺖ‪.‬‬‫ﻧﻤﻮﻧﻪﻫﺎﯾﯽ ﺍﺯ ﻭﺍﮐﻨﺶﻫﺎﯼ ﻣﺘﺎﺗﺰﯾﺰ ﮐﻪ ﺑﻪ ﮐﻤﮏ ﮐﺎﺗﺎﻟﯿﺰﻭﺭ ﺭﻭﺗﻨﯿﻮﻡ )*‪ (Ru‬ﺍﻧﺠﺎﻡ ﻣﯽﺷﻮﻧﺪ ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩﺍﻧﺪ‪.‬‬‫‪B.1‬‬‫ﺳﺎﺧﺘﺎﺭ ﺣﺪﻭﺍﺳﻂ ‪ I‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪) .‬ﻧﯿﺎﺯﯼ ﺑﻪ ﻧﻤﺎﯾﺶ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﻧﯿﺴﺖ(‬‫‪3pt‬‬

‫‪Q8-5‬‬‫)‪persian (Iran‬‬‫‪B.2‬‬‫‪IChO‬‬‫‪IRN-2 C-8 Q-5‬‬‫ﺑﺎ ﺷﺮﻭﻉ ﺍﺯ ﭘﯿﺶﻣﺎﺩﻩ ﻓﻌﺎﻝ ﻧﻮﺭﯼ ‪ ، J‬ﻭ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻫﻤﺎﻥ ﺗﻮﺍﻟﯽ ﻭﺍﮐﻨﺶ ‪ ،‬ﻣﺸﺘﻖ ﻓﻌﺎﻝ ﻧﻮﺭﯼ ﺳﻮﻣﺎﻧﻦ ‪ K‬ﺑﺪﺳﺖ‬‫ﻣﯽﺁﯾﺪ‪ .‬ﻣﺮﺍﮐﺰ ﻓﻀﺎﯾﯽ ﺩﺭ ‪ J‬ﺩﺭ ﺣﯿﻦ ﻭﺍﮐﻨﺶ ﻣﺘﺎﺗﺰﯾﺰ ﺩﭼﺎﺭ ﻭﺍﺭﻭﻧﮕﯽ ﻧﻤﯽﺷﻮﻧﺪ‪ .‬ﺳﺎﺧﺘﺎﺭ ‪ K‬ﺭﺍ ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ‬‫ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺭﺳﻢ ﮐﻨﯿﺪ‪.‬‬‫‪4pt‬‬IChO
IRN-2 C-8 A-1

A8-1
persian (Iran)‫ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺁﻟﯽ ﺩﯾﻨﺎﻣﯿﮏ ﻭ ﮐﺎﯾﺮﺍﻟﯿﺘﻪ ﺁﻥﻫﺎ‬A ‫ﻗﺴﻤﺖ‬(9 pt) A.1
pt) 3) Cpt) 3) Bpt) 3) A(3 pt) A.2

IChO
IRN-2 C-8 A-2

A8-2
persian (Iran)(7 pt) A.3

IChO
IRN-2 C-8 A-3

A8-3
persian (Iran)B ‫ﻗﺴﻤﺖ‬(3 pt) B.1(4 pt) B.2

IRN-2 C-9 C-1

IRN-2 C-9 C
Ehsan Naderi Donig

IChO
Problem 9
Cover sheet

Please return this cover sheet together with all the related question sheets.

‫‪Q9-1‬‬‫‪IChO‬‬‫‪IRN-2 C-9 Q-1‬‬‫)‪persian (Iran‬‬‫ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﮐﭙﺴﻮﻝ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﻭ ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﺩﻭﺳﺖ ﻧﺪﺍﺭﺩ‬‫‪Question‬‬‫‪Points‬‬‫‪A.1‬‬‫‪13‬‬‫‪10% of the total‬‬‫‪A.2‬‬‫‪2‬‬‫‪A.3‬‬‫‪2‬‬‫‪A.4‬‬‫‪3‬‬‫‪A.5‬‬‫‪3‬‬‫‪Total‬‬‫‪23‬‬‫‪Score‬‬‫ﺑﭽﻪﻫﺎﯼ ﺧﻮﺏ ﺍﯾﻦ ﮐﺎﺭ ﺭﺍ ﻧﻤﯽﮐﻨﻨﺪ‪ ،‬ﺍﻣﺎ ﺍﮔﺮ ﺗﻮﭖ ﺗﻨﯿﺲ ﺭﺍ ﺑﺎﺯ ﮐﻨﯿﺪ‪ ،‬ﻣﯽﺗﻮﺍﻧﯿﺪ ﺁﻥ ﺭﺍ ﺑﻪ ﺩﻭ ﻗﻄﻌﻪ ‪ U‬ﺷﮑﻞ ﺗﮑﻪ ﺗﮑﻪ ﮐﻨﯿﺪ‪.‬‬‫ﺑﺮ ﺍﺳﺎﺱ ﺍﯾﻦ ﺍﯾﺪﻩ‪ ،‬ﺗﺮﮐﯿﺒﺎﺕ ‪ 1‬ﻭ ‪ 2‬ﺑﻪﺻﻮﺭﺕ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ‪ U‬ﺷﮑﻞ ﺑﺎ ﺍﻧﺪﺍﺯﻩﻫﺎﯼ ﻣﺨﺘﻠﻒ ﺳﻨﺘﺰ ﺷﺪﻧﺪ‪ .‬ﺗﺮﮐﯿﺐ ‪ 3‬ﺑﺮﺍﯼ ﻣﻘﺎﯾﺴﻪ ﺑﺎ ‪ 1‬ﺗﻬﯿﻪ ﺷﺪ ﻭ ﺭﻓﺘﺎﺭ‬‫ﮐﭙﺴﻮﻝﺳﺎﺯﯼ ﺍﯾﻦ ﺗﺮﮐﯿﺒﺎﺕ ﻣﻮﺭﺩ ﺑﺮﺭﺳﯽ ﻗﺮﺍﺭ ﮔﺮﻓﺖ‪.‬‬‫‪Q9-2‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-9 Q-2‬‬‫ﺭﻭﺵ ﺳﻨﺘﺰ ‪ 2‬ﺩﺭ ﭘﺎﯾﯿﻦ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪ .‬ﺩﺭﺻﺪ ﺟﺮﻣﯽ ﻋﻨﺎﺻﺮ ﺩﺭ ﺗﺮﮐﯿﺐ ‪ 9‬ﺑﻪ ﺻﻮﺭﺕ ﺯﯾﺮ ﺍﺳﺖ‪:‬‬‫‪9: C; 40.49%, H; 1.70%, and O; 17.98%‬‬

‫‪Q9-3‬‬‫)‪persian (Iran‬‬‫‪A.1‬‬‫‪IChO‬‬‫‪IRN-2 C-9 Q-3‬‬‫ﺳﺎﺧﺘﺎﺭﻫﺎﯼ ﺗﺮﮐﯿﺒﺎﺕ ‪ 9–4‬ﺭﺍ ﺭﺳﻢ ﮐﻨﯿﺪ‪ .‬ﻣﯽﺗﻮﺍﻧﯿﺪ ﺍﺯ ﻧﻤﺎﯾﺶ ﺷﯿﻤﯽ ﻓﻀﺎﯾﯽ ﺻﺮﻑﻧﻈﺮ ﮐﻨﯿﺪ‪ .‬ﺍﺯ ”‪ ”PMB‬ﺑﻪ‬‫ﺟﺎﯼ ﺭﺳﻢ ﺳﺎﺧﺘﺎﺭ ﮐﺎﻣﻞ ﮔﺮﻭﻩ ‪ p-methoxybenzyl‬ﮐﻪ ﺩﺭ ﺷﻤﺎﯼ ﺑﺎﻻ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺘﻔﺎﺩﻩ ﮐﻨﯿﺪ‪.‬‬‫‪13pt‬‬‫ﺩﺭ ﻃﯿﻒ ﺟﺮﻣﯽ ‪ ،1‬ﻗﻠﻪ ﻳﻮﻧﯽ ﻣﺮﺑﻮﻁ ﺑﻪ ﺩﻳﻤﺮ ﺁﻥ ﯾﻌﻨﯽ ‪ 1 2‬ﺑﻪ ﻭﺿﻮﺡ ﻣﺸﺎﻫﺪﻩ ﺷﺪ‪ ،‬ﺩﺭ ﺣﺎﻟﯽ ﮐﻪ ﻗﻠﻪ ﻳﻮﻧﯽ ﺑﺮﺍﯼ ‪ 3 2‬ﺩﺭ ﻃﻴﻒ ‪ 3‬ﻣﺸﺎﻫﺪﻩ ﻧﺸﺪ‪ .‬ﺩﺭ ﻃﯿﻒ‬‫ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﻣﺤﻠﻮﻟﯽ ﺍﺯ ‪ ،112‬ﺗﻤﺎﻡ ﭘﺮﻭﺗﻮﻥ ﻫﺎﯼ ‪ NH‬ﮐﻪ ﺍﺯ ﺗﺮﮐﯿﺐ ‪ 1‬ﻣﯽﺁﯾﻨﺪ ﺑﻪ ﺻﻮﺭﺕ ﻣﻌﺎﺩﻝ ﺷﯿﻤﯿﺎﯾﯽ ﻣﺸﺎﻫﺪﻩ ﺷﺪﻧﺪ ﻭ ﺟﺎﺑﺠﺎﯾﯽ ﺷﯿﻤﯿﺎﯾﯽ ﺁﻧﻬﺎ‬‫ﺑﻪ ﻃﻮﺭ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪﺍﯼ ﺑﺎ ﭘﺮﻭﺗﻮﻥﻫﺎﯼ ‪ NH‬ﺩﺭ ﺗﺮﮐﯿﺐ ‪ 3‬ﻣﺘﻔﺎﻭﺕ ﺍﺳﺖ‪ .‬ﺍﯾﻦ ﺩﺍﺩﻩﻫﺎ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ ﮐﻪ ﭘﯿﻮﻧﺪﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻧﯽ ﺑﯿﻦ ‪ NH‬ﺍﺯ ﺗﺮﮐﯿﺐ ‪1‬‬‫ﻭ ﺍﺗﻢﻫﺎﯼ ‪ X‬ﺍﺯ ﻣﻮﻟﮑﻮﻝ ﺩﯾﮕﺮ ‪ 1‬ﺍﯾﺠﺎﺩ ﻣﯽﺷﻮﻧﺪ ﺗﺎ ﮐﭙﺴﻮﻝ ﺩﯾﻤﺮﯼ ﺗﺸﮑﯿﻞ ﺷﻮﺩ‪.‬‬‫‪A.2‬‬‫ﺩﻭﺭ ﻫﻤﻪ ﺍﺗﻢ)ﻫﺎﯼ( ﻣﻨﺎﺳﺐ ‪ X‬ﺩﺭ ‪ 1‬ﺩﺍﯾﺮﻩ ﺑﮑﺸﯿﺪ‪.‬‬‫‪2pt‬‬‫‪A.3‬‬‫ﺗﻌﺪﺍﺩ ﭘﯿﻮﻧﺪﻫﺎﯼ ﻫﯿﺪﺭﻭﮊﻧﯽ ﺭﺍ ﺩﺭ ﮐﭙﺴﻮﻝ ﺩﯾﻤﺮﯼ ) ‪ (112‬ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫‪2pt‬‬

‫‪Q9-4‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-9 Q-4‬‬‫ﮐﭙﺴﻮﻝ ﺩﯾﻤﺮﯼ ‪ (112 ) 1‬ﺩﺍﺭﺍﯼ ﯾﮏ ﻓﻀﺎﯼ ﺩﺍﺧﻠﯽ ﺍﺳﺖ ﮐﻪ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﮐﻮﭼﮏ ﻣﻨﺎﺳﺐ ‪ Z‬ﻣﯽﺗﻮﺍﻧﺪ ﺩﺭ ﺁﻥ ﻣﺤﺼﻮﺭ ﺷﻮﺩ‪ .‬ﺍﯾﻦ ﭘﺪﯾﺪﻩ ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺯﯾﺮ ﺑﯿﺎﻥ‬‫ﻣﯽﺷﻮﺩ‪:‬‬‫ﺛﺎﺑﺖ ﺗﻌﺎﺩﻝ ﻣﺤﺼﻮﺭ ﺷﺪﻥ )ﮐﭙﺴﻮﻟﻪ ﺷﺪﻥ( ‪ Z‬ﺩﺭ ‪ 1 2‬ﺩﺭ ﭘﺎﯾﯿﻦ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪:‬‬‫ﻣﺤﺼﻮﺭ ﺷﺪﻥ )ﮐﭙﺴﻮﻟﻪ ﺷﺪﻥ( ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺩﺭ ﺩﺍﺧﻞ ﮐﭙﺴﻮﻝ ﺭﺍ ﻣﯽﺗﻮﺍﻥ ﺑﻪ ﮐﻤﮏ ﻃﯿﻒﺳﻨﺠﯽ ‪ NMR‬ﺩﻧﺒﺎﻝ ﮐﺮﺩ‪ .‬ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ‪ 1 2 ،‬ﺩﺭ ‪C6 D6‬‬
‫ﺳﯿﮕﻨﺎﻝﻫﺎﯼ ﻣﺨﺘﻠﻔﯽ ﺩﺭ ﻃﯿﻒ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺩﺭ ﻗﺒﻞ ﻭ ﺑﻌﺪ ﺍﺯ ﺍﺿﺎﻓﻪ ﮐﺮﺩﻥ ‪ CH4‬ﻣﯽﺩﻫﺪ‪.‬‬‫ﺗﺮﮐﯿﺐ ‪ 2‬ﻫﻢ ﯾﮏ ﮐﭙﺴﻮﻝ ﺳﺨﺖ ﻭ ﺑﺰﺭﮔﺘﺮ ﺩﯾﻤﺮﯼ ﺗﺸﮑﯿﻞ ﻣﯽﺩﻫﺪ ) ‪ .(222‬ﻃﯿﻒ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺑﺮﺍﯼ ‪ 2 2‬ﺩﺭ ‪ ،C6 D5 F ، C6 D6‬ﻭ ﻣﺨﻠﻮﻁ ﺣﻼﻝﻫﺎﯼ‬‫‪ C6 D6 / C6 D5 F‬ﺑﺎ ﺛﺎﺑﺖ ﻧﮕﻪ ﺩﺍﺷﺘﻦ ﺳﺎﯾﺮ ﺷﺮﺍﯾﻂ ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼ ﺷﺪ‪ .‬ﺟﺎﺑﺠﺎﯾﯽﻫﺎﯼ ﺷﯿﻤﯿﺎﯾﯽ ﺑﺮﺍﯼ ﭘﺮﻭﺗﻮﻥ ‪ Ha‬ﺍﺯ ‪ 2‬ﺩﺭ ﺣﻼﻝﻫﺎﯼ ﺑﺎﻻ ﺩﺭ ﺯﯾﺮ ﺧﻼﺻﻪ‬‫ﺷﺪﻩ ﺍﺳﺖ‪ ،‬ﻭ ﻫﯿﭻ ﺳﯿﮕﻨﺎﻝ ﺩﯾﮕﺮﯼ ﺍﺯ ‪ Ha‬ﺩﺭ ‪ ،2‬ﺑﻪ ﺟﺰ ﺳﯿﮕﻨﺎﻝﻫﺎﯼ ﻟﯿﺴﺖ ﺷﺪﻩ ﻣﺸﺎﻫﺪﻩ ﻧﻤﯽﺷﻮﺩ‪ .‬ﻓﺮﺽ ﮐﻨﯿﺪ ﮐﻪ ﻓﻀﺎﯼ ﺩﺍﺧﻠﯽ ﮐﭙﺴﻮﻝ ﻫﻤﯿﺸﻪ ﺑﺎ‬‫ﺑﯿﺸﺘﺮﯾﻦ ﺗﻌﺪﺍﺩ ﻣﻤﮑﻦ ﺍﺯ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ﺣﻼﻝ ﭘﺮ ﻣﯽﺷﻮﺩ ﻭ ﻫﺮ ﺳﯿﮕﻨﺎﻝ ﻣﺮﺑﻮﻁ ﺑﻪ ﯾﮏ ﮔﻮﻧﻪ ﺍﺯ ﮐﭙﺴﻮﻝ ﭘﺮ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫ﺣﻼﻝ‬‫‪C6 D6‬‬‫‪C6 D5 F‬‬‫‪C6 D5 F / C6 D6‬‬‫‪A.4‬‬‫‪𝛿(ppm) of Ha‬‬‫‪4.60‬‬‫‪4.71‬‬‫‪4.82 ,4.71 ,4.60‬‬‫ﺑﺮﺍﯼ ﻫﺮ ﮔﻮﻧﻪ ‪ 2 2‬ﮐﻪ ﯾﮏ ﺳﯿﮕﻨﺎﻝ ﺑﺮﺍﯼ ‪ Ha‬ﻣﯽﺩﻫﺪ ﺗﻌﺪﺍﺩ ﻣﻮﻟﮑﻮﻝﻫﺎﯼ ‪C6 D6‬ﻭ ‪ C6 D5 F‬ﻣﺤﺼﻮﺭ ﺷﺪﻩ ﺩﺭ ﺁﻥ ﺭﺍ‬‫ﺗﻌﯿﯿﻦ ﮐﻨﯿﺪ‪.‬‬‫‪3pt‬‬

‫‪Q9-5‬‬‫‪IChO‬‬‫‪IRN-2 C-9 Q-5‬‬‫)‪persian (Iran‬‬‫ﺍﻧﺪﺍﺯﻩ ﮔﯿﺮﯼﻫﺎﯼ ﭘﺮﻭﺗﻮﻥ ‪ NMR‬ﺩﺭ ‪ C6 D6‬ﻧﺸﺎﻥ ﺩﺍﺩ ﮐﻪ ‪ 2 2‬ﻣﯽﺗﻮﺍﻧﺪ ﯾﮏ ﻣﻮﻟﮑﻮﻝ ﺍﺯ ‪-1‬ﺁﺩﺍﻣﺎﻧﺘﺎﻥ ﮐﺮﺑﻮﮐﺴﯿﻠﯿﮏ ﺍﺳﯿﺪ )‪ (AdA‬ﺭﺍ ﺩﺭ ﺧﻮﺩ ﺟﺎﯼ ﺩﻫﺪ ﻭ‬‫ﺛﺎﺑﺖﻫﺎﯼ ﺗﺠﻤﻊ ) ‪ (𝐾a‬ﮐﻪ ﺩﺭ ﭘﺎﯾﯿﻦ ﺑﯿﺎﻥ ﺷﺪﻩ ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﻣﺨﺘﻠﻒ ﺗﻌﯿﯿﻦ ﺷﺪﻧﺪ‪ [solvent@222 ] .‬ﮔﻮﻧﻪﺍﯼ ﺷﺎﻣﻞ ﯾﮏ ﯾﺎ ﭼﻨﺪ ﻣﻮﻟﮑﻮﻝ ﺣﻼﻝ ﺭﺍ ﻧﺸﺎﻥ‬‫ﻣﯽﺩﻫﺪ‪.‬‬‫ﺑﻪ ﻃﻮﺭ ﻣﺸﺎﺑﻪ‪ ،‬ﻣﻘﺎﺩﯾﺮ ‪ 𝐾a‬ﺑﺮﺍﯼ ‪ CH4‬ﻭ ‪ 1 2‬ﺑﻪ ﺻﻮﺭﺗﯽ ﮐﻪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ )‪ (2‬ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﺩﺭ ﺩﻣﺎﻫﺎﯼ ﻣﺨﺘﻠﻒ ﺩﺭ ‪ C6 D6‬ﺑﻪ ﮐﻤﮏ ﺍﻧﺪﺍﺯﻩﮔﯿﺮﯼﻫﺎﯼ‬‫ﭘﺮﻭﺗﻮﻥ‪ NMR‬ﺗﻌﯿﯿﻦ ﺷﺪﻧﺪ‪ .‬ﻧﻤﻮﺩﺍﺭﻫﺎﯼ ﺩﻭ ﺛﺎﺑﺖ ﺗﺠﻤﻊ ) ﺑﻪ ﺻﻮﺭﺕ ‪ ln 𝐾a‬ﺩﺭ ﻣﻘﺎﺑﻞ ‪ 1‬ﺑﻪ ﺭﻭﯼ ‪ ( T‬ﺩﺭ ﺯﯾﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ‪.‬‬‫‪A.5‬‬‫ﮔﺰﺑﻨﻪ ﺻﺤﯿﺢ ) ‪ A‬ﯾﺎ ‪ ( B‬ﺭﺍ ﺑﺮﺍﯼ ﻫﺮ ﯾﮏ ﺍﺯ ﻣﻮﺍﺭﺩ )‪ (5)–(1‬ﺩﺭ ﭘﺎﺭﺍﮔﺮﺍﻑ ﺯﯾﺮ ﺍﻧﺘﺨﺎﺏ ﮐﻨﯿﺪ‪.‬‬‫ﻫﯿﭻ ﻣﻮﻟﮑﻮﻝ ‪ C6 D6‬ﺩﺭ ‪ 1 2‬ﻣﺤﺼﻮﺭ ﻧﻤﯽﺷﻮﺩ‪ .‬ﺩﺭ ﺧﻂ ‪ ،II‬ﺗﻐﯿﯿﺮ ﺁﻧﺘﺮﻭﭘﯽ )‪ (1) ، (ΔS‬ﺍﺳﺖ ﻭ ﺗﻐﯿﯿﺮ ﺁﻧﺘﺎﻟﭙﯽ‬‫)‪ (2) ، (ΔH‬ﺍﺳﺖ ‪ ،‬ﮐﻪ ﻧﺸﺎﻥ ﻣﯽﺩﻫﺪ ﺩﺭ ﺧﻂ ‪ II‬ﻧﯿﺮﻭﯼ ﭘﯿﺶﺑﺮﻧﺪﻩ ﺑﺮﺍﯼ ﮐﭙﺴﻮﻝﺳﺎﺯﯼ )‪ (3‬ﺍﺳﺖ ‪ .‬ﺑﻨﺎﺑﺮﺍﯾﻦ ‪ ،‬ﺧﻂ‬‫‪ I‬ﻣﺮﺑﻮﻁ ﺑﻪ )‪ (4‬ﻭ ﺧﻂ ‪ II‬ﻣﺮﺑﻮﻁ ﺑﻪ ) ‪ (5‬ﺍﺳﺖ‪.‬‬
‫‪A‬‬

‫‪B‬‬

‫)‪(1‬‬‫ﻣﺜﺒﺖ‬‫ﻣﻨﻔﯽ‬‫)‪(3‬‬‫𝑆‪Δ‬‬‫𝐻‪Δ‬‬‫‪ 1 2‬ﻭ ‪CH4‬‬‫‪ 2 2‬ﻭ ‪AdA‬‬‫)‪(2‬‬‫)‪(4‬‬‫)‪(5‬‬‫ﻣﺜﺒﺖ‬‫‪ 1 2‬ﻭ ‪CH4‬‬‫ﻣﻨﻔﯽ‬‫‪ 2 2‬ﻭ ‪AdA‬‬‫‪3pt‬‬

‫‪A9-1‬‬‫)‪persian (Iran‬‬‫‪IChO‬‬‫‪IRN-2 C-9 A-1‬‬‫ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﮐﭙﺴﻮﻝ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﻭ ﭼﯿﺰﻫﺎﯾﯽ ﮐﻪ ﺩﻭﺳﺖ ﻧﺪﺍﺭﺩ‬‫‪(13 pt) A.1‬‬‫‪pt) 2) 4‬‬‫‪pt) 3) 5‬‬‫‪pt) 2) 6‬‬‫‪pt) 2) 7‬‬‫‪pt) 2) 8‬‬‫‪pt) 2) 9‬‬

IChO

A9-2

IRN-2 C-9 A-2

persian (Iran)(2 pt) A.2(2 pt) A.3(3 pt) A.4
C6 D5 F ‫ﺗﻌﺪﺍﺩ‬C6 D6 ‫ﺗﻌﺪﺍﺩ‬𝛿(ppm) of Ha
4.60 ppm

4.71 ppm

4.82 ppm(3 pt) A.5
(1) ∶

(2) ∶

(3) ∶

(4) ∶

(5) ∶