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This document discusses the concept of Hess's law in thermodynamics. It explains that Hess's law states that the enthalpy change of a reaction is the same whether the reaction occurs in one step or multiple steps. It provides examples of calculating enthalpy changes of reactions using Hess's law. Important notes are provided on applying Hess's law, such as reversing the sign of enthalpy changes when reversing reactions or multiplying/dividing reactions.

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‫اﻟﻔﺻل‬ ‫اﻻول‬ ‫اﻟﺛرﻣوداﯾﻧﻣك‬ ( ‫اﻟدرس‬ ‫اﻷول‬ ) ‫اﻟﺛرﻣوداﯾﻧﻣك‬ ‫طرﯾﻘﺔ‬ ‫ﻗﺎﻧون‬ ‫ھﯾس‬ : /‫ﺳؤال‬ ‫ﻟﻣﺎذا‬ ‫ﯾﺗم‬ ‫اﻟﻠﺟوء‬ ‫إﻟﻰ‬ ‫طرﯾﻘﺔ‬ ‫ھﯾس‬ ‫ﻟﺣﺳﺎب‬ Δ ‫𝐻؟‬𝑟 ° ‫ﺟواب‬ / ‫ﱠ‬‫ﻷن‬ ‫ﺑﻌض‬ ‫اﻟﻣرﻛﺑﺎت‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫ﻻ‬ ‫ﯾﻣﻛن‬ ‫ﺗﺻﻧﯾﻌﮭﺎ‬ ً‫ة‬‫ﻣﺑﺎﺷر‬ ‫ﻣن‬ ‫ﻋﻧﺎﺻرھﺎ‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورة‬ (‫)ﻋﻠل‬ ‫ﺑﺳﺑب‬ : ‫ﱠ‬‫أن‬ 1 ( ‫ﺑﻌض‬ ‫اﻟﺗﻔﺎﻋﻼت‬ ‫ﺗﺳﯾر‬ ‫ﺑﺑطء‬ .‫ﺷدﯾد‬ 2 ( ‫ن‬ ّ‫ﺗﻛو‬ ‫ﻣرﻛﺑﺎت‬ ‫ﺟﺎﻧﺑﯾﺔ‬ ‫ﻏﯾر‬ ‫ﻣرﻏوب‬ .‫ﻓﯾﮭﺎ‬ ‫ھﯾس؟‬ ‫ﻗﺎﻧون‬ ‫ﻧص‬ ‫ھو‬ ‫ﻣﺎ‬ :‫ﯾﻧص‬ ‫)ﻋﻧد‬ ‫ﺗﺣوﯾل‬ ‫اﻟﻣﺗﻔﺎﻋﻼت‬ ‫إﻟﻰ‬ ‫ﻧواﺗﺞ‬ ‫ﱠ‬‫ﻓﺈن‬ ‫اﻟﺗﻐﯾر‬ ‫ﻓﻲ‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ھو‬ ‫ﻧﻔﺳﮫ‬ ‫ﺳواء‬ ‫ﺗم‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻓﻲ‬ ‫ﺧطوة‬ ‫واﺣدة‬ .(‫اﻟﺧطوات‬ ‫ﻣن‬ ‫ﺳﻠﺳﻠﺔ‬ ‫ﻋﺑر‬ ‫أو‬ / ‫ﻣﺛﺎل‬ 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 (𝑔) ⟶𝐶𝑂2 𝑔 ( ) ∆𝐻𝑟 =− 393. 5 𝐾𝐽/𝑚𝑜𝑙 ‫وﻋﻧد‬ ‫ﺗﻛوﯾن‬ ‫ﺑﺄﻛﺛر‬ ‫ﻣن‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓﺄن‬ ‫ھو‬ ‫ﻧﻔﺳﮫ‬ 𝐶𝑂2 ∆𝐻𝑟 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 1 2 𝑂2 (𝑔) ⟶𝐶𝑂 𝑔 ( ) ∆𝐻𝑟 =− 110. 5 𝐾𝐽/𝑚𝑜𝑙 𝐶𝑂 𝑔 ( ) + 1 2 𝑂2 (𝑔) ⟶𝐶𝑂2 𝑔 ( ) ∆𝐻𝑟 =− 283 𝐾𝐽/𝑚𝑜𝑙 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 (𝑔) ⟶𝐶𝑂2 𝑔 ( ) ∆𝐻𝑟 =− 393. 5 𝐾𝐽/𝑚𝑜𝑙
‫ھﯾس‬ ‫ﻣﺳﺎﺋل‬ ‫ﺣل‬ ‫ﺣول‬ ‫ﻣﮭﻣﺔ‬ ‫ﻣﻼﺣظﺎت‬ : 1 ( ‫ﻋﻧد‬ ‫ﻗﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻧﻘﻠب‬ ‫إﺷﺎرة‬ ‫ﻗﯾﻣﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ . 2 ( ‫ﻋﻧد‬ ‫ﺿرب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫أو‬ ‫ﻗﺳﻣﺗﮭﺎ‬ ‫ﻋﻠﻰ‬ ‫ﻣﻌﺎﻣل‬ ‫ﻋددي‬ ‫ﻓﺗﺿرب‬ ‫أو‬ ‫ﺗﻘﺳم‬ ‫ﻗﯾﻣﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﺑﻧﻔس‬ .‫اﻟﻌدد‬ 3 ( ‫ھﻧﺎك‬ ‫ﻧوﻋﺎن‬ ‫ﻣن‬ :‫اﻟﻣﻌﺎدﻻت‬ :‫اﻷول‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟطﻠب‬ (‫)اﻟﮭدف‬ ‫وﺗﺳﻣﻰ‬ ‫ﺑﺎﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ :‫واﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ .‫اﻟﻔرﻋﯾﺔ‬ 4 ( ‫ﻧﻣر‬ ‫ﻋﻠﻰ‬ ‫ﻣواد‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﯾﺔ‬ ً‫ء‬‫اﺑﺗدا‬ ‫ﻣن‬ ‫أول‬ ،‫ﻣﺎدة‬ ‫وﻟﯾس‬ ً‫ﺎ‬‫ﺷرط‬ ‫أن‬ ‫ﻧﻣر‬ ‫ﻋﻠﻰ‬ ‫ﺟﻣﯾﻊ‬ ‫ﻣواد‬ ‫ھذه‬ .‫اﻟﻣﻌﺎدﻟﺔ‬ 5 ( ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟﻔرﻋﯾﺔ‬ ‫اﻟﺗﻲ‬ ‫ﺗم‬ ‫اﻟﺗﻌﺎﻣل‬ ‫ﻣﻌﮭﺎ‬ ‫ﻻ‬ ‫ﯾﺗم‬ ‫اﻟرﺟوع‬ ‫إﻟﯾﮭﺎ‬ .ً‫ﺛﺎﻧﯾﺔ‬ 6 ( ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻷوﻟﻰ‬ :‫اﻟذھﺑﯾﺔ‬ ‫إذا‬ ‫ﺗﻛررت‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫أﻛﺛر‬ ‫ﻣن‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓرﻋﯾﺔ‬ ‫ﻓﯾﺗم‬ ‫ﺗﺟﺎوزھﺎ‬ ‫ﻟﻠﻣﺎدة‬ ‫اﻟﺗﻲ‬ ‫ﺗﻠﯾﮭﺎ‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ .‫اﻟرﺋﯾﺳﯾﺔ‬ 7 ( ‫اﻟﻣﻼﺣظﺔ‬ :‫اﻟذھﺑﯾﺔ‬ ‫إذا‬ ‫ﻛﺎﻧت‬ ‫اﻟﻣﺎدة‬ ‫اﻟﻣﺗﻛررة‬ ‫ھﻲ‬ ‫آﺧر‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﯾﺔ‬ ‫ﻓﺳﻧﺿطر‬ ‫ﻟﻠﺗﻌﺎﻣل‬ ‫ﻣﻌﮭﺎ‬ ‫ﺑﺷرط‬ .‫اﻟﺗﻐﯾﯾر‬ ‫ﺑﻌد‬ ‫اﻟﻔرﻋﯾﺔ‬ ‫اﻟﻣﻌﺎدﻻت‬ ‫ﻓﻲ‬ ‫ﻣوﻻﺗﮭﺎ‬ ‫ﻋدد‬ ‫ﺣﺳﺎب‬ ‫ﻣﺛﺎل‬ 1 / ‫اﺣﺳب‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﻣرﻛب‬ ‫ﻣن‬ ‫ﻋﻧﺎﺻره‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﺑﺄﺛﺑت‬ :‫ﺻورھﺎ‬ 𝐶𝑆2 (𝐿) Δ = ? 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 2𝑆‫ﻣﻌﯿﻨﻲ‬ → 𝐶𝑆2 (𝐿) 𝐻𝑓 ° :‫اﻵﺗﯾﺔ‬ ‫اﻟﺣرارﯾﺔ‬ ‫اﻟﻣﻌﺎدﻻت‬ ‫أﻋطﯾت‬ ‫إذا‬ = - 394 KJ/mol 1 ( )𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 𝑔 ( ) →𝐶𝑂2 (𝑔) ∆𝐻𝑟 ° = - 296 KJ/mol (2) 𝑆‫ﻣﻌﯿﻨﻲ‬ + 𝑂2 𝑔 ( ) →𝑆𝑂2 (𝑔) ∆𝐻𝑟 °
= - 1072 KJ/mol 3 ( )𝐶𝑆2 (𝐿) + 3𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝑆𝑂2 (𝑔) ∆𝐻𝑟 ° /‫اﻟﺣل‬ ‫ﺗﺑﻘﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 1 ( ‫ﻛﻣﺎ‬ ،‫ھﻲ‬ ‫وﻧﺿرب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 2 ( × 2 ، ‫وﺗﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 3 ( : = - 394 (1) 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 → 𝐶𝑂2 ∆𝐻𝑟 ° KJ/mol = - 592 (2)2𝑆 + 2𝑂2 → 2𝑆𝑂2 ∆𝐻𝑟 ° KJ/mol = + 1072 KJ/mol 3 ( )𝐶𝑂2 (𝑔) + 2𝑆𝑂2 (𝑔) → 𝐶𝑆2 (𝐿) + 3𝑂2 (𝑔) ∆𝐻𝑟 ° Δ = = 86 KJ/mol 𝐶 ‫ﻛﺮاﻓﯿﺖ‬ + 𝑆‫ﻣﻌﯿﻨﻲ‬ → 𝐶𝑆2 (𝐿) 𝐻𝑓 ° ∆𝐻𝑟 ° ‫اﻻول‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . ‫ﯾﺗم‬ ‫اﻟﻠﺟوء‬ ‫إﻟﻰ‬ ‫طرﯾﻘﺔ‬ ‫ھﯾس‬ ‫ﻟﺣﺳﺎب‬ Δ ‫ﱠ‬‫ﻷن‬ ‫ﺑﻌض‬ ‫اﻟﻣرﻛﺑﺎت‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫ﯾﻣﻛن‬ ‫ﺗﺻﻧﯾﻌﮭﺎ‬ 𝐻𝑟 ° ‫ﺻورة‬ ‫ﺑﺄﺛﺑت‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﻋﻧﺎﺻرھﺎ‬ ‫ﻣن‬ ً‫ة‬‫ﻣﺑﺎﺷر‬ ) T, F ( ‫اﻟﺟواب‬ / F 2 . :‫اﻟﻣﻌﺎدﻻت‬ ‫ﻣن‬ ‫ﻧوﻋﺎن‬ ‫ھﻧﺎك‬ ) A :‫اﻷول‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟطﻠب‬ (‫)اﻟﮭدف‬ ‫وﺗﺳﻣﻰ‬ ‫ﺑﺎﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ :‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ .‫اﻟﻔرﻋﯾﺔ‬ ) B :‫اﻷول‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟطﻠب‬ (‫)اﻟﮭدف‬ ‫وﺗﺳﻣﻰ‬ ‫ﺑﺎﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ :‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ .‫اﻟرﺋﯾﺳﯾﺔ‬ ‫اﻟﺟواب‬ / A 3 . ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻷوﻟﻰ‬ :‫اﻟذھﺑﯾﺔ‬ ‫اذا‬ ‫وﺟدت‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﯾﺔ‬ ‫وھﻲ‬ ‫ﻣﺗﻛررة‬ ‫ﻓﻲ‬ ‫اﻛﺛر‬ ‫ﻣن‬ ‫اﻟﺗﻲ‬ ‫اﻟﻣﺎدة‬ ‫اﻟﻰ‬ ‫اﻟﻣﺎدة‬ ‫ھذه‬ ‫ﺗﺟﺎوز‬ ‫ﻓﯾﺗم‬ ‫ﻓرﻋﯾﺔ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﺔ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺗﻠﯾﮭﺎ‬ ) T, F (
‫اﻟﺟواب‬ / T ( ‫اﻟدرس‬ ‫اﻟﺛﺎﻧﻲ‬ ) ‫ﺣﺳﺎﺑﺎت‬ ‫ﺑﺎﺳﺗﺧدام‬ ‫ﻗﺎﻧون‬ ‫ھﯾس‬ ‫س‬ 1 / ‫إذا‬ ‫أﻋطﯾت‬ ‫اﻟﻣﻌﺎدﻻت‬ ‫اﻟﺣرارﯾﺔ‬ ‫اﻵﺗﯾﺔ‬ ‫ﻋﻧد‬ ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) STP :( Δ = 25 KJ 1 ( )𝐹𝑒𝑂(𝑠) + 𝐻2 (𝑔) → 𝐹𝑒(𝑠) + 𝐻2 𝑂(𝑔) 𝐻𝑟 ° Δ = 318 KJ 2 ( )3𝐹𝑒𝑂(𝑠) + 1 2 𝑂2 (𝑔) → 𝐹𝑒3 𝑂4 (𝑠) 𝐻𝑟 ° Δ = - 242 KJ 3 ( )𝐻2 (𝑔) + 1 2 𝑂2 (𝑔) → 𝐻2 𝑂(𝑔) 𝐻𝑟 ° ‫اﺣﺳب‬ Δ ‫ﻟﻠﺗﻔﺎﻋل‬ :‫اﻟﺗﺎﻟﻲ‬ 𝐻𝑟 ° Δ =? 3𝐹𝑒(𝑠) + 4𝐻2 (𝑔) → 𝐹𝑒3 𝑂4 (𝑠) + 4𝐻2 (𝑔) 𝐻𝑟 °
/‫اﻟﺣل‬ ‫ﻧﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 1 ( ‫وﻧﺿرﺑﮭﺎ‬ × 3 ‫وﺗﺑﻘﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 2 ( ‫ﻛﻣﺎ‬ ‫ھﻲ‬ ‫وﻧﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 3 :( = -75KJ 1 ( ) 3𝐹𝑒 𝑠 ( ) + 3𝐻2 𝑂 ⟶3𝐹𝑒𝑂 𝑠 ( ) + 3𝐻2 ∆𝐻𝑟 ° 2 ( )𝐹𝑒𝑂 𝑆 ( ) + 1 2 𝑂2 ⟶𝐹𝑒3 𝑂4 𝑆 ( ) ∆𝐻𝑟 ° = − 318𝐾𝐽 3 ( )𝐻2 𝑂 𝑔 ( ) ⟶𝐻2 𝑔 ( ) + 1 2 𝑂2(𝑔) ∆𝐻𝑟 ° = + 242𝐾𝐽 Δ = - 151 KJ2 3𝐹𝑒(𝑠) + 4𝐻2 𝑂(𝑔) → 𝐹𝑒3 𝑂4 + 4𝐻 𝐻𝑟 ° /‫ﻣﻼﺣظﺔ‬ ‫ھﻧﺎ‬ ‫ﻧطﺑق‬ ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻟذھﺑﯾﺔ‬ ‫اﻷوﻟﻰ‬ ‫ﺑﻌدم‬ ‫ﻣوازﻧﺔ‬ ‫ﮫ‬‫ﱠ‬‫ﻧ‬‫ﻷ‬ ‫ﻣﺗﻛرر‬ ‫أﻛﺛر‬ ‫ﻣن‬ ‫ﻣرة‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ ،‫اﻟﻔرﻋﯾﺔ‬ 𝐻2 𝑂 ‫وﻛذﻟك‬ ‫ﻧطﺑق‬ ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻟذھﺑﯾﺔ‬ ‫اﻟﺛﺎﻧﯾﺔ‬ ‫ﺑﺿرورة‬ ‫ﻣوازﻧﺔ‬ ‫رﻏم‬ ‫ﺗﻛراره‬ ‫وﻟﻛن‬ ‫ﺑﻌد‬ ‫ﺣﺳﺎب‬ ‫ﻣوﻻﺗﮫ‬ ‫اﻟﺳﺎﺑﻘﺔ‬ ‫ﺑﻌد‬ 𝐻2 ‫اﻟﺗﻐﯾﯾر‬ . ‫س‬ 2 / ‫إذا‬ ‫ﻋﻠﻣت‬ ‫ﱠ‬‫أن‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﺣﺗراق‬ ّ‫ل‬‫ﻛ‬ ‫ﻣن‬ ‫ﻏﺎز‬ CO ‫و‬ ‫و‬ ‫ﺑوﺣدات‬ KJ/mol ‫ھﻲ‬ ‫ﻋﻠﻰ‬ :‫اﻟﺗواﻟﻲ‬ -) 𝐻2 𝐶𝐻3 𝑂𝐻 727 -/ 286 -/ 284 ،( ‫اﺣﺳب‬ Δ ‫ﺑﺎﺳﺗﺧدام‬ ‫ﻗﺎﻧون‬ ‫ھﯾس‬ ‫ﻟﻠﺗﻔﺎﻋل‬ :‫اﻵﺗﻲ‬ 𝐻𝑟 ° 𝐶𝑂 + 2𝐻2 → 𝐶𝐻3 𝑂𝐻 .‫ﺑﺎﻟﺳؤال‬ ‫ﺗﻌط‬ ‫ﻟم‬ ‫إذا‬ ‫اﻟﻔرﻋﯾﺔ‬ ‫اﻻﺣﺗراق‬ ‫ﻣﻌﺎدﻻت‬ ‫ﻧﻛﺗب‬ ‫ﻣن‬ ‫ﻧﺣن‬ ‫ﺑطرﯾﻘﺔ‬ ‫ﻧﺣﻠﮫ‬ ‫أن‬ ‫ﯾﻣﻛن‬ ‫ﱠ‬‫ﻷن‬ ،‫ھﯾس‬ ‫ﺑطرﯾﻘﺔ‬ ‫اﻟﺳؤال‬ ‫ﺣددﻧﺎ‬ .(‫)اﻟﺳﻣﯾﺷن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻧﺛﺎﻟﺑﻲ‬ = - 284 KJ/mol (1) 𝐶𝑂 + 1 2 𝑂2 → 𝐶𝑂2 ∆𝐻𝑟 ° = - 286 KJ/mol 2 ( )𝐻2 + 1 2 𝑂2 → 𝐻2 𝑂 ∆𝐻𝑟 ° = - 727 KJ/mol 3 ( )𝐶𝐻3 𝑂𝐻 + 3 2 𝑂2 → 𝐶𝑂2 + 2𝐻2 𝑂 ∆𝐻𝑟 ° ‫وﻟﻠﺣﺻول‬ ‫ﻋﻠﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ ‫ﺗﺑﻘﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 1 ( ‫ﻛﻣﺎ‬ ،‫ھﻲ‬ ‫وﻧﺿرب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟﻔرﻋﯾﺔ‬ ) 2 ( 2 × ، ‫وﻧﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟﻔرﻋﯾﺔ‬ ) 3 :( = - 284 KJ/mol (1) 𝐶𝑂 + 1 2 𝑂2 → 𝐶𝑂2 ∆𝐻𝑟 ° = - 572 KJ/mol 2 ( )2𝐻2 + 𝑂2 → 2𝐻2 𝑂 ∆𝐻𝑟 °
= + 727 KJ/mol 3 ( )𝐶𝑂2 + 2𝐻2 𝑂 → 𝐶𝐻3 𝑂𝐻 + 3 2 𝑂2 ∆𝐻𝑟 ° Δ = - 129 KJ/mol 𝐶𝑂 + 2𝐻2 → 𝐶𝐻3 𝑂𝐻 𝐻𝑟 ° /‫ﻣﻼﺣظﺔ‬ ‫ﯾﻣﻛن‬ ‫ﺣل‬ ‫اﻟﺳؤال‬ ‫ﺑطرﯾﻘﺔ‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) Σ ( ‫ﺑﻌد‬ ‫أن‬ ‫ﻧﺣول‬ ‫ﻗﯾﻣﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﻟﻠﻣﯾﺛﺎﻧول‬ ‫ﻣن‬ ‫اﻹﺷﺎرة‬ ‫ﺑﻘﻠب‬ ‫وذﻟك‬ (‫)اﻟﻣﺎص‬ ‫اﻟﺗﻛوﯾن‬ ‫إﻟﻰ‬ (‫)اﻟﺑﺎﻋث‬ ‫اﻻﺣﺗراق‬ ‫ﻟﻠﻣوﺟب‬ ‫اﻟﺳﺎﻟب‬ ‫ﻣن‬ . ‫س‬ 3 / ‫اﺣﺳب‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ( )𝐻𝑃𝑂3 = - 180KJ 𝑃4 𝑂10 + 4𝐻𝑁𝑂3 →4𝐻𝑃𝑂3 + 2𝑁2 𝑂5 ∆𝐻𝑟 ° ‫ﻋﻠﻣت‬ ‫اذا‬ Δ = -) 2984 ( KJ/mol 𝐻𝑓 𝑃4 𝑂10 ° Δ = -) 174 ( KJ/mol 𝐻𝑓 𝐻𝑁𝑂3 °
Δ =(-43)KJ/mol ?? 𝐻𝑓 𝑁2 𝑂5 ° Δ𝐻𝑟 ° = ∑ 𝑛∆ 𝐻𝑓(𝑃) ° − ∑ 𝑛 ∆𝐻𝑓 𝑅 ( ) ° Δ𝐻𝑟 ° = [4Δ𝐻𝑓(𝐻𝑃𝑂3 ) ° + 2Δ𝐻𝑓(𝑁2 𝑂5 ) ° ] − [Δ𝐻𝑓(𝑃4 𝑂10 ) ° + 4 Δ𝐻𝑓(𝐻𝑁𝑂3 ) ° ] − 180 = 4Δ𝐻𝑓 𝐻𝑃𝑂3 ( ) ° + 2 − 43 ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − − 2984 ( ) + 4 − 174 ( ) [ ] − 180 = 4Δ𝐻𝑓 𝐻𝑃𝑂3 ( ) ° − 86 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − (− 3680)] - 4∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° = 3594 + 180⟹4∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° =− 3594 − 180 ∴∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° = −3774 4 ⟹∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° =− 943. 5 𝐾𝐽/𝑚𝑜𝑙 (‫اﻟﺛﺎﻟث‬ ‫)اﻟدرس‬ ‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬
: ‫اﻹﻧﺗروﺑﻲ‬ ‫»ھﻲ‬ ‫داﻟﺔ‬ ‫ﺣﺎﻟﺔ‬ ‫ﺛرﻣوداﯾﻧﻣﯾﻛﯾﺔ‬ ‫ﺗﻌﺗﺑر‬ ‫ﻣﻘﯾﺎس‬ ‫ﻟدرﺟﺔ‬ ،‫اﻟﻌﺷواﺋﯾﺔ‬ ‫أو‬ ‫ﻻ‬ ‫اﻧﺗظﺎم‬ ،‫اﻟﻧظﺎم‬ ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) S ،( ‫وﺗﻘﺎس‬ ‫ﺑوﺣدة‬ ) J/K mol .«( ‫اﻻﻧﺗروﺑﻲ؟‬ ‫اﻟﺣﺎﻟﺔ‬ ‫داﻟﺔ‬ ‫ﻓﺎﺋدة‬ ‫ﻣﺎ‬ /‫ج‬ ‫ﯾﺳﺗﻔﺎد‬ ‫ﻣﻧﮭﺎ‬ ‫ﻟﻠﺗﻧﺑؤ‬ ‫ﺑﺗﻠﻘﺎﺋﯾﺔ‬ ،‫اﻟﺗﻔﺎﻋل‬ ‫ﻓﻛﻠﻣﺎ‬ ‫ﻛﺎن‬ ‫اﻟﻼﻧظﺎم‬ ً‫ﻼ‬‫ﻗﻠﯾ‬ ‫)اﻟﻌﺷواﺋﯾﺔ‬ (‫أﻛﺑر‬ ‫ﻛﺎﻧت‬ ‫ﻗﯾﻣﺔ‬ ‫أﻛﺑر‬ ‫اﻻﻧﺗروﺑﻲ‬ ) ΔS = (+ ‫واﻟﻌﻛس‬ ‫ﺻﺣﯾﺢ‬ :‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫اﻻﻧﺗروﺑﻲ‬ ‫ﺣﺳﺎب‬ ∆𝑆𝑟 о = ∑𝑛 𝑆(𝑃) о − ∑𝑛 𝑆(𝑅) о ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ) ΔG :( ‫وﯾﻣﻛن‬ ‫ﺣﺳﺎب‬ ‫اﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻔﺎﻋل‬ ‫ﺑﺎﻻﻋﺗﻣﺎد‬ ‫ﻋﻠﻰ‬ ‫اﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻟﻠﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻣن‬ ‫ﺧﻼل‬ ‫اﺳﺗﺧدام‬ ∆𝐺𝑓 ° (‫)اﻟﺳﻣﯾﺷن‬ ‫اﻟﻣﺟﻣوع‬ ‫ﻗﺎﻧون‬ Δ =𝐺𝑟 ° ∑ 𝑛 ∆𝐺𝑓(𝑃) ° − ∑ 𝑛 ∆𝐺𝑓(𝑅) ° :‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬ ‫ﻟﻧﺎ‬ ‫ﺗﺗﯾﺢ‬ ‫ﺛرﻣوداﯾﻧﻣﯾﻛﯾﺔ‬ ‫ﺣﺎﻟﺔ‬ ‫داﻟﺔ‬ ‫»ھﻲ‬ ‫اﻟﺗﻲ‬ ‫اﻟﻌظﻣﻰ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﺗﻣﺛل‬ ‫وھﻲ‬ ،‫اﻟﺗﻔﺎﻋل‬ ‫ﺑﺗﻠﻘﺎﺋﯾﺔ‬ ‫اﻟﺗﻧﺑؤ‬ ‫ﯾﻣﻛن‬ ،‫واﻻﻧﺗروﺑﻲ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﻓﻲ‬ ‫اﻟﺗﻐﯾر‬ ‫ﻗﯾﺎس‬ ‫ﻣن‬ ‫ﻋﻠﯾﮭﺎ‬ ‫اﻟﺣﺻول‬ ‫اﻟﺗﺎﻟﯾﺔ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﺑﺣﺳب‬ : ∆𝐺 о = ∆𝐻 о − 𝑇∆𝑆 о ΔG = ‫ﺳﺎﻟﺑﺔ‬ ‫)اﻟﺗﻔﺎﻋل‬ ‫أو‬ ‫اﻟﺗﻐﯾر‬ ‫اﻟﻔﯾزﯾﺎﺋﻲ‬ .(‫ﺗﻠﻘﺎﺋﻲ‬ ΔG = ‫ﻣوﺟﺑﺔ‬ ‫)اﻟﺗﻔﺎﻋل‬ ‫أو‬ ‫اﻟﺗﻐﯾر‬ ‫اﻟﻔﯾزﯾﺎﺋﻲ‬ ‫ﻏﯾر‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ) ΔG = ‫ﺻﻔر‬ ‫)اﻟﺗﻔﺎﻋل‬ ‫أو‬ ‫اﻟﺗﻐﯾر‬ ‫اﻟﻔﯾزﯾﺎﺋﻲ‬ ‫ﻓﻲ‬ ‫ﺣﺎﻟﺔ‬ .(‫اﺗزان‬ ‫ﻣﻼﺣظﺎت‬ ‫ﻣﮭﻣﺔ‬ : 1 . Δ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورھﺎ‬ = .‫ﺻﻔر‬ 𝐻𝑓 ° ‫ﻣﺛﺎل‬ / / / / :( ... ‫اﻟﺦ‬ .( 𝐴𝑙(𝑠) 𝐶𝑙2 (𝑔) 𝑁2 (𝑔) 𝑂2 (𝑔) 𝐻2 (𝑔)
2 . ‫أﻓﺿل‬ ‫طرﯾﻘﺔ‬ ‫ﻻﺳﺗﺧدام‬ ‫اﻟﺳﻣﯾﺷن‬ ‫ھو‬ ‫إﻧﮭﺎء‬ ‫ﻧﺎﺗﺞ‬ ّ‫ل‬‫ﻛ‬ ‫طرف‬ ‫ﺑدﻗﺔ‬ ‫ﺛم‬ ‫ﻧﻘوم‬ ‫ﺑﻌﻣﻠﯾﺔ‬ ‫طرح‬ ‫اﻟطرف‬ ‫اﻷول‬ ‫ﻣن‬ .‫اﻟﺛﺎﻧﻲ‬ 3 . ‫رﺑﻣﺎ‬ ‫ﺗﻛون‬ Δ ‫ﻣﻌﻠوﻣﺔ‬ ‫و‬ Δ ‫ﻟﻣﻌظم‬ ‫ﻣواد‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻣﻌﻠوﻣﺔ‬ ‫وﯾﻛون‬ ‫اﻟﻣطﻠوب‬ Δ ‫ﻟﻣﺎدة‬ ‫ﻣﺟﮭوﻟﺔ‬ .‫واﺣدة‬ 𝐺𝑟 ° 𝐺𝑓 ° 𝐺𝑓 ° 4 . ‫إذا‬ ‫ﻛﺎن‬ ‫اﻟطرف‬ ‫اﻟﺛﺎﻧﻲ‬ ‫ﻧﺎﺗﺟﮫ‬ ‫ﺳﺎﻟب‬ ‫ﻓﺗﺻﺑﺢ‬ ‫اﻟﻌﻣﻠﯾﺔ‬ :‫اﻟﺣﺳﺎﺑﯾﺔ‬ ‫اﻟﺛﺎﻧﻲ‬ ‫اﻟطرف‬ + ‫اﻷول‬ ‫اﻟطرف‬ ‫س‬ 1 / ‫ﻋرف‬ ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ‫ﻟﻠﺗﻛوﯾن‬ ،‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫وﻣﺎ‬ ‫ھﻲ‬ ‫وﺣداﺗﮭﺎ؟‬ »/ ‫اﻟﺟواب‬ ‫ﺗﻛوﯾن‬ ‫ﻋﻧد‬ ‫اﻟﺣرة‬ ‫اﻟطﺎﻗﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺗﻐﯾﯾر‬ ‫ﻣﻘدار‬ ‫ھﻲ‬ ‫ﺑﺄﺛﺑت‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﻋﻧﺎﺻرھﺎ‬ ‫ﻣن‬ ‫ﻣرﻛب‬ ‫أي‬ ‫ﻣن‬ ‫واﺣد‬ ‫ﻣول‬ ،‫ﺻورة‬ ‫وﻋﻧد‬ ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) 1 atm/25 ( ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) ( ‫ووﺣداﺗﮭﺎ‬ ) KJ/mol)» s ℃ 𝐷𝐺𝑓 ° ‫ﻣﺛﺎل‬ 2 / ‫اﺣﺳب‬ ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺗﺎﻟﻲ‬ ‫ﻋﻧد‬ ‫اﻟظروف‬ ،‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﺛم‬ ‫ن‬ّ‫ﯾ‬‫ﺑ‬ ‫ھل‬ ‫ﯾﺟري‬ ‫اﻟﺗﻔﺎﻋل‬ ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯾ‬ ‫أم‬ :‫اﻟظروف‬ ‫ھذه‬ ‫ﻋﻧد‬ ،‫ﻻ‬ 2𝐶6 𝐻6 (𝐿) + 15𝑂2 (𝑔) → 12𝐶𝑂2 (𝑔) + 6𝐻2 𝑂(𝐿) :‫ﻋﻠﻣت‬ ‫إذا‬ = (- 394) KJ/mol, ∆𝐺𝑓𝐶𝑂2 ° = (- 237) KJ/mol ∆𝐺𝑓𝐻2 𝑂 ° = (173) KJ/mol ∆𝐺𝑓𝐶6 𝐻6 ° /‫اﻟﺣل‬ Δ = 𝐺𝑟 ° ∑ 𝑛 ∆𝐺𝑓(𝑃) ° − ∑ 𝑛 ∆𝐺𝑓(𝑅) ° Δ = ] 12 + 6 [ - ] + 15 [ 𝐺𝑟 ° ∆𝐺𝑓𝐶𝑂2 ° ∆𝐺𝑓𝐻2 𝑂 ° 2∆𝐺𝑓𝐶6 𝐻6 ° ∆𝐺𝑓𝑂2 ° = Δ = [(- 4728) + (- 1422)] - (346) 𝐺𝑟 ° Δ = - 6496 KJ/mol = ‫)ﺳﺎﻟﺒﺔ‬ , ( ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﱠ‬‫ﻷن‬ ∴ 𝐺𝑟 ° ∆𝐺°𝑟
‫ﺗﻣرﯾن‬ 1 / ‫اﺣﺳب‬ ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺗﺎﻟﻲ‬ ‫ﻋﻧد‬ ‫اﻟظروف‬ ،‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ن‬ّ‫ﯾ‬‫وﺑ‬ ‫ھل‬ ‫ﯾﺟري‬ ‫اﻟﺗﻔﺎﻋل‬ ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯾ‬ ‫أم‬ ‫ﻻ؟‬ 2𝑁𝑂(𝑔) + 𝑂2 (𝑔) →2𝑁𝑂2 (𝑔) :‫ﻋﻠﻣت‬ ‫إذا‬ ∆𝐺𝑓𝑁𝑂 ° = 87 𝐾𝐽 𝑚𝑜𝑙 ∆𝐺𝑓𝑁𝑂2 ° = 52 𝐾𝐽 𝑚𝑜𝑙 /‫اﻟﺣل‬ Δ = ] 2 ) [( - ] 2 ) ( + [ 𝐺𝑟 ° ∆𝐺𝑓𝑁𝑂2 ° ∆𝐺𝑓𝑁𝑂 ° ∆𝐺𝑓𝑂2 ° = ] 2 ) 52 [( - ] 2 ) 87 ( + 0 [ Δ = (104) - (174) ⇒ Δ = - 70 KJ/mol ‫ﺗﻠﻘﺎﺋﻲ‬ ∴ 𝐺𝑟 ° 𝐺𝑟 ° ‫ﻣﻼﺣظﺔ‬ / ‫ﻟﻠﻌﻨﺎﺻﺮ‬ ‫اﻟﺤﺮة‬ ‫ﺑﺎﺛﺒﺖ‬ ‫ﺻﻮرة‬ = ‫ﺻﻔﺮ‬ ∆𝐺 °
‫اﻟﺛﺎﻟث‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . Δ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورھﺎ‬ = .‫ﺻﻔر‬ 𝐺𝑓 ° ) T, F ( ‫اﻟﺟواب‬ / T 2 . ‫أﻓﺿل‬ ‫طرﯾﻘﺔ‬ ‫ﻻﺳﺗﺧدام‬ ‫اﻟﺳﻣﯾﺷن‬ ‫ھو‬ ‫إﻧﮭﺎء‬ ‫ﻧﺎﺗﺞ‬ ّ‫ل‬‫ﻛ‬ ‫طرف‬ ‫ﺑدﻗﺔ‬ ‫ﺛم‬ ‫ﻧﻘوم‬ ‫ﺑﻌﻣﻠﯾﺔ‬ ‫ﺟﻣﻊ‬ .‫اﻟﺛﺎﻧﻲ‬ ‫ﻣن‬ ‫اﻷول‬ ‫اﻟطرف‬ ) T, F ( ‫اﻟﺟواب‬ / F 3 . ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻛوﯾن‬ ‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬ : ) A ‫ھﻲ‬ ‫ﻣﻘدار‬ ‫اﻟﺗﻐﯾﯾر‬ ‫ﻓﻲ‬ ‫اﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻋﻧد‬ ‫ﺗﻛوﯾن‬ ‫ﻣول‬ ‫واﺣد‬ ‫ﻣن‬ ‫أي‬ ‫ﻣرﻛب‬ ‫ﻣن‬ ‫ﻋﻧﺎﺻرھﺎ‬ ،‫ﺻورة‬ ‫ﺑﺄﺛﺑت‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ) B ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) 1 atm/50 ( ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) ( ‫ووﺣداﺗﮭﺎ‬ ) KJ/mol .( ∆𝐺𝑓 ° ) C ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) 1 atm/25 ( ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) ( ‫ووﺣداﺗﮭﺎ‬ ) KJ/mol .( ∆𝐺𝑓 ° ‫اﻟﺟواب‬ / A & C
( ‫اﻟراﺑﻊ‬ ‫)اﻟدرس‬ ‫اﻟﺗﻔﺎﻋﻼت‬ ‫ﺳﯾر‬ ‫وﺗﺟﺎه‬ ‫ﻛﺑس‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗطﺑﯾﻘﺎت‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫اﻟﺗﻔﺎﻋﻼت‬ ‫ﺳﯾر‬ ‫واﺗﺟﺎه‬ ‫ﻛﺑس‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗطﺑﯾﻘﺎت‬ ∆𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ‫ﺳؤال‬ / ‫اﻟﺗﻔﺎﻋل؟‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫ﯾﺿﻣﻧﺎن‬ ‫ﻋﺎﻣﻠﯾن‬ ‫أﻛﺛر‬ ‫ھﻣﺎ‬ ‫ﻣﺎ‬ /‫ج‬ ‫ﱠ‬‫إن‬ ‫اﻟﺘﻐﯿﺮ‬ ‫ﻓﻲ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ ) Δ ( ‫ﯾﺘﻀﻤﻦ‬ ‫ﻋﺎﻣﻠﯿﻦ‬ ‫ﻣﮭﻤﯿﻦ‬ ‫ﯾﺆﺛﺮان‬ ‫ﻋﻠﻰ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ،‫اﻟﺘﻔﺎﻋﻞ‬ :‫وھﻤﺎ‬ 𝐺𝑟 ° 1 . 𝐷𝐻𝑟 о 2 . 𝐷𝑆𝑟 ° ‫واذا‬ ‫ﻛﺎﻧﺖ‬ ‫اﺷﺎرة‬ ‫و‬ ‫ﻣﺘﺸﺎﺑﮭﺔ‬ ‫ﺗﻜﻮن‬ ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫ھﻲ‬ ‫اﻟﻤﺘﺤﻜﻢ‬ ‫اﻟﺮﺋﯿﺴﻲ‬ ‫ﻓﻲ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ 𝐷𝐻𝑟 о 𝐷𝑆𝑟 о /‫ﻋﻠل‬ ‫ﺗﻠﻘﺎﺋﯾﺔ؟‬ ‫أﻛﺛر‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻛﺎن‬ ‫ﻋﺎﻟﯾﺔ‬ ‫اﻻﻧﺗروﺑﻲ‬ ‫ﻛﺎﻧت‬ ‫ﻛﻠﻣﺎ‬ /‫ج‬ (∆𝐺°𝑟 = ∆𝐻°𝑟 − 𝑇∆𝑆°𝑟 ) ‫ﱠ‬‫ﻷن‬ ‫اﻻﻧﺘﺮوﺑﻲ‬ ‫ﺗﻘﻊ‬ ‫ﺿﻤﻦ‬ ‫اﻟﺤﺪ‬ ) ‫اﻟﺴﺎﻟﺐ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫ﻛﻠﻤﺎ‬ ‫ﺗﻜﻮن‬ ‫ﻗﯿﻤﺔ‬ ‫ھﺬا‬ ‫اﻟﺤﺪ‬ ‫أﻋﻠﻰ‬ ‫ﻣﻦ‬ ) > 𝑇∆𝑆°𝑟 ) ∆𝐻°𝑟 ∆𝐻°𝑟 ( ‫ﻛﺎﻧﺖ‬ ‫ﻗﯿﻤﺔ‬ = ،‫ﺳﺎﻟﺒﺔ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ . 𝑇∆𝑆°𝑟 𝐷𝐺°𝑟 ‫ﻣﻼﺣظﺔ‬ / ‫ﺣﯿﻨﻤﺎ‬ ‫ﺗﻜﻮن‬ ‫إﺷﺎرات‬ ) ΔH, ΔS ( ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻓﻠﯿﺲ‬ ‫ھﻨﺎك‬ ‫ﺗﺄﺛﯿﺮ‬ ‫ﻟﺪرﺟﺔ‬ ،‫اﻟﺤﺮارة‬ ‫وﻻﺣﻆ‬ ‫ﱠ‬‫أن‬ ‫إﺷﺎرة‬ ) ΔH ( ‫ﺗﺸﺎﺑﮫ‬ ‫إﺷﺎرة‬ ) ΔG ( ‫ﻓﻲ‬ .‫اﻟﻐﺎﻟﺐ‬
:‫ﻣﮭﻣﺔ‬ ‫ﻣﻼﺣظﺎت‬ ΔG .1 = ‫ﺳﺎﻟﺒﺔ‬ (-) ‫اﻟﺘﻔﺎﻋﻞ‬ ،‫ﺗﻠﻘﺎﺋﻲ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ (+) ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ 2 . :‫ﻋﻤﻠﯿﺎت‬ ‫)اﻻﻧﺼﮭﺎر‬ / ‫اﻟﺘﺒﺨﺮ‬ (‫)اﻟﻐﻠﯿﺎن‬ / ‫اﻟﺘﺴﺎﻣﻲ‬ (... ΔS = (+) ‫ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ΔH = (+) ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ 3 . ‫ﻋﻤﻠﯿﺎت‬ ‫)اﻻﻧﺠﻤﺎد‬ ‫)اﻟﺘﺒﻠﻮر‬ / ‫اﻟﺘﻜﺜﯿﻒ‬ (‫)اﻟﻨﺪى‬ / ‫اﻟﺘﺼﻠﺐ‬ (... ΔS = (-) ‫ﺗﻘﻞ‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ΔH = (-) ‫ﺑﺎﻋﺚ‬ ‫ﻟﻠﺤﺮارة‬ 4 . ‫ﺣﯿﻨﻤﺎ‬ :‫ﯾﻘﻮل‬ ‫)ﯾﺴﺒﺐ‬ ‫اﻧﺨﻔﺎض‬ ‫درﺟﺔ‬ ‫ﺣﺮارة‬ :(‫اﻟﻤﺤﻠﻮل‬ ‫ﺑﺎﻋﺜﺔ‬ ‫واﻟﻌﻜﺲ‬ ‫ﻣﺎﺻﺔ‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ) ΔH > TΔS ( 5 . ‫ذوﺑﺎن‬ ‫ﺻﻠﺐ‬ ‫أو‬ ،‫ﺳﺎﺋﻞ‬ ‫ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ +) = ΔS ( ‫وﻋﻠﻰ‬ ‫اﻷﻏﻠﺐ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ +) = ΔH ( 6 . ‫ذوﺑﺎن‬ ،‫ﻏﺎز‬ ‫ﺗﻘﻞ‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ -) = ΔS ،( ‫وھﻲ‬ :‫ﻋﻤﻠﯿﺎت‬ ‫ﺑﺎﻋﺜﺔ‬ -) = ΔH ( 7 . ‫ﻋﻤﻠﯿﺎت‬ ‫)اﻟﺘﺤﻠﻞ‬ / ‫اﻟﺘﻔﻜﻚ‬ / :(‫اﻟﺘﺄﯾﻦ‬ ‫)ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ +) = ΔS ( ‫واﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ +) = ΔH ( 8 . :‫ﻋﻤﻠﯿﺎت‬ ‫)اﻻﺗﺤﺎد‬ / ‫اﻻﻧﺪﻣﺎج‬ :(... ‫ﺗﻘﻞ‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ -) = ΔH ( 9 . ‫أھم‬ :‫ﻣﻼﺣظﺔ‬ ‫)إن‬ ‫ﻛﺎﻧﺖ‬ ‫اﻟﻌﺒﺎرات‬ ‫ﻣﻨﻔﯿﺔ‬ ‫ﻓﺎﻟﺘﻔﺎﻋﻞ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ +) = ΔG ،( ‫وﻧﺮﻓﻊ‬ ‫اﻟﻨﻔﻲ‬ ً‫ﺎ‬‫ﻣﺆﻗﺘ‬ ‫ﺛﻢ‬ ‫ﻧﺤﻞ‬ ‫اﻟﻤﺴﺄﻟﺔ‬ ‫ﻋﻠﻰ‬ .(‫اﻟﻘﺎدﻣﺔ‬ ‫اﻟﻤﺴﺎﺋﻞ‬ ‫ﻣﻦ‬ ‫ذﻟﻚ‬ ‫ﺳﯿﺘﻮﺿﺢ‬ ‫ﻛﻤﺎ‬ ،‫ﻛﺒﺲ‬ ‫ﻋﻼﻗﺔ‬ ‫ﺿﻮء‬
‫اﻟراﺑﻊ‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . ‫اﻟﺗﻔﺎﻋل‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫ﯾﺿﻣﻧﺎن‬ ‫ﻋﺎﻣﻠﯾن‬ ‫أﻛﺛر‬ ‫ھﻣﺎ‬ ‫ﻣﺎ‬ ) A Δ𝐻°𝑟 & Δ𝐶°𝑟 ) B ∆𝐻°𝑟 & Δ𝑆°𝑟 ) C ∆𝐺°𝑟 & Δ𝑆°𝑟 (‫)ﺧﯾﺎرات‬ ‫اﻟﺳؤال‬ ‫ﻧوع‬ ‫اﻟﺟواب‬ / B 2 . ‫اﻟﺗﺻﻠب‬ / (‫)اﻟﻧدى‬ ‫اﻟﺗﻛﺛﯾف‬ / ‫)اﻟﺗﺑﻠور‬ ‫)اﻻﻧﺟﻣﺎد‬ ‫ﻋﻣﻠﯾﺎت‬ (... ΔS = (+) ‫ﺗزداد‬ ‫اﻟﻌﺷواﺋﯾﺔ‬ ΔH = (-) ‫ﺑﺎﻋث‬ ‫ﻟﻠﺣرارة‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 3 . ،‫ﻏﺎز‬ ‫ذوﺑﺎن‬ ‫ﺗﻘل‬ ‫اﻟﻌﺷواﺋﯾﺔ‬ -) = ΔS ،( ‫وھﻲ‬ :‫ﻋﻣﻠﯾﺎت‬ ‫ﺑﺎﻋﺛﺔ‬ -) = ΔH ( ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / T (‫اﻟﺧﺎﻣس‬ ‫)اﻟدرس‬ ‫ﻛﺑس‬ ‫ﻋﻼﻗﺔ‬ ‫وﻓق‬ ‫ﺗﻌﺎﻟﯾل‬ ‫ﻋﻠل‬ ‫ﻣﺎ‬ ‫ﯾﺄﺗﻲ‬ ‫ﻋﻠﻰ‬ ‫وﻓق‬ ‫ﻋﻼﻗﺔ‬ :‫ﻛﺑس‬ ‫س‬/ 3
1 . ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫اﻟﺟﻠﯾد‬ ‫اﻧﺻﮭﺎر‬ ‫ﻋﻣﻠﯾﺔ‬ /‫ج‬ - ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﺳﺎﻟﺒﺔ‬ ‫)اﻟﺘﻔﺎﻋﻞ‬ (‫ﺗﻠﻘﺎﺋﻲ‬ - ‫اﻧﺼﮭﺎر‬ ΔS = ‫ﻣﻮﺟﺒﺔ‬ ‫)ﺗﺰداد‬ (‫اﻟﻌﺸﻮاﺋﯿﺔ‬ - ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻣﺎﺻﺔ‬ ΔH = ‫ﻣﻮﺟﺒﺔ‬ ΔG = ΔH - TΔS + (-) + = - ΔH < TΔS 2 . ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫اﻷوﻟﯾﺔ‬ ‫ﻋﻧﺎﺻره‬ ‫إﻟﻰ‬ ‫اﻟﻣﺎء‬ ‫ﯾﺗﺣﻠل‬ ‫ﻻ‬ /‫ج‬ - ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ - ‫ﯾﺘﺤﻠﻞ‬ ΔS = ‫ﻣﻮﺟﺒﺔ‬ ‫)ﺗﺰداد‬ (‫اﻟﻌﺸﻮاﺋﯿﺔ‬ - ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻣﺎﺻﺔ‬ ‫ﻟﻠﺤﺮارة‬ ΔH = ‫ﻣﻮﺟﺒﺔ‬ ΔG = ΔH - TΔS + (-) + = + ΔH > TΔS 3 . ‫ﯾذوب‬ ‫ﻏﺎز‬ ‫ﻓﻲ‬ ‫اﻟﻣﺎء‬ ‫وﯾﺑﻌث‬ ‫ﺣرارة‬ ‫ﻋﺎﻟﯾﺔ‬ ‫؟‬ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯾ‬ 𝑆𝑂2 /‫ج‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ ‫ذوﺑﺎن‬ ‫ﻏﺎز‬ ΔS = ‫ﺳﺎﻟﺒﺔ‬ ‫)ﺗﻘﻞ‬ (‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﺑﺎﻋﺜﺔ‬ ΔH = ‫ﺳﺎﻟﺒﺔ‬ ΔG = ΔH - TΔS - (+) - = -
ΔH > TΔS 4 . ‫ﻻ‬ ‫ﺗﺗﻔﻛك‬ ‫ﻛﺎرﺑوﻧﺎت‬ ‫اﻟﻛﺎﻟﺳﯾوم‬ ‫؟‬ً‫ﺎ‬‫اﻋﺗﯾﺎدﯾ‬ /‫ج‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ ‫ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ΔS = ‫ﻣﻮﺟﺒﺔ‬ ΔG = ΔH - TΔS + (-) + = + ΔH > TΔS :‫ﻣﮭﻤﺔ‬ ‫ﻣﻼﺣﻈﺎت‬
1 ‫ﻋﻨﺪﻣﺎ‬- ‫ﺗﻜﻮن‬ ‫اﺷﺎرة‬ ) ( ‫ﻣﺘﺸﺎﺑﮭﺔ‬ ∆𝐻, ∆𝑆 a) ∆𝐻 =− ∆𝑆 =− ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 − ( ) − − ( ) − ( ) (+) ‫ھﻨﺎ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﺗﻜﻮن‬ (‫)ﺗﺒﺮﯾﺪ‬ ∆𝐻 > 𝑇∆𝑆 b) ∆𝐻 =+ ∆𝑆 =+ ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 + ( ) − + ( ) + ( ) (−) ‫ھﻨﺎ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﺗﻜﻮن‬ (‫)ﺗﺴﺨﯿﻦ‬ 𝑇∆𝑆 > ∆𝐻 ‫اﻟﺧﺎﻣس‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬
1 . ‫؟‬ ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫اﻟﺟﻠﯾد‬ ‫اﻧﺻﮭﺎر‬ ‫ﻋﻣﻠﯾﺔ‬ ) A ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ΔG = ‫ﺳﺎﻟﺑﺔ‬ ‫)اﻟﺗﻔﺎﻋل‬ (‫ﺗﻠﻘﺎﺋﻲ‬ - ‫اﻧﺻﮭﺎر‬ ΔS = ‫ﻣوﺟﺑﺔ‬ ‫)ﺗزداد‬ (‫اﻟﻌﺷواﺋﯾﺔ‬ - ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﻣﺎﺻﺔ‬ ΔH = ‫ﺳﺎﻟﺑﺔ‬ ΔG = ΔH - TΔS + (-) + = - ΔH > TΔS ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 2 . ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫اﻷوﻟﯾﺔ‬ ‫ﻋﻧﺎﺻره‬ ‫إﻟﻰ‬ ‫اﻟﻣﺎء‬ ‫ﯾﺗﺣﻠل‬ ‫ﻻ‬ :‫ﺑﺳﺑب‬ ‫وذﻟك‬ ) A ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ΔG = ‫ﻣوﺟﺑﺔ‬ - ‫ﯾﺗﺣﻠل‬ ΔS = ‫ﻣوﺟﺑﺔ‬ ‫)ﺗزداد‬ (‫اﻟﻌﺷواﺋﯾﺔ‬ ) B ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﻏﯾر‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ΔG = ‫ﻣوﺟﺑﺔ‬ - ‫ﯾﺗﺣﻠل‬ ΔS = ‫ﻣوﺟﺑﺔ‬ ‫)ﺗزداد‬ (‫اﻟﻌﺷواﺋﯾﺔ‬ - ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﻣﺎﺻﺔ‬ ‫ﻟﻠﺣرارة‬ ΔH = ‫ﻣوﺟﺑﺔ‬ ( ‫اﻟﺳﺎدس‬ ‫)اﻟدرس‬ ‫ﻛﺑس‬ ‫ﻋﻼﻗﺔ‬ ‫ﺣﺳﺎﺑﺎت‬ :‫ﻛﺑس‬ ‫ﻋﻼﻗﺔ‬ ‫ﺣﺳﺎﺑﺎت‬
‫ﯾﻤﻜﻦ‬ ‫اﺳﺘﺨﺪام‬ ‫ﻋﻼﻗﺔ‬ ‫ﻛﺒﺲ‬ ‫ﻟﺤﺴﺎب‬ ‫ﻗﯿﻤﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ Δ ‫اﻟﻘﯿﺎﺳﯿﺔ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫ﻣﻌﺮﻓﺔ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫أو‬ ‫ﻋﺪم‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫وذﻟﻚ‬ 𝐺𝑟 ° ‫إذا‬ ‫ﺗﻢ‬ ‫ﻗﯿﺎس‬ ‫ﻗﯿﻢ‬ Δ ‫و‬ ، ‫ﻋﻨﺪ‬ ‫اﻟﻈﺮوف‬ ‫اﻟﻘﯿﺎﺳﯿﺔ‬ ) 1 atm/25 ( ، ‫وذﻟﻚ‬ ‫ﻣﻦ‬ ‫ﺧﻼل‬ ‫اﻟﻌﻼﻗﺔ‬ :‫اﻟﺘﺎﻟﯿﺔ‬ 𝐻𝑟 ° ∆𝑆𝑟 ° ℃ ∆𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° ‫ﻣﻼﺣظﺔ‬ 1 / ‫ﯾﺠﺐ‬ ‫اﻟﺘﻤﯿﯿﺰ‬ ‫ﺑﯿﻦ‬ Δ ‫و‬ ‫وﺑﯿﻦ‬ Δ ‫و‬ Δ . 𝐺𝑟 ° ∆𝐺𝑓 ° 𝐻𝑟 ° 𝐻𝑓 ° ‫ﻣﻼﺣظﺔ‬ 2 / ‫ﻗﯿﻢ‬ ‫اﻻﻧﺜﺎﻟﺒﻲ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ ‫ﻟﻠﻌﻨﺎﺻﺮ‬ ‫اﻟﺤﺮة‬ ‫ﺑﺄﺛﺒﺖ‬ ‫ﺻﻮرة‬ ‫ﺗﺴﺎوي‬ ‫ﺻﻔﺮ‬ ‫ﻣﻼﺣظﺔ‬ 3 / ‫ﯾﺠﺐ‬ ‫ﺗﻮﺣﯿﺪ‬ ‫وﺣﺪﺗﻲ‬ ‫و‬ Δ ‫ﻗﺒﻞ‬ ‫ﺗﻄﺒﯿﻖ‬ ‫ﻋﻼﻗﺔ‬ .‫ﻛﺒﺲ‬ ∆𝑆𝑟 ° 𝐺𝑟 ° ‫ﻣﻼﺣظﺔ‬ 4 / ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ (‫)اﻻﻧﺘﺮوﺑﻲ‬ ‫ﻟﻠﺠﺰﯾﺌﺎت‬ ‫ﺗﻌﺘﻤﺪ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻄﺎﻗﺔ‬ ،‫اﻟﺤﺮﻛﯿﺔ‬ ‫وھﻲ‬ ‫ﺣﺮﻛﺎت‬ ‫ﺻﻐﯿﺮة‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ ‫ﻟﻘﯿﻢ‬ ‫اﻻﻧﺜﺎﻟﺒﻲ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ ‫ﻟﺬﻟﻚ‬ ‫ﺗﻜﻮن‬ ‫وﺣﺪة‬ ΔS ‫ﺻﻐﯿﺮة‬ ‫ﺑﺎﻟﺠﻮل‬ ) J ( ‫ووﺣﺪات‬ ΔH ‫و‬ ΔG ‫ﻛﺒﯿﺮة‬ ‫ﺑﺎﻟـ‬ ) KJ .( ‫س‬ 4 / ‫ﺟد‬ ‫ﻗﯾﻣﺔ‬ ) ( ‫ﻟﻠﺗﻔﺎﻋل‬ :‫اﻟﺗﺎﻟﻲ‬ ∆𝐺°𝑟 2𝐶𝑂(𝑔) + 𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) ‫اﻟﻣﻌﻠوﻣﺎت‬ ‫أﻋطﯾت‬ ‫إذا‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫اﻟظروف‬ ‫ﺗﺣت‬ ‫ﯾﺟري‬ ‫اﻟذي‬ :‫اﻵﺗﯾﺔ‬ Δ = -) 110.5 ( KJ/mol , 𝐻𝑓𝐶𝑂 ° Δ = -) 393.5 ( KJ/mol 𝐻𝑓𝐶𝑂2 ° = 198 J/K mol , 𝑆𝐶𝑂 ° = 214 J/K mol , 𝑆𝐶𝑂2 ° = 205 J/K mol 𝑆𝑂2 ° /‫اﻟﺣل‬ 1 - Δ 𝐻𝑟 ° = 2∆𝐻𝑓𝐶𝑂2 ° ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 2∆𝐻𝑓𝐶𝑂 ° ( )+ ∆𝐻𝑓𝑂2 ° ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦
Δ 𝐻𝑟 ° = 2 × − 393. 5 ( ) [ ] − 2 × − 110. 5 ( ) + 0 [ ] Δ 𝐻𝑟 ° = − 787 ( ) + 221 ( ) ∆𝐻𝑟 ° = − 566 𝐾𝐽/𝑚𝑜𝑙 2. ∆𝑆𝑟 ° = 2𝑆𝐶𝑂2 ° ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 2𝑆𝐶𝑂 ° + 𝑆𝑂2 ° ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ∆𝑆𝑟 ° = [2 × 214] − [2 × 198) + 205] ∆𝑆𝑟 ° = − 173 𝐽/𝐾 𝑚𝑜𝑙 = -0.173 KJ/K mol ‫ﻧﻮﺣﺪ‬ ‫اﻟﻮﺣﺪات‬ ∆𝑆𝑟 ° = −173 1000 3. Δ ∴ 𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇 ∆𝑆𝑟 ° , 𝑇𝑘 = 𝑡℃ + 273 𝑇𝑘 = 25 + 273 = 298𝐾 = -566 - (298) (-0.173) Δ ∴ 𝐺𝑟 ° = − 514. 446 𝐾𝐽 𝑚𝑜𝑙 (‫ﺗﻠﻘﺎﺋﻲ‬ ‫)اﻟﺘﻔﺎﻋﻞ‬ ‫س‬ 5 / ‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺗﺎﻟﻲ‬ 2𝐻2 𝑔 ( ) + 𝑂2 (𝑔) → 2𝐻2 𝑂(𝑔) (J/K.mol) ‫ﻟﻠﺗﻔﺎﻋل‬ ‫ﺑوﺣدة‬ ‫اﺣﺳب‬ ‫ﻗﯾﻣﺔ‬ ∆𝑆𝑟 ° :‫ﻋﻠﻣت‬ ‫إذا‬ Δ𝐻𝑓𝐻2 𝑂 ° = − 242 ( )𝐾𝐽/𝑚𝑜𝑙) Δ𝐺𝑓𝐻2 𝑂 ° = − 228 ( )𝐾𝐽/𝑚𝑜𝑙)
/‫اﻟﺣل‬ Δ 𝐻𝑟 ° = ∑ 𝑛∆𝐻𝑓(𝑃) ° − ∑ 𝑛 ∆𝐻𝑓(𝑅) ° = ] 2Δ [𝐻𝑓𝐻2 𝑂 ° − [2∆𝐻𝑓𝐻2 ° + ∆𝐻𝑓𝑂2 ° ] Δ 𝐻𝑟 ° = [2(− 242)] − [0]⟹∆𝐻𝑟 ° = − 484 𝐾𝐽/𝑚𝑜𝑙 Δ 𝐺𝑟 ° = ∑ 𝑛 ∆𝐺𝑓(𝑃) ° ∑ 𝑛 ∆𝐺𝑓(𝑅) ° = ] 2Δ [ − ] 0 + 0 [ 𝐺𝑓𝐻2 𝑂 ° T = 25 + 273 T = 298 K Δ 𝐺𝑟 ° = [2(− 228)]⟹∆𝐺𝑟 ° = − 456 𝐾𝐽/𝑚𝑜𝑙 Δ 𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° - 456 = - 484 - ) 298 ( ∆𝑆𝑟 ° ⟹ ∆𝑆𝑟 ° = −28 298 = − 0. 094 𝐾𝐽/𝐾 𝑚𝑜𝑙 ∴ ∆𝑆𝑟 ° = − 0. 094 × 1000⟹ ∆𝑆𝑟 ° = − 94 𝐽/𝐾 𝑚𝑜𝑙 ‫اﻟﺳﺎدس‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬
1 . Δ𝐺𝑟 ° = Δ𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / T 2 . ) A ‫ﯾﺟب‬ ‫اﻟﺗﻣﯾﯾز‬ ‫ﺑﯾن‬ Δ ‫و‬ ‫وﺑﯾن‬ Δ ‫و‬ Δ . 𝐺𝑟 ° ∆𝐺°𝑓 𝐻𝑟 ° 𝐻𝑓 ° ) B ‫ﻗﯾم‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫واﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورة‬ ‫ﺗﺳﺎوي‬ 1 ) C ‫ﻗﯾم‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫واﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورة‬ ‫ﺗﺳﺎوي‬ ‫ﺻﻔر‬ (‫)ﺧﯾﺎرات‬ ‫اﻟﺳؤال‬ ‫ﻧوع‬ ‫اﻟﺟواب‬ / A& C 3 . ‫اﻟطﺎﻗﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻌﺗﻣد‬ ‫ﻻ‬ ‫ﻟﻠﺟزﯾﺋﺎت‬ (‫)اﻻﻧﺗروﺑﻲ‬ ‫اﻟﻌﺷواﺋﯾﺔ‬ ‫ﺻﻐﯾرة‬ ‫ﺣرﻛﺎت‬ ‫وھﻲ‬ ،‫اﻟﺣرﻛﯾﺔ‬ ‫اﻟﺣرة‬ ‫واﻟطﺎﻗﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﻟﻘﯾم‬ ‫ﺑﺎﻟﻧﺳﺑﺔ‬ ‫ﻟذﻟك‬ ‫ﺗﻛون‬ ‫وﺣدة‬ ΔS ‫ﺻﻐﯾرة‬ ‫ﺑﺎﻟﺟول‬ ) J ( ‫ووﺣدات‬ ΔH ‫و‬ ΔG ‫ﻛﺑﯾرة‬ ‫ﺑﺎﻟـ‬ ) KJ .( ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F ( ‫اﻟﺳﺎﺑﻊ‬ ‫)اﻟدرس‬ ‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬ ‫ﺗطﺑﯾﻘﺎت‬
‫س‬ 6 / ‫ﺟﺪ‬ ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫اﻟﺘﻲ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻋﻨﺪھﺎ‬ ‫اﻟﺘﻔﺎﻋﻼت‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ،‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫إذا‬ :‫ﻋﻠﻤﺖ‬ ‫ﻓرع‬ ) A /( ∆𝐻𝑟 = (+ 126) 𝐾𝐽/𝑚𝑜𝑙 ∆𝑆𝑟 = (+ 48) 𝐽/𝐾 𝑚𝑜𝑙 /‫اﻟﺣل‬ ΔG ‫ﻧﻔﺘﺮض‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬ ‫ﺣﺎﻟﺔ‬ ،‫اﺗﺰان‬ ، ‫ﺻﻔﺮ‬ = ∴ ‫وﻧﻮﺣﺪ‬ ‫اﻟﻮﺣﺪات‬ ∆𝐻 ∆𝑆 = 𝑇⟹ 0 = ∆𝐻 − ∆𝑆 ∆𝑆𝑟 = 48 1000 ⟹ ∆𝑆𝑟 = + 0. 048 𝐾𝐽/𝐾 𝑚𝑜𝑙 ∴ 𝑇 = 126 0.048 ⟹ 𝑇 = 2625 𝐾 /‫ﻣﻼﺣظﺔ‬ ‫إذا‬ ‫ﻛﺎﻧﺖ‬ ‫إﺷﺎرة‬ ΔS ‫و‬ ΔH = ‫ﻣﻮﺟﺒﺔ‬ ‫ﻓﯿﺠﺐ‬ ‫ﺗﺴﺨﯿﻦ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻷﻛﺜﺮ‬ ‫ﻣﻦ‬ ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫اﻟﻤﺤﺴﻮﺑﺔ‬ ‫ﻋﻨﺪ‬ ‫اﻻﺗﺰان‬ . ‫وﻋﻠﯿﮫ‬ ‫ﯾﺠﺐ‬ ‫ﺗﺴﺨﯿﻦ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻷﻛﺜﺮ‬ ‫ﻣﻦ‬ ) 2625 K ( ‫ﻟﻜﻲ‬ ‫ﯾﺼﺒﺢ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ .‫ﺗﻠﻘﺎﺋﻲ‬ (B) ‫ﻓرع‬/ ∆𝑆𝑟 = (− 105) 𝐽/𝐾 𝑚𝑜𝑙, ∆𝐻𝑟 = (− 12) 𝐾𝐽/𝑚𝑜𝑙 ‫ﻧﻔﺘﺮض‬ ΔG = ‫ﺻﻔﺮ‬ ‫)ﺣﺎﻟﺔ‬ (‫اﻻﺗﺰان‬ ∴ 𝑇 = ∆𝐻 ∆𝑆 , ∆𝑆 = −105 1000 ⟹ ∆𝑆 = − 0. 105 𝐾𝐽/𝐾 𝑚𝑜𝑙 𝑇 = −12 −0.105 ⟹ 𝑇 = 114. 3 𝐾 ‫وﻋﻠﯿﮫ‬ ‫ﯾﺠﺐ‬ ‫ﺗﺒﺮﯾﺪ‬ ‫إﻧﺎء‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻷﻗﻞ‬ ‫ﻣﻦ‬ ) 114.3 K ( ‫ﻟﯿﺼﺒﺢ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ .‫ﺗﻠﻘﺎﺋﻲ‬ ‫داﺋﻤﺎ‬ ‫ﻓﻲ‬ ‫اﻟﺴﺆال‬ ‫ﻣﺘﺸﺎﺑﮭﺔ‬ ΔS ‫و‬ ΔH ‫ﺳﺘﻌﻄﻰ‬ ‫أﺷﺎرات‬ ‫ﻗﯿﻢ‬ ‫ﻣﻼﺣظﺔ‬/ = ‫ﺳﺎﻟﺒﺔ‬ ‫)ﻧﺒﺮداﻟﺘﻔﺎﻋﻞ‬ ) ΔS ‫و‬ ΔH ‫اذا‬ ‫ﻛﺎﻧﺖ‬
‫س‬ 7 : ‫ﺗﺘﻔﻜﻚ‬ ‫ﻛﺎرﺑﻮﻧﺎت‬ ‫اﻟﻜﺎﻟﺴﯿﻮم‬ ‫ﺣﺴﺐ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫اﻵﺗﯿﺔ‬ : 𝐶𝑎𝐶𝑂3 (𝑠) ∆→ 𝐶𝑎𝑂(𝑠) + 𝐶𝑂2 (𝑔) (160 J/K mol) =Δ𝑆°𝑟 ‫ﻓﺈذا‬ ‫ﻋﻠﻤﺖ‬ Δ ‫ﻟﻜﻞ‬ ‫ﻣﻦ‬ ‫و‬ CaO ‫و‬ ‫ھﻲ‬ ‫ﻋﻠﻰ‬ :‫اﻟﺘﻮاﻟﻲ‬ 𝐻𝑓 ° 𝐶𝑂2 𝐶𝑎𝐶𝑂3 -) 1207 / - 635 / - 393.5 ( ‫ﺑﻮﺣﺪات‬ KJ/mol ، :‫ﺟﺪ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫ﺛﻢ‬ ‫أرﺳﻢ‬ ‫ﻣﺨﻄﻂ‬ ‫اﻟﻄﺎﻗﺔ‬ Δ 1) 𝐻𝑟 ° ‫ﻟﻠﺘﻔﺎﻋﻞ‬Δ 2) 𝐺𝑟 ° 3) ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫اﻟﺘﻲ‬ ‫ﯾﺼﺒﺢ‬ ‫ﻋﻨﺪھﺎ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﯿﺎ‬ /‫اﻟﺣل‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ 1 ( ∆𝐻𝑟 ° = ∑ 𝑛∆ 𝐻𝑓(𝑃) ° − ∑ 𝑛 ∆𝐻𝑓 𝑅 ( ) ° ∆𝐻𝑟 ° = [(− 635) + (− 393. 5)] − [− 1207] ∆𝐻𝑟 ° = (− 1028) + (1207) ∆𝐻𝑟 ° = 178. 5 𝐾𝐽/𝑚𝑜𝑙 ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ 2 ( T = 25 + 273 = 298 K
∆𝑆𝑟 ° = 160 1000 = 0. 16 𝐾𝐽/𝐾 𝑚𝑜𝑙 ∴ ∆𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° ∆𝐺𝑟 ° = 178. 5 − (298) (0. 16) ∆𝐺𝑟 ° = 130. 82 𝐾𝐽/𝑚𝑜𝑙 3) (0 = Δ ‫ﻧﻔﺘﺮض‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺘﺰن‬ 𝐺𝑟 ° ) 0 = ∆𝐻𝑟 ° − 𝑇∆𝑆⟹𝑇 = ∆𝐻𝑟 ° ∆𝑆𝑟 ° ⟹ 𝑇 = 178.5 0.16 = 115. 6 𝐾 ‫ﻣﻮﺟﺒﺔ‬Δ ‫و‬ ‫ﺑﻤﺎ‬ ‫اﻧﮫ‬ ‫ﻗﯿﻤﺘﻲ‬ 𝐻𝑟 ° ∆𝑆𝑟 ° ‫ﻟﻜﻲ‬ ‫ﯾﺼﺒﺢ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ .ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯿ‬ 1115.6 K ً‫ا‬‫إذ‬ ‫ﯾﺠﺐ‬ ‫ﺗﺴﺨﯿﻦ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫إﻟﻰ‬ ‫أﻛﺜﺮ‬ ‫ﻣﻦ‬
‫اﻟﺳﺎﺑﻊ‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . ‫إذا‬ ‫ﻛﺎﻧت‬ ‫إﺷﺎرة‬ ΔS ‫و‬ ΔH = ‫ﺳﺎﻟﺑﺔ‬ ‫ﻓﯾﺟب‬ ‫ﺗﺳﺧﯾن‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻷﻛﺛر‬ ‫ﻣن‬ ‫درﺟﺔ‬ ‫اﻟﺣرارة‬ .‫اﻻﺗزان‬ ‫ﻋﻧد‬ ‫اﻟﻣﺣﺳوﺑﺔ‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 2 . ‫اذا‬ ‫ﻛﺎﻧت‬ S Δ ‫و‬ ΔH = ‫ﺳﺎﻟﺑﺔ‬ (‫)ﻧﺑرداﻟﺗﻔﺎﻋل‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 3 . ‫ﯾﺟب‬ ‫ﺗﺳﺧﯾن‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻷﻛﺛر‬ ‫ﻣن‬ ) 2625 K ( ‫ﻟﻛﻲ‬ ‫ﯾﺻﺑﺢ‬ ‫اﻟﺗﻔﺎﻋل‬ .‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / T

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byTessa Mero
 
AI Trends in Creative Operations 2024 by Artwork Flow.pdf
bymarketingartwork
 

Iiiimmpoooo

  • 1.
    ‫اﻟﻔﺻل‬ ‫اﻻول‬ ‫اﻟﺛرﻣوداﯾﻧﻣك‬ ( ‫اﻟدرس‬ ‫اﻷول‬ ) ‫اﻟﺛرﻣوداﯾﻧﻣك‬ ‫طرﯾﻘﺔ‬ ‫ﻗﺎﻧون‬ ‫ھﯾس‬: /‫ﺳؤال‬ ‫ﻟﻣﺎذا‬ ‫ﯾﺗم‬ ‫اﻟﻠﺟوء‬ ‫إﻟﻰ‬ ‫طرﯾﻘﺔ‬ ‫ھﯾس‬ ‫ﻟﺣﺳﺎب‬ Δ ‫𝐻؟‬𝑟 ° ‫ﺟواب‬ / ‫ﱠ‬‫ﻷن‬ ‫ﺑﻌض‬ ‫اﻟﻣرﻛﺑﺎت‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫ﻻ‬ ‫ﯾﻣﻛن‬ ‫ﺗﺻﻧﯾﻌﮭﺎ‬ ً‫ة‬‫ﻣﺑﺎﺷر‬ ‫ﻣن‬ ‫ﻋﻧﺎﺻرھﺎ‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورة‬ (‫)ﻋﻠل‬ ‫ﺑﺳﺑب‬ : ‫ﱠ‬‫أن‬ 1 ( ‫ﺑﻌض‬ ‫اﻟﺗﻔﺎﻋﻼت‬ ‫ﺗﺳﯾر‬ ‫ﺑﺑطء‬ .‫ﺷدﯾد‬ 2 ( ‫ن‬ ّ‫ﺗﻛو‬ ‫ﻣرﻛﺑﺎت‬ ‫ﺟﺎﻧﺑﯾﺔ‬ ‫ﻏﯾر‬ ‫ﻣرﻏوب‬ .‫ﻓﯾﮭﺎ‬ ‫ھﯾس؟‬ ‫ﻗﺎﻧون‬ ‫ﻧص‬ ‫ھو‬ ‫ﻣﺎ‬ :‫ﯾﻧص‬ ‫)ﻋﻧد‬ ‫ﺗﺣوﯾل‬ ‫اﻟﻣﺗﻔﺎﻋﻼت‬ ‫إﻟﻰ‬ ‫ﻧواﺗﺞ‬ ‫ﱠ‬‫ﻓﺈن‬ ‫اﻟﺗﻐﯾر‬ ‫ﻓﻲ‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ھو‬ ‫ﻧﻔﺳﮫ‬ ‫ﺳواء‬ ‫ﺗم‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻓﻲ‬ ‫ﺧطوة‬ ‫واﺣدة‬ .(‫اﻟﺧطوات‬ ‫ﻣن‬ ‫ﺳﻠﺳﻠﺔ‬ ‫ﻋﺑر‬ ‫أو‬ / ‫ﻣﺛﺎل‬ 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 (𝑔) ⟶𝐶𝑂2 𝑔 ( ) ∆𝐻𝑟 =− 393. 5 𝐾𝐽/𝑚𝑜𝑙 ‫وﻋﻧد‬ ‫ﺗﻛوﯾن‬ ‫ﺑﺄﻛﺛر‬ ‫ﻣن‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓﺄن‬ ‫ھو‬ ‫ﻧﻔﺳﮫ‬ 𝐶𝑂2 ∆𝐻𝑟 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 1 2 𝑂2 (𝑔) ⟶𝐶𝑂 𝑔 ( ) ∆𝐻𝑟 =− 110. 5 𝐾𝐽/𝑚𝑜𝑙 𝐶𝑂 𝑔 ( ) + 1 2 𝑂2 (𝑔) ⟶𝐶𝑂2 𝑔 ( ) ∆𝐻𝑟 =− 283 𝐾𝐽/𝑚𝑜𝑙 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 (𝑔) ⟶𝐶𝑂2 𝑔 ( ) ∆𝐻𝑟 =− 393. 5 𝐾𝐽/𝑚𝑜𝑙
  • 2.
    ‫ھﯾس‬ ‫ﻣﺳﺎﺋل‬ ‫ﺣل‬‫ﺣول‬ ‫ﻣﮭﻣﺔ‬ ‫ﻣﻼﺣظﺎت‬ : 1 ( ‫ﻋﻧد‬ ‫ﻗﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻧﻘﻠب‬ ‫إﺷﺎرة‬ ‫ﻗﯾﻣﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ . 2 ( ‫ﻋﻧد‬ ‫ﺿرب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫أو‬ ‫ﻗﺳﻣﺗﮭﺎ‬ ‫ﻋﻠﻰ‬ ‫ﻣﻌﺎﻣل‬ ‫ﻋددي‬ ‫ﻓﺗﺿرب‬ ‫أو‬ ‫ﺗﻘﺳم‬ ‫ﻗﯾﻣﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﺑﻧﻔس‬ .‫اﻟﻌدد‬ 3 ( ‫ھﻧﺎك‬ ‫ﻧوﻋﺎن‬ ‫ﻣن‬ :‫اﻟﻣﻌﺎدﻻت‬ :‫اﻷول‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟطﻠب‬ (‫)اﻟﮭدف‬ ‫وﺗﺳﻣﻰ‬ ‫ﺑﺎﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ :‫واﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ .‫اﻟﻔرﻋﯾﺔ‬ 4 ( ‫ﻧﻣر‬ ‫ﻋﻠﻰ‬ ‫ﻣواد‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﯾﺔ‬ ً‫ء‬‫اﺑﺗدا‬ ‫ﻣن‬ ‫أول‬ ،‫ﻣﺎدة‬ ‫وﻟﯾس‬ ً‫ﺎ‬‫ﺷرط‬ ‫أن‬ ‫ﻧﻣر‬ ‫ﻋﻠﻰ‬ ‫ﺟﻣﯾﻊ‬ ‫ﻣواد‬ ‫ھذه‬ .‫اﻟﻣﻌﺎدﻟﺔ‬ 5 ( ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟﻔرﻋﯾﺔ‬ ‫اﻟﺗﻲ‬ ‫ﺗم‬ ‫اﻟﺗﻌﺎﻣل‬ ‫ﻣﻌﮭﺎ‬ ‫ﻻ‬ ‫ﯾﺗم‬ ‫اﻟرﺟوع‬ ‫إﻟﯾﮭﺎ‬ .ً‫ﺛﺎﻧﯾﺔ‬ 6 ( ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻷوﻟﻰ‬ :‫اﻟذھﺑﯾﺔ‬ ‫إذا‬ ‫ﺗﻛررت‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫أﻛﺛر‬ ‫ﻣن‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﻓرﻋﯾﺔ‬ ‫ﻓﯾﺗم‬ ‫ﺗﺟﺎوزھﺎ‬ ‫ﻟﻠﻣﺎدة‬ ‫اﻟﺗﻲ‬ ‫ﺗﻠﯾﮭﺎ‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ .‫اﻟرﺋﯾﺳﯾﺔ‬ 7 ( ‫اﻟﻣﻼﺣظﺔ‬ :‫اﻟذھﺑﯾﺔ‬ ‫إذا‬ ‫ﻛﺎﻧت‬ ‫اﻟﻣﺎدة‬ ‫اﻟﻣﺗﻛررة‬ ‫ھﻲ‬ ‫آﺧر‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﯾﺔ‬ ‫ﻓﺳﻧﺿطر‬ ‫ﻟﻠﺗﻌﺎﻣل‬ ‫ﻣﻌﮭﺎ‬ ‫ﺑﺷرط‬ .‫اﻟﺗﻐﯾﯾر‬ ‫ﺑﻌد‬ ‫اﻟﻔرﻋﯾﺔ‬ ‫اﻟﻣﻌﺎدﻻت‬ ‫ﻓﻲ‬ ‫ﻣوﻻﺗﮭﺎ‬ ‫ﻋدد‬ ‫ﺣﺳﺎب‬ ‫ﻣﺛﺎل‬ 1 / ‫اﺣﺳب‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﻣرﻛب‬ ‫ﻣن‬ ‫ﻋﻧﺎﺻره‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﺑﺄﺛﺑت‬ :‫ﺻورھﺎ‬ 𝐶𝑆2 (𝐿) Δ = ? 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 2𝑆‫ﻣﻌﯿﻨﻲ‬ → 𝐶𝑆2 (𝐿) 𝐻𝑓 ° :‫اﻵﺗﯾﺔ‬ ‫اﻟﺣرارﯾﺔ‬ ‫اﻟﻣﻌﺎدﻻت‬ ‫أﻋطﯾت‬ ‫إذا‬ = - 394 KJ/mol 1 ( )𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 𝑔 ( ) →𝐶𝑂2 (𝑔) ∆𝐻𝑟 ° = - 296 KJ/mol (2) 𝑆‫ﻣﻌﯿﻨﻲ‬ + 𝑂2 𝑔 ( ) →𝑆𝑂2 (𝑔) ∆𝐻𝑟 °
  • 3.
    = - 1072KJ/mol 3 ( )𝐶𝑆2 (𝐿) + 3𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝑆𝑂2 (𝑔) ∆𝐻𝑟 ° /‫اﻟﺣل‬ ‫ﺗﺑﻘﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 1 ( ‫ﻛﻣﺎ‬ ،‫ھﻲ‬ ‫وﻧﺿرب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 2 ( × 2 ، ‫وﺗﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 3 ( : = - 394 (1) 𝐶‫ﻛﺮاﻓﯿﺖ‬ + 𝑂2 → 𝐶𝑂2 ∆𝐻𝑟 ° KJ/mol = - 592 (2)2𝑆 + 2𝑂2 → 2𝑆𝑂2 ∆𝐻𝑟 ° KJ/mol = + 1072 KJ/mol 3 ( )𝐶𝑂2 (𝑔) + 2𝑆𝑂2 (𝑔) → 𝐶𝑆2 (𝐿) + 3𝑂2 (𝑔) ∆𝐻𝑟 ° Δ = = 86 KJ/mol 𝐶 ‫ﻛﺮاﻓﯿﺖ‬ + 𝑆‫ﻣﻌﯿﻨﻲ‬ → 𝐶𝑆2 (𝐿) 𝐻𝑓 ° ∆𝐻𝑟 ° ‫اﻻول‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . ‫ﯾﺗم‬ ‫اﻟﻠﺟوء‬ ‫إﻟﻰ‬ ‫طرﯾﻘﺔ‬ ‫ھﯾس‬ ‫ﻟﺣﺳﺎب‬ Δ ‫ﱠ‬‫ﻷن‬ ‫ﺑﻌض‬ ‫اﻟﻣرﻛﺑﺎت‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫ﯾﻣﻛن‬ ‫ﺗﺻﻧﯾﻌﮭﺎ‬ 𝐻𝑟 ° ‫ﺻورة‬ ‫ﺑﺄﺛﺑت‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﻋﻧﺎﺻرھﺎ‬ ‫ﻣن‬ ً‫ة‬‫ﻣﺑﺎﺷر‬ ) T, F ( ‫اﻟﺟواب‬ / F 2 . :‫اﻟﻣﻌﺎدﻻت‬ ‫ﻣن‬ ‫ﻧوﻋﺎن‬ ‫ھﻧﺎك‬ ) A :‫اﻷول‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟطﻠب‬ (‫)اﻟﮭدف‬ ‫وﺗﺳﻣﻰ‬ ‫ﺑﺎﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ :‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ .‫اﻟﻔرﻋﯾﺔ‬ ) B :‫اﻷول‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟطﻠب‬ (‫)اﻟﮭدف‬ ‫وﺗﺳﻣﻰ‬ ‫ﺑﺎﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ :‫اﻟﺛﺎﻧﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ .‫اﻟرﺋﯾﺳﯾﺔ‬ ‫اﻟﺟواب‬ / A 3 . ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻷوﻟﻰ‬ :‫اﻟذھﺑﯾﺔ‬ ‫اذا‬ ‫وﺟدت‬ ‫ﻣﺎدة‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﯾﺔ‬ ‫وھﻲ‬ ‫ﻣﺗﻛررة‬ ‫ﻓﻲ‬ ‫اﻛﺛر‬ ‫ﻣن‬ ‫اﻟﺗﻲ‬ ‫اﻟﻣﺎدة‬ ‫اﻟﻰ‬ ‫اﻟﻣﺎدة‬ ‫ھذه‬ ‫ﺗﺟﺎوز‬ ‫ﻓﯾﺗم‬ ‫ﻓرﻋﯾﺔ‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫اﻟرﺋﯾﺳﺔ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻓﻲ‬ ‫ﺗﻠﯾﮭﺎ‬ ) T, F (
  • 4.
    ‫اﻟﺟواب‬ / T ( ‫اﻟدرس‬ ‫اﻟﺛﺎﻧﻲ‬ ) ‫ﺣﺳﺎﺑﺎت‬ ‫ﺑﺎﺳﺗﺧدام‬ ‫ﻗﺎﻧون‬ ‫ھﯾس‬ ‫س‬ 1 / ‫إذا‬ ‫أﻋطﯾت‬ ‫اﻟﻣﻌﺎدﻻت‬ ‫اﻟﺣرارﯾﺔ‬ ‫اﻵﺗﯾﺔ‬ ‫ﻋﻧد‬ ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) STP :( Δ= 25 KJ 1 ( )𝐹𝑒𝑂(𝑠) + 𝐻2 (𝑔) → 𝐹𝑒(𝑠) + 𝐻2 𝑂(𝑔) 𝐻𝑟 ° Δ = 318 KJ 2 ( )3𝐹𝑒𝑂(𝑠) + 1 2 𝑂2 (𝑔) → 𝐹𝑒3 𝑂4 (𝑠) 𝐻𝑟 ° Δ = - 242 KJ 3 ( )𝐻2 (𝑔) + 1 2 𝑂2 (𝑔) → 𝐻2 𝑂(𝑔) 𝐻𝑟 ° ‫اﺣﺳب‬ Δ ‫ﻟﻠﺗﻔﺎﻋل‬ :‫اﻟﺗﺎﻟﻲ‬ 𝐻𝑟 ° Δ =? 3𝐹𝑒(𝑠) + 4𝐻2 (𝑔) → 𝐹𝑒3 𝑂4 (𝑠) + 4𝐻2 (𝑔) 𝐻𝑟 °
  • 5.
    /‫اﻟﺣل‬ ‫ﻧﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 1 ( ‫وﻧﺿرﺑﮭﺎ‬ × 3 ‫وﺗﺑﻘﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 2 ( ‫ﻛﻣﺎ‬ ‫ھﻲ‬ ‫وﻧﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 3 :( = -75KJ 1 ( )3𝐹𝑒 𝑠 ( ) + 3𝐻2 𝑂 ⟶3𝐹𝑒𝑂 𝑠 ( ) + 3𝐻2 ∆𝐻𝑟 ° 2 ( )𝐹𝑒𝑂 𝑆 ( ) + 1 2 𝑂2 ⟶𝐹𝑒3 𝑂4 𝑆 ( ) ∆𝐻𝑟 ° = − 318𝐾𝐽 3 ( )𝐻2 𝑂 𝑔 ( ) ⟶𝐻2 𝑔 ( ) + 1 2 𝑂2(𝑔) ∆𝐻𝑟 ° = + 242𝐾𝐽 Δ = - 151 KJ2 3𝐹𝑒(𝑠) + 4𝐻2 𝑂(𝑔) → 𝐹𝑒3 𝑂4 + 4𝐻 𝐻𝑟 ° /‫ﻣﻼﺣظﺔ‬ ‫ھﻧﺎ‬ ‫ﻧطﺑق‬ ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻟذھﺑﯾﺔ‬ ‫اﻷوﻟﻰ‬ ‫ﺑﻌدم‬ ‫ﻣوازﻧﺔ‬ ‫ﮫ‬‫ﱠ‬‫ﻧ‬‫ﻷ‬ ‫ﻣﺗﻛرر‬ ‫أﻛﺛر‬ ‫ﻣن‬ ‫ﻣرة‬ ‫ﻓﻲ‬ ‫اﻟﻣﻌﺎدﻻت‬ ،‫اﻟﻔرﻋﯾﺔ‬ 𝐻2 𝑂 ‫وﻛذﻟك‬ ‫ﻧطﺑق‬ ‫اﻟﻣﻼﺣظﺔ‬ ‫اﻟذھﺑﯾﺔ‬ ‫اﻟﺛﺎﻧﯾﺔ‬ ‫ﺑﺿرورة‬ ‫ﻣوازﻧﺔ‬ ‫رﻏم‬ ‫ﺗﻛراره‬ ‫وﻟﻛن‬ ‫ﺑﻌد‬ ‫ﺣﺳﺎب‬ ‫ﻣوﻻﺗﮫ‬ ‫اﻟﺳﺎﺑﻘﺔ‬ ‫ﺑﻌد‬ 𝐻2 ‫اﻟﺗﻐﯾﯾر‬ . ‫س‬ 2 / ‫إذا‬ ‫ﻋﻠﻣت‬ ‫ﱠ‬‫أن‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﺣﺗراق‬ ّ‫ل‬‫ﻛ‬ ‫ﻣن‬ ‫ﻏﺎز‬ CO ‫و‬ ‫و‬ ‫ﺑوﺣدات‬ KJ/mol ‫ھﻲ‬ ‫ﻋﻠﻰ‬ :‫اﻟﺗواﻟﻲ‬ -) 𝐻2 𝐶𝐻3 𝑂𝐻 727 -/ 286 -/ 284 ،( ‫اﺣﺳب‬ Δ ‫ﺑﺎﺳﺗﺧدام‬ ‫ﻗﺎﻧون‬ ‫ھﯾس‬ ‫ﻟﻠﺗﻔﺎﻋل‬ :‫اﻵﺗﻲ‬ 𝐻𝑟 ° 𝐶𝑂 + 2𝐻2 → 𝐶𝐻3 𝑂𝐻 .‫ﺑﺎﻟﺳؤال‬ ‫ﺗﻌط‬ ‫ﻟم‬ ‫إذا‬ ‫اﻟﻔرﻋﯾﺔ‬ ‫اﻻﺣﺗراق‬ ‫ﻣﻌﺎدﻻت‬ ‫ﻧﻛﺗب‬ ‫ﻣن‬ ‫ﻧﺣن‬ ‫ﺑطرﯾﻘﺔ‬ ‫ﻧﺣﻠﮫ‬ ‫أن‬ ‫ﯾﻣﻛن‬ ‫ﱠ‬‫ﻷن‬ ،‫ھﯾس‬ ‫ﺑطرﯾﻘﺔ‬ ‫اﻟﺳؤال‬ ‫ﺣددﻧﺎ‬ .(‫)اﻟﺳﻣﯾﺷن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻧﺛﺎﻟﺑﻲ‬ = - 284 KJ/mol (1) 𝐶𝑂 + 1 2 𝑂2 → 𝐶𝑂2 ∆𝐻𝑟 ° = - 286 KJ/mol 2 ( )𝐻2 + 1 2 𝑂2 → 𝐻2 𝑂 ∆𝐻𝑟 ° = - 727 KJ/mol 3 ( )𝐶𝐻3 𝑂𝐻 + 3 2 𝑂2 → 𝐶𝑂2 + 2𝐻2 𝑂 ∆𝐻𝑟 ° ‫وﻟﻠﺣﺻول‬ ‫ﻋﻠﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ،‫اﻟرﺋﯾﺳﯾﺔ‬ ‫ﺗﺑﻘﻰ‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ) 1 ( ‫ﻛﻣﺎ‬ ،‫ھﻲ‬ ‫وﻧﺿرب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟﻔرﻋﯾﺔ‬ ) 2 ( 2 × ، ‫وﻧﻘﻠب‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫اﻟﻔرﻋﯾﺔ‬ ) 3 :( = - 284 KJ/mol (1) 𝐶𝑂 + 1 2 𝑂2 → 𝐶𝑂2 ∆𝐻𝑟 ° = - 572 KJ/mol 2 ( )2𝐻2 + 𝑂2 → 2𝐻2 𝑂 ∆𝐻𝑟 °
  • 6.
    = + 727KJ/mol 3 ( )𝐶𝑂2 + 2𝐻2 𝑂 → 𝐶𝐻3 𝑂𝐻 + 3 2 𝑂2 ∆𝐻𝑟 ° Δ = - 129 KJ/mol 𝐶𝑂 + 2𝐻2 → 𝐶𝐻3 𝑂𝐻 𝐻𝑟 ° /‫ﻣﻼﺣظﺔ‬ ‫ﯾﻣﻛن‬ ‫ﺣل‬ ‫اﻟﺳؤال‬ ‫ﺑطرﯾﻘﺔ‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) Σ ( ‫ﺑﻌد‬ ‫أن‬ ‫ﻧﺣول‬ ‫ﻗﯾﻣﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﻟﻠﻣﯾﺛﺎﻧول‬ ‫ﻣن‬ ‫اﻹﺷﺎرة‬ ‫ﺑﻘﻠب‬ ‫وذﻟك‬ (‫)اﻟﻣﺎص‬ ‫اﻟﺗﻛوﯾن‬ ‫إﻟﻰ‬ (‫)اﻟﺑﺎﻋث‬ ‫اﻻﺣﺗراق‬ ‫ﻟﻠﻣوﺟب‬ ‫اﻟﺳﺎﻟب‬ ‫ﻣن‬ . ‫س‬ 3 / ‫اﺣﺳب‬ ‫اﻧﺛﺎﻟﺑﻲ‬ ‫اﻟﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ( )𝐻𝑃𝑂3 = - 180KJ 𝑃4 𝑂10 + 4𝐻𝑁𝑂3 →4𝐻𝑃𝑂3 + 2𝑁2 𝑂5 ∆𝐻𝑟 ° ‫ﻋﻠﻣت‬ ‫اذا‬ Δ = -) 2984 ( KJ/mol 𝐻𝑓 𝑃4 𝑂10 ° Δ = -) 174 ( KJ/mol 𝐻𝑓 𝐻𝑁𝑂3 °
  • 7.
    Δ =(-43)KJ/mol ?? 𝐻𝑓𝑁2 𝑂5 ° Δ𝐻𝑟 ° = ∑ 𝑛∆ 𝐻𝑓(𝑃) ° − ∑ 𝑛 ∆𝐻𝑓 𝑅 ( ) ° Δ𝐻𝑟 ° = [4Δ𝐻𝑓(𝐻𝑃𝑂3 ) ° + 2Δ𝐻𝑓(𝑁2 𝑂5 ) ° ] − [Δ𝐻𝑓(𝑃4 𝑂10 ) ° + 4 Δ𝐻𝑓(𝐻𝑁𝑂3 ) ° ] − 180 = 4Δ𝐻𝑓 𝐻𝑃𝑂3 ( ) ° + 2 − 43 ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − − 2984 ( ) + 4 − 174 ( ) [ ] − 180 = 4Δ𝐻𝑓 𝐻𝑃𝑂3 ( ) ° − 86 ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − (− 3680)] - 4∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° = 3594 + 180⟹4∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° =− 3594 − 180 ∴∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° = −3774 4 ⟹∆𝐻𝑓 𝐻𝑃𝑂3 ( ) ° =− 943. 5 𝐾𝐽/𝑚𝑜𝑙 (‫اﻟﺛﺎﻟث‬ ‫)اﻟدرس‬ ‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬
  • 8.
    : ‫اﻹﻧﺗروﺑﻲ‬ ‫»ھﻲ‬ ‫داﻟﺔ‬ ‫ﺣﺎﻟﺔ‬ ‫ﺛرﻣوداﯾﻧﻣﯾﻛﯾﺔ‬ ‫ﺗﻌﺗﺑر‬ ‫ﻣﻘﯾﺎس‬ ‫ﻟدرﺟﺔ‬ ،‫اﻟﻌﺷواﺋﯾﺔ‬ ‫أو‬ ‫ﻻ‬ ‫اﻧﺗظﺎم‬ ،‫اﻟﻧظﺎم‬ ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) S ،( ‫وﺗﻘﺎس‬ ‫ﺑوﺣدة‬ ) J/K mol .«( ‫اﻻﻧﺗروﺑﻲ؟‬‫اﻟﺣﺎﻟﺔ‬ ‫داﻟﺔ‬ ‫ﻓﺎﺋدة‬ ‫ﻣﺎ‬ /‫ج‬ ‫ﯾﺳﺗﻔﺎد‬ ‫ﻣﻧﮭﺎ‬ ‫ﻟﻠﺗﻧﺑؤ‬ ‫ﺑﺗﻠﻘﺎﺋﯾﺔ‬ ،‫اﻟﺗﻔﺎﻋل‬ ‫ﻓﻛﻠﻣﺎ‬ ‫ﻛﺎن‬ ‫اﻟﻼﻧظﺎم‬ ً‫ﻼ‬‫ﻗﻠﯾ‬ ‫)اﻟﻌﺷواﺋﯾﺔ‬ (‫أﻛﺑر‬ ‫ﻛﺎﻧت‬ ‫ﻗﯾﻣﺔ‬ ‫أﻛﺑر‬ ‫اﻻﻧﺗروﺑﻲ‬ ) ΔS = (+ ‫واﻟﻌﻛس‬ ‫ﺻﺣﯾﺢ‬ :‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫اﻻﻧﺗروﺑﻲ‬ ‫ﺣﺳﺎب‬ ∆𝑆𝑟 о = ∑𝑛 𝑆(𝑃) о − ∑𝑛 𝑆(𝑅) о ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ) ΔG :( ‫وﯾﻣﻛن‬ ‫ﺣﺳﺎب‬ ‫اﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻔﺎﻋل‬ ‫ﺑﺎﻻﻋﺗﻣﺎد‬ ‫ﻋﻠﻰ‬ ‫اﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻟﻠﺗﻛوﯾن‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻣن‬ ‫ﺧﻼل‬ ‫اﺳﺗﺧدام‬ ∆𝐺𝑓 ° (‫)اﻟﺳﻣﯾﺷن‬ ‫اﻟﻣﺟﻣوع‬ ‫ﻗﺎﻧون‬ Δ =𝐺𝑟 ° ∑ 𝑛 ∆𝐺𝑓(𝑃) ° − ∑ 𝑛 ∆𝐺𝑓(𝑅) ° :‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬ ‫ﻟﻧﺎ‬ ‫ﺗﺗﯾﺢ‬ ‫ﺛرﻣوداﯾﻧﻣﯾﻛﯾﺔ‬ ‫ﺣﺎﻟﺔ‬ ‫داﻟﺔ‬ ‫»ھﻲ‬ ‫اﻟﺗﻲ‬ ‫اﻟﻌظﻣﻰ‬ ‫اﻟطﺎﻗﺔ‬ ‫ﺗﻣﺛل‬ ‫وھﻲ‬ ،‫اﻟﺗﻔﺎﻋل‬ ‫ﺑﺗﻠﻘﺎﺋﯾﺔ‬ ‫اﻟﺗﻧﺑؤ‬ ‫ﯾﻣﻛن‬ ،‫واﻻﻧﺗروﺑﻲ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﻓﻲ‬ ‫اﻟﺗﻐﯾر‬ ‫ﻗﯾﺎس‬ ‫ﻣن‬ ‫ﻋﻠﯾﮭﺎ‬ ‫اﻟﺣﺻول‬ ‫اﻟﺗﺎﻟﯾﺔ‬ ‫اﻟﻌﻼﻗﺔ‬ ‫ﺑﺣﺳب‬ : ∆𝐺 о = ∆𝐻 о − 𝑇∆𝑆 о ΔG = ‫ﺳﺎﻟﺑﺔ‬ ‫)اﻟﺗﻔﺎﻋل‬ ‫أو‬ ‫اﻟﺗﻐﯾر‬ ‫اﻟﻔﯾزﯾﺎﺋﻲ‬ .(‫ﺗﻠﻘﺎﺋﻲ‬ ΔG = ‫ﻣوﺟﺑﺔ‬ ‫)اﻟﺗﻔﺎﻋل‬ ‫أو‬ ‫اﻟﺗﻐﯾر‬ ‫اﻟﻔﯾزﯾﺎﺋﻲ‬ ‫ﻏﯾر‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ) ΔG = ‫ﺻﻔر‬ ‫)اﻟﺗﻔﺎﻋل‬ ‫أو‬ ‫اﻟﺗﻐﯾر‬ ‫اﻟﻔﯾزﯾﺎﺋﻲ‬ ‫ﻓﻲ‬ ‫ﺣﺎﻟﺔ‬ .(‫اﺗزان‬ ‫ﻣﻼﺣظﺎت‬ ‫ﻣﮭﻣﺔ‬ : 1 . Δ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورھﺎ‬ = .‫ﺻﻔر‬ 𝐻𝑓 ° ‫ﻣﺛﺎل‬ / / / / :( ... ‫اﻟﺦ‬ .( 𝐴𝑙(𝑠) 𝐶𝑙2 (𝑔) 𝑁2 (𝑔) 𝑂2 (𝑔) 𝐻2 (𝑔)
  • 9.
    2 . ‫أﻓﺿل‬ ‫طرﯾﻘﺔ‬ ‫ﻻﺳﺗﺧدام‬ ‫اﻟﺳﻣﯾﺷن‬ ‫ھو‬ ‫إﻧﮭﺎء‬ ‫ﻧﺎﺗﺞ‬ ّ‫ل‬‫ﻛ‬ ‫طرف‬ ‫ﺑدﻗﺔ‬ ‫ﺛم‬ ‫ﻧﻘوم‬ ‫ﺑﻌﻣﻠﯾﺔ‬ ‫طرح‬ ‫اﻟطرف‬ ‫اﻷول‬ ‫ﻣن‬ .‫اﻟﺛﺎﻧﻲ‬ 3 . ‫رﺑﻣﺎ‬ ‫ﺗﻛون‬ Δ ‫ﻣﻌﻠوﻣﺔ‬ ‫و‬ Δ ‫ﻟﻣﻌظم‬ ‫ﻣواد‬ ‫اﻟﻣﻌﺎدﻟﺔ‬ ‫ﻣﻌﻠوﻣﺔ‬ ‫وﯾﻛون‬ ‫اﻟﻣطﻠوب‬ Δ ‫ﻟﻣﺎدة‬ ‫ﻣﺟﮭوﻟﺔ‬ .‫واﺣدة‬ 𝐺𝑟 ° 𝐺𝑓 ° 𝐺𝑓 ° 4 . ‫إذا‬ ‫ﻛﺎن‬ ‫اﻟطرف‬ ‫اﻟﺛﺎﻧﻲ‬ ‫ﻧﺎﺗﺟﮫ‬ ‫ﺳﺎﻟب‬ ‫ﻓﺗﺻﺑﺢ‬ ‫اﻟﻌﻣﻠﯾﺔ‬ :‫اﻟﺣﺳﺎﺑﯾﺔ‬ ‫اﻟﺛﺎﻧﻲ‬ ‫اﻟطرف‬+ ‫اﻷول‬ ‫اﻟطرف‬ ‫س‬ 1 / ‫ﻋرف‬ ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ‫ﻟﻠﺗﻛوﯾن‬ ،‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫وﻣﺎ‬ ‫ھﻲ‬ ‫وﺣداﺗﮭﺎ؟‬ »/ ‫اﻟﺟواب‬ ‫ﺗﻛوﯾن‬ ‫ﻋﻧد‬ ‫اﻟﺣرة‬ ‫اﻟطﺎﻗﺔ‬ ‫ﻓﻲ‬ ‫اﻟﺗﻐﯾﯾر‬ ‫ﻣﻘدار‬ ‫ھﻲ‬ ‫ﺑﺄﺛﺑت‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ‫ﻋﻧﺎﺻرھﺎ‬ ‫ﻣن‬ ‫ﻣرﻛب‬ ‫أي‬ ‫ﻣن‬ ‫واﺣد‬ ‫ﻣول‬ ،‫ﺻورة‬ ‫وﻋﻧد‬ ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) 1 atm/25 ( ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) ( ‫ووﺣداﺗﮭﺎ‬ ) KJ/mol)» s ℃ 𝐷𝐺𝑓 ° ‫ﻣﺛﺎل‬ 2 / ‫اﺣﺳب‬ ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺗﺎﻟﻲ‬ ‫ﻋﻧد‬ ‫اﻟظروف‬ ،‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﺛم‬ ‫ن‬ّ‫ﯾ‬‫ﺑ‬ ‫ھل‬ ‫ﯾﺟري‬ ‫اﻟﺗﻔﺎﻋل‬ ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯾ‬ ‫أم‬ :‫اﻟظروف‬ ‫ھذه‬ ‫ﻋﻧد‬ ،‫ﻻ‬ 2𝐶6 𝐻6 (𝐿) + 15𝑂2 (𝑔) → 12𝐶𝑂2 (𝑔) + 6𝐻2 𝑂(𝐿) :‫ﻋﻠﻣت‬ ‫إذا‬ = (- 394) KJ/mol, ∆𝐺𝑓𝐶𝑂2 ° = (- 237) KJ/mol ∆𝐺𝑓𝐻2 𝑂 ° = (173) KJ/mol ∆𝐺𝑓𝐶6 𝐻6 ° /‫اﻟﺣل‬ Δ = 𝐺𝑟 ° ∑ 𝑛 ∆𝐺𝑓(𝑃) ° − ∑ 𝑛 ∆𝐺𝑓(𝑅) ° Δ = ] 12 + 6 [ - ] + 15 [ 𝐺𝑟 ° ∆𝐺𝑓𝐶𝑂2 ° ∆𝐺𝑓𝐻2 𝑂 ° 2∆𝐺𝑓𝐶6 𝐻6 ° ∆𝐺𝑓𝑂2 ° = Δ = [(- 4728) + (- 1422)] - (346) 𝐺𝑟 ° Δ = - 6496 KJ/mol = ‫)ﺳﺎﻟﺒﺔ‬ , ( ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﱠ‬‫ﻷن‬ ∴ 𝐺𝑟 ° ∆𝐺°𝑟
  • 10.
    ‫ﺗﻣرﯾن‬ 1 / ‫اﺣﺳب‬ ‫طﺎﻗﺔ‬ ‫ﻛﺑس‬ ‫اﻟﺣرة‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺗﺎﻟﻲ‬ ‫ﻋﻧد‬ ‫اﻟظروف‬ ،‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ن‬ّ‫ﯾ‬‫وﺑ‬ ‫ھل‬ ‫ﯾﺟري‬ ‫اﻟﺗﻔﺎﻋل‬ ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯾ‬ ‫أم‬ ‫ﻻ؟‬ 2𝑁𝑂(𝑔) + 𝑂2 (𝑔) →2𝑁𝑂2 (𝑔) :‫ﻋﻠﻣت‬ ‫إذا‬ ∆𝐺𝑓𝑁𝑂 ° =87 𝐾𝐽 𝑚𝑜𝑙 ∆𝐺𝑓𝑁𝑂2 ° = 52 𝐾𝐽 𝑚𝑜𝑙 /‫اﻟﺣل‬ Δ = ] 2 ) [( - ] 2 ) ( + [ 𝐺𝑟 ° ∆𝐺𝑓𝑁𝑂2 ° ∆𝐺𝑓𝑁𝑂 ° ∆𝐺𝑓𝑂2 ° = ] 2 ) 52 [( - ] 2 ) 87 ( + 0 [ Δ = (104) - (174) ⇒ Δ = - 70 KJ/mol ‫ﺗﻠﻘﺎﺋﻲ‬ ∴ 𝐺𝑟 ° 𝐺𝑟 ° ‫ﻣﻼﺣظﺔ‬ / ‫ﻟﻠﻌﻨﺎﺻﺮ‬ ‫اﻟﺤﺮة‬ ‫ﺑﺎﺛﺒﺖ‬ ‫ﺻﻮرة‬ = ‫ﺻﻔﺮ‬ ∆𝐺 °
  • 11.
    ‫اﻟﺛﺎﻟث‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . Δ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورھﺎ‬ = .‫ﺻﻔر‬𝐺𝑓 ° ) T, F ( ‫اﻟﺟواب‬ / T 2 . ‫أﻓﺿل‬ ‫طرﯾﻘﺔ‬ ‫ﻻﺳﺗﺧدام‬ ‫اﻟﺳﻣﯾﺷن‬ ‫ھو‬ ‫إﻧﮭﺎء‬ ‫ﻧﺎﺗﺞ‬ ّ‫ل‬‫ﻛ‬ ‫طرف‬ ‫ﺑدﻗﺔ‬ ‫ﺛم‬ ‫ﻧﻘوم‬ ‫ﺑﻌﻣﻠﯾﺔ‬ ‫ﺟﻣﻊ‬ .‫اﻟﺛﺎﻧﻲ‬ ‫ﻣن‬ ‫اﻷول‬ ‫اﻟطرف‬ ) T, F ( ‫اﻟﺟواب‬ / F 3 . ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫ﻟﻠﺗﻛوﯾن‬ ‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬ : ) A ‫ھﻲ‬ ‫ﻣﻘدار‬ ‫اﻟﺗﻐﯾﯾر‬ ‫ﻓﻲ‬ ‫اﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻋﻧد‬ ‫ﺗﻛوﯾن‬ ‫ﻣول‬ ‫واﺣد‬ ‫ﻣن‬ ‫أي‬ ‫ﻣرﻛب‬ ‫ﻣن‬ ‫ﻋﻧﺎﺻرھﺎ‬ ،‫ﺻورة‬ ‫ﺑﺄﺛﺑت‬ ‫اﻷﺳﺎﺳﯾﺔ‬ ) B ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) 1 atm/50 ( ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) ( ‫ووﺣداﺗﮭﺎ‬ ) KJ/mol .( ∆𝐺𝑓 ° ) C ‫اﻟظروف‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ) 1 atm/25 ( ‫وﯾرﻣز‬ ‫ﻟﮭﺎ‬ ) ( ‫ووﺣداﺗﮭﺎ‬ ) KJ/mol .( ∆𝐺𝑓 ° ‫اﻟﺟواب‬ / A & C
  • 12.
    ( ‫اﻟراﺑﻊ‬ ‫)اﻟدرس‬ ‫اﻟﺗﻔﺎﻋﻼت‬‫ﺳﯾر‬ ‫وﺗﺟﺎه‬ ‫ﻛﺑس‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗطﺑﯾﻘﺎت‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫اﻟﻛﯾﻣﯾﺎﺋﯾﺔ‬ ‫اﻟﺗﻔﺎﻋﻼت‬ ‫ﺳﯾر‬ ‫واﺗﺟﺎه‬ ‫ﻛﺑس‬ ‫ﻣﻌﺎدﻟﺔ‬ ‫ﺗطﺑﯾﻘﺎت‬ ∆𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ‫ﺳؤال‬ / ‫اﻟﺗﻔﺎﻋل؟‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫ﯾﺿﻣﻧﺎن‬ ‫ﻋﺎﻣﻠﯾن‬ ‫أﻛﺛر‬ ‫ھﻣﺎ‬ ‫ﻣﺎ‬ /‫ج‬ ‫ﱠ‬‫إن‬ ‫اﻟﺘﻐﯿﺮ‬ ‫ﻓﻲ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ ) Δ ( ‫ﯾﺘﻀﻤﻦ‬ ‫ﻋﺎﻣﻠﯿﻦ‬ ‫ﻣﮭﻤﯿﻦ‬ ‫ﯾﺆﺛﺮان‬ ‫ﻋﻠﻰ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ،‫اﻟﺘﻔﺎﻋﻞ‬ :‫وھﻤﺎ‬ 𝐺𝑟 ° 1 . 𝐷𝐻𝑟 о 2 . 𝐷𝑆𝑟 ° ‫واذا‬ ‫ﻛﺎﻧﺖ‬ ‫اﺷﺎرة‬ ‫و‬ ‫ﻣﺘﺸﺎﺑﮭﺔ‬ ‫ﺗﻜﻮن‬ ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫ھﻲ‬ ‫اﻟﻤﺘﺤﻜﻢ‬ ‫اﻟﺮﺋﯿﺴﻲ‬ ‫ﻓﻲ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ 𝐷𝐻𝑟 о 𝐷𝑆𝑟 о /‫ﻋﻠل‬ ‫ﺗﻠﻘﺎﺋﯾﺔ؟‬ ‫أﻛﺛر‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻛﺎن‬ ‫ﻋﺎﻟﯾﺔ‬ ‫اﻻﻧﺗروﺑﻲ‬ ‫ﻛﺎﻧت‬ ‫ﻛﻠﻣﺎ‬ /‫ج‬ (∆𝐺°𝑟 = ∆𝐻°𝑟 − 𝑇∆𝑆°𝑟 ) ‫ﱠ‬‫ﻷن‬ ‫اﻻﻧﺘﺮوﺑﻲ‬ ‫ﺗﻘﻊ‬ ‫ﺿﻤﻦ‬ ‫اﻟﺤﺪ‬ ) ‫اﻟﺴﺎﻟﺐ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫ﻛﻠﻤﺎ‬ ‫ﺗﻜﻮن‬ ‫ﻗﯿﻤﺔ‬ ‫ھﺬا‬ ‫اﻟﺤﺪ‬ ‫أﻋﻠﻰ‬ ‫ﻣﻦ‬ ) > 𝑇∆𝑆°𝑟 ) ∆𝐻°𝑟 ∆𝐻°𝑟 ( ‫ﻛﺎﻧﺖ‬ ‫ﻗﯿﻤﺔ‬ = ،‫ﺳﺎﻟﺒﺔ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ . 𝑇∆𝑆°𝑟 𝐷𝐺°𝑟 ‫ﻣﻼﺣظﺔ‬ / ‫ﺣﯿﻨﻤﺎ‬ ‫ﺗﻜﻮن‬ ‫إﺷﺎرات‬ ) ΔH, ΔS ( ‫ﻣﺨﺘﻠﻔﺔ‬ ‫ﻓﻠﯿﺲ‬ ‫ھﻨﺎك‬ ‫ﺗﺄﺛﯿﺮ‬ ‫ﻟﺪرﺟﺔ‬ ،‫اﻟﺤﺮارة‬ ‫وﻻﺣﻆ‬ ‫ﱠ‬‫أن‬ ‫إﺷﺎرة‬ ) ΔH ( ‫ﺗﺸﺎﺑﮫ‬ ‫إﺷﺎرة‬ ) ΔG ( ‫ﻓﻲ‬ .‫اﻟﻐﺎﻟﺐ‬
  • 13.
    :‫ﻣﮭﻣﺔ‬ ‫ﻣﻼﺣظﺎت‬ ΔG .1 = ‫ﺳﺎﻟﺒﺔ‬ (-) ‫اﻟﺘﻔﺎﻋﻞ‬ ،‫ﺗﻠﻘﺎﺋﻲ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ (+) ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ 2 . :‫ﻋﻤﻠﯿﺎت‬ ‫)اﻻﻧﺼﮭﺎر‬ / ‫اﻟﺘﺒﺨﺮ‬ (‫)اﻟﻐﻠﯿﺎن‬ / ‫اﻟﺘﺴﺎﻣﻲ‬ (... ΔS = (+) ‫ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ΔH = (+) ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ 3 . ‫ﻋﻤﻠﯿﺎت‬ ‫)اﻻﻧﺠﻤﺎد‬ ‫)اﻟﺘﺒﻠﻮر‬ / ‫اﻟﺘﻜﺜﯿﻒ‬ (‫)اﻟﻨﺪى‬ / ‫اﻟﺘﺼﻠﺐ‬ (... ΔS = (-) ‫ﺗﻘﻞ‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ΔH = (-) ‫ﺑﺎﻋﺚ‬ ‫ﻟﻠﺤﺮارة‬ 4 . ‫ﺣﯿﻨﻤﺎ‬ :‫ﯾﻘﻮل‬ ‫)ﯾﺴﺒﺐ‬ ‫اﻧﺨﻔﺎض‬ ‫درﺟﺔ‬ ‫ﺣﺮارة‬ :(‫اﻟﻤﺤﻠﻮل‬ ‫ﺑﺎﻋﺜﺔ‬‫واﻟﻌﻜﺲ‬ ‫ﻣﺎﺻﺔ‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ) ΔH > TΔS ( 5 . ‫ذوﺑﺎن‬ ‫ﺻﻠﺐ‬ ‫أو‬ ،‫ﺳﺎﺋﻞ‬ ‫ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ +) = ΔS ( ‫وﻋﻠﻰ‬ ‫اﻷﻏﻠﺐ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ +) = ΔH ( 6 . ‫ذوﺑﺎن‬ ،‫ﻏﺎز‬ ‫ﺗﻘﻞ‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ -) = ΔS ،( ‫وھﻲ‬ :‫ﻋﻤﻠﯿﺎت‬ ‫ﺑﺎﻋﺜﺔ‬ -) = ΔH ( 7 . ‫ﻋﻤﻠﯿﺎت‬ ‫)اﻟﺘﺤﻠﻞ‬ / ‫اﻟﺘﻔﻜﻚ‬ / :(‫اﻟﺘﺄﯾﻦ‬ ‫)ﺗﺰداد‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ +) = ΔS ( ‫واﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ +) = ΔH ( 8 . :‫ﻋﻤﻠﯿﺎت‬ ‫)اﻻﺗﺤﺎد‬ / ‫اﻻﻧﺪﻣﺎج‬ :(... ‫ﺗﻘﻞ‬ ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ -) = ΔH ( 9 . ‫أھم‬ :‫ﻣﻼﺣظﺔ‬ ‫)إن‬ ‫ﻛﺎﻧﺖ‬ ‫اﻟﻌﺒﺎرات‬ ‫ﻣﻨﻔﯿﺔ‬ ‫ﻓﺎﻟﺘﻔﺎﻋﻞ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ +) = ΔG ،( ‫وﻧﺮﻓﻊ‬ ‫اﻟﻨﻔﻲ‬ ً‫ﺎ‬‫ﻣﺆﻗﺘ‬ ‫ﺛﻢ‬ ‫ﻧﺤﻞ‬ ‫اﻟﻤﺴﺄﻟﺔ‬ ‫ﻋﻠﻰ‬ .(‫اﻟﻘﺎدﻣﺔ‬ ‫اﻟﻤﺴﺎﺋﻞ‬ ‫ﻣﻦ‬ ‫ذﻟﻚ‬ ‫ﺳﯿﺘﻮﺿﺢ‬ ‫ﻛﻤﺎ‬ ،‫ﻛﺒﺲ‬ ‫ﻋﻼﻗﺔ‬ ‫ﺿﻮء‬
  • 14.
    ‫اﻟراﺑﻊ‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . ‫اﻟﺗﻔﺎﻋل‬‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫ﯾﺿﻣﻧﺎن‬ ‫ﻋﺎﻣﻠﯾن‬ ‫أﻛﺛر‬ ‫ھﻣﺎ‬ ‫ﻣﺎ‬ ) A Δ𝐻°𝑟 & Δ𝐶°𝑟 ) B ∆𝐻°𝑟 & Δ𝑆°𝑟 ) C ∆𝐺°𝑟 & Δ𝑆°𝑟 (‫)ﺧﯾﺎرات‬ ‫اﻟﺳؤال‬ ‫ﻧوع‬ ‫اﻟﺟواب‬ / B 2 . ‫اﻟﺗﺻﻠب‬ / (‫)اﻟﻧدى‬ ‫اﻟﺗﻛﺛﯾف‬ / ‫)اﻟﺗﺑﻠور‬ ‫)اﻻﻧﺟﻣﺎد‬ ‫ﻋﻣﻠﯾﺎت‬ (... ΔS = (+) ‫ﺗزداد‬ ‫اﻟﻌﺷواﺋﯾﺔ‬ ΔH = (-) ‫ﺑﺎﻋث‬ ‫ﻟﻠﺣرارة‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 3 . ،‫ﻏﺎز‬ ‫ذوﺑﺎن‬ ‫ﺗﻘل‬ ‫اﻟﻌﺷواﺋﯾﺔ‬ -) = ΔS ،( ‫وھﻲ‬ :‫ﻋﻣﻠﯾﺎت‬ ‫ﺑﺎﻋﺛﺔ‬ -) = ΔH ( ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / T (‫اﻟﺧﺎﻣس‬ ‫)اﻟدرس‬ ‫ﻛﺑس‬ ‫ﻋﻼﻗﺔ‬ ‫وﻓق‬ ‫ﺗﻌﺎﻟﯾل‬ ‫ﻋﻠل‬ ‫ﻣﺎ‬ ‫ﯾﺄﺗﻲ‬ ‫ﻋﻠﻰ‬ ‫وﻓق‬ ‫ﻋﻼﻗﺔ‬ :‫ﻛﺑس‬ ‫س‬/ 3
  • 15.
    1 . ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬‫اﻟﺟﻠﯾد‬ ‫اﻧﺻﮭﺎر‬ ‫ﻋﻣﻠﯾﺔ‬ /‫ج‬ - ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﺳﺎﻟﺒﺔ‬ ‫)اﻟﺘﻔﺎﻋﻞ‬ (‫ﺗﻠﻘﺎﺋﻲ‬ - ‫اﻧﺼﮭﺎر‬ ΔS = ‫ﻣﻮﺟﺒﺔ‬ ‫)ﺗﺰداد‬ (‫اﻟﻌﺸﻮاﺋﯿﺔ‬ - ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻣﺎﺻﺔ‬ ΔH = ‫ﻣﻮﺟﺒﺔ‬ ΔG = ΔH - TΔS + (-) + = - ΔH < TΔS 2 . ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫اﻷوﻟﯾﺔ‬ ‫ﻋﻧﺎﺻره‬ ‫إﻟﻰ‬ ‫اﻟﻣﺎء‬ ‫ﯾﺗﺣﻠل‬ ‫ﻻ‬ /‫ج‬ - ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ - ‫ﯾﺘﺤﻠﻞ‬ ΔS = ‫ﻣﻮﺟﺒﺔ‬ ‫)ﺗﺰداد‬ (‫اﻟﻌﺸﻮاﺋﯿﺔ‬ - ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﻣﺎﺻﺔ‬ ‫ﻟﻠﺤﺮارة‬ ΔH = ‫ﻣﻮﺟﺒﺔ‬ ΔG = ΔH - TΔS + (-) + = + ΔH > TΔS 3 . ‫ﯾذوب‬ ‫ﻏﺎز‬ ‫ﻓﻲ‬ ‫اﻟﻣﺎء‬ ‫وﯾﺑﻌث‬ ‫ﺣرارة‬ ‫ﻋﺎﻟﯾﺔ‬ ‫؟‬ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯾ‬ 𝑆𝑂2 /‫ج‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ΔG = ‫ﻣﻮﺟﺒﺔ‬ ‫ذوﺑﺎن‬ ‫ﻏﺎز‬ ΔS = ‫ﺳﺎﻟﺒﺔ‬ ‫)ﺗﻘﻞ‬ (‫اﻟﻌﺸﻮاﺋﯿﺔ‬ ‫اﻟﻌﻤﻠﯿﺔ‬ ‫ﺑﺎﻋﺜﺔ‬ ΔH = ‫ﺳﺎﻟﺒﺔ‬ ΔG = ΔH - TΔS - (+) - = -
  • 16.
  • 17.
    1 ‫ﻋﻨﺪﻣﺎ‬- ‫ﺗﻜﻮن‬ ‫اﺷﺎرة‬ ) ( ‫ﻣﺘﺸﺎﺑﮭﺔ‬ ∆𝐻, ∆𝑆 a)∆𝐻 =− ∆𝑆 =− ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 − ( ) − − ( ) − ( ) (+) ‫ھﻨﺎ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﺗﻜﻮن‬ (‫)ﺗﺒﺮﯾﺪ‬ ∆𝐻 > 𝑇∆𝑆 b) ∆𝐻 =+ ∆𝑆 =+ ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 + ( ) − + ( ) + ( ) (−) ‫ھﻨﺎ‬ ‫ﯾﻜﻮن‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﻋﻨﺪﻣﺎ‬ ‫ﺗﻜﻮن‬ (‫)ﺗﺴﺨﯿﻦ‬ 𝑇∆𝑆 > ∆𝐻 ‫اﻟﺧﺎﻣس‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬
  • 18.
    1 . ‫؟‬ ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬‫ﺗﻠﻘﺎﺋﯾﺔ‬ ‫اﻟﺟﻠﯾد‬ ‫اﻧﺻﮭﺎر‬ ‫ﻋﻣﻠﯾﺔ‬ ) A ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ΔG = ‫ﺳﺎﻟﺑﺔ‬ ‫)اﻟﺗﻔﺎﻋل‬ (‫ﺗﻠﻘﺎﺋﻲ‬ - ‫اﻧﺻﮭﺎر‬ ΔS = ‫ﻣوﺟﺑﺔ‬ ‫)ﺗزداد‬ (‫اﻟﻌﺷواﺋﯾﺔ‬ - ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﻣﺎﺻﺔ‬ ΔH = ‫ﺳﺎﻟﺑﺔ‬ ΔG = ΔH - TΔS + (-) + = - ΔH > TΔS ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 2 . ‫اﻻﻋﺗﯾﺎدﯾﺔ‬ ‫ﺑﺎﻟظروف‬ ‫اﻷوﻟﯾﺔ‬ ‫ﻋﻧﺎﺻره‬ ‫إﻟﻰ‬ ‫اﻟﻣﺎء‬ ‫ﯾﺗﺣﻠل‬ ‫ﻻ‬ :‫ﺑﺳﺑب‬ ‫وذﻟك‬ ) A ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ΔG = ‫ﻣوﺟﺑﺔ‬ - ‫ﯾﺗﺣﻠل‬ ΔS = ‫ﻣوﺟﺑﺔ‬ ‫)ﺗزداد‬ (‫اﻟﻌﺷواﺋﯾﺔ‬ ) B ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﻏﯾر‬ ‫ﺗﻠﻘﺎﺋﯾﺔ‬ ΔG = ‫ﻣوﺟﺑﺔ‬ - ‫ﯾﺗﺣﻠل‬ ΔS = ‫ﻣوﺟﺑﺔ‬ ‫)ﺗزداد‬ (‫اﻟﻌﺷواﺋﯾﺔ‬ - ‫اﻟﻌﻣﻠﯾﺔ‬ ‫ﻣﺎﺻﺔ‬ ‫ﻟﻠﺣرارة‬ ΔH = ‫ﻣوﺟﺑﺔ‬ ( ‫اﻟﺳﺎدس‬ ‫)اﻟدرس‬ ‫ﻛﺑس‬ ‫ﻋﻼﻗﺔ‬ ‫ﺣﺳﺎﺑﺎت‬ :‫ﻛﺑس‬ ‫ﻋﻼﻗﺔ‬ ‫ﺣﺳﺎﺑﺎت‬
  • 19.
    ‫ﯾﻤﻜﻦ‬ ‫اﺳﺘﺨﺪام‬ ‫ﻋﻼﻗﺔ‬ ‫ﻛﺒﺲ‬ ‫ﻟﺤﺴﺎب‬ ‫ﻗﯿﻤﺔ‬ ‫اﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ Δ ‫اﻟﻘﯿﺎﺳﯿﺔ‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫ﻣﻌﺮﻓﺔ‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫أو‬ ‫ﻋﺪم‬ ‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫وذﻟﻚ‬ 𝐺𝑟 ° ‫إذا‬ ‫ﺗﻢ‬ ‫ﻗﯿﺎس‬ ‫ﻗﯿﻢ‬ Δ ‫و‬ ، ‫ﻋﻨﺪ‬ ‫اﻟﻈﺮوف‬ ‫اﻟﻘﯿﺎﺳﯿﺔ‬ ) 1 atm/25 ( ، ‫وذﻟﻚ‬ ‫ﻣﻦ‬ ‫ﺧﻼل‬ ‫اﻟﻌﻼﻗﺔ‬ :‫اﻟﺘﺎﻟﯿﺔ‬ 𝐻𝑟 ° ∆𝑆𝑟 ° ℃ ∆𝐺𝑟 ° =∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° ‫ﻣﻼﺣظﺔ‬ 1 / ‫ﯾﺠﺐ‬ ‫اﻟﺘﻤﯿﯿﺰ‬ ‫ﺑﯿﻦ‬ Δ ‫و‬ ‫وﺑﯿﻦ‬ Δ ‫و‬ Δ . 𝐺𝑟 ° ∆𝐺𝑓 ° 𝐻𝑟 ° 𝐻𝑓 ° ‫ﻣﻼﺣظﺔ‬ 2 / ‫ﻗﯿﻢ‬ ‫اﻻﻧﺜﺎﻟﺒﻲ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ ‫ﻟﻠﻌﻨﺎﺻﺮ‬ ‫اﻟﺤﺮة‬ ‫ﺑﺄﺛﺒﺖ‬ ‫ﺻﻮرة‬ ‫ﺗﺴﺎوي‬ ‫ﺻﻔﺮ‬ ‫ﻣﻼﺣظﺔ‬ 3 / ‫ﯾﺠﺐ‬ ‫ﺗﻮﺣﯿﺪ‬ ‫وﺣﺪﺗﻲ‬ ‫و‬ Δ ‫ﻗﺒﻞ‬ ‫ﺗﻄﺒﯿﻖ‬ ‫ﻋﻼﻗﺔ‬ .‫ﻛﺒﺲ‬ ∆𝑆𝑟 ° 𝐺𝑟 ° ‫ﻣﻼﺣظﺔ‬ 4 / ‫اﻟﻌﺸﻮاﺋﯿﺔ‬ (‫)اﻻﻧﺘﺮوﺑﻲ‬ ‫ﻟﻠﺠﺰﯾﺌﺎت‬ ‫ﺗﻌﺘﻤﺪ‬ ‫ﻋﻠﻰ‬ ‫اﻟﻄﺎﻗﺔ‬ ،‫اﻟﺤﺮﻛﯿﺔ‬ ‫وھﻲ‬ ‫ﺣﺮﻛﺎت‬ ‫ﺻﻐﯿﺮة‬ ‫ﺑﺎﻟﻨﺴﺒﺔ‬ ‫ﻟﻘﯿﻢ‬ ‫اﻻﻧﺜﺎﻟﺒﻲ‬ ‫واﻟﻄﺎﻗﺔ‬ ‫اﻟﺤﺮة‬ ‫ﻟﺬﻟﻚ‬ ‫ﺗﻜﻮن‬ ‫وﺣﺪة‬ ΔS ‫ﺻﻐﯿﺮة‬ ‫ﺑﺎﻟﺠﻮل‬ ) J ( ‫ووﺣﺪات‬ ΔH ‫و‬ ΔG ‫ﻛﺒﯿﺮة‬ ‫ﺑﺎﻟـ‬ ) KJ .( ‫س‬ 4 / ‫ﺟد‬ ‫ﻗﯾﻣﺔ‬ ) ( ‫ﻟﻠﺗﻔﺎﻋل‬ :‫اﻟﺗﺎﻟﻲ‬ ∆𝐺°𝑟 2𝐶𝑂(𝑔) + 𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) ‫اﻟﻣﻌﻠوﻣﺎت‬ ‫أﻋطﯾت‬ ‫إذا‬ ‫اﻟﻘﯾﺎﺳﯾﺔ‬ ‫اﻟظروف‬ ‫ﺗﺣت‬ ‫ﯾﺟري‬ ‫اﻟذي‬ :‫اﻵﺗﯾﺔ‬ Δ = -) 110.5 ( KJ/mol , 𝐻𝑓𝐶𝑂 ° Δ = -) 393.5 ( KJ/mol 𝐻𝑓𝐶𝑂2 ° = 198 J/K mol , 𝑆𝐶𝑂 ° = 214 J/K mol , 𝑆𝐶𝑂2 ° = 205 J/K mol 𝑆𝑂2 ° /‫اﻟﺣل‬ 1 - Δ 𝐻𝑟 ° = 2∆𝐻𝑓𝐶𝑂2 ° ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 2∆𝐻𝑓𝐶𝑂 ° ( )+ ∆𝐻𝑓𝑂2 ° ( ) ⎡ ⎢ ⎣ ⎤ ⎥ ⎦
  • 20.
    Δ 𝐻𝑟 ° = 2 ×− 393. 5 ( ) [ ] − 2 × − 110. 5 ( ) + 0 [ ] Δ 𝐻𝑟 ° = − 787 ( ) + 221 ( ) ∆𝐻𝑟 ° = − 566 𝐾𝐽/𝑚𝑜𝑙 2. ∆𝑆𝑟 ° = 2𝑆𝐶𝑂2 ° ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ − 2𝑆𝐶𝑂 ° + 𝑆𝑂2 ° ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ∆𝑆𝑟 ° = [2 × 214] − [2 × 198) + 205] ∆𝑆𝑟 ° = − 173 𝐽/𝐾 𝑚𝑜𝑙 = -0.173 KJ/K mol ‫ﻧﻮﺣﺪ‬ ‫اﻟﻮﺣﺪات‬ ∆𝑆𝑟 ° = −173 1000 3. Δ ∴ 𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇 ∆𝑆𝑟 ° , 𝑇𝑘 = 𝑡℃ + 273 𝑇𝑘 = 25 + 273 = 298𝐾 = -566 - (298) (-0.173) Δ ∴ 𝐺𝑟 ° = − 514. 446 𝐾𝐽 𝑚𝑜𝑙 (‫ﺗﻠﻘﺎﺋﻲ‬ ‫)اﻟﺘﻔﺎﻋﻞ‬ ‫س‬ 5 / ‫ﻟﻠﺗﻔﺎﻋل‬ ‫اﻟﺗﺎﻟﻲ‬ 2𝐻2 𝑔 ( ) + 𝑂2 (𝑔) → 2𝐻2 𝑂(𝑔) (J/K.mol) ‫ﻟﻠﺗﻔﺎﻋل‬ ‫ﺑوﺣدة‬ ‫اﺣﺳب‬ ‫ﻗﯾﻣﺔ‬ ∆𝑆𝑟 ° :‫ﻋﻠﻣت‬ ‫إذا‬ Δ𝐻𝑓𝐻2 𝑂 ° = − 242 ( )𝐾𝐽/𝑚𝑜𝑙) Δ𝐺𝑓𝐻2 𝑂 ° = − 228 ( )𝐾𝐽/𝑚𝑜𝑙)
  • 21.
    /‫اﻟﺣل‬ Δ 𝐻𝑟 ° = ∑ 𝑛∆𝐻𝑓(𝑃) ° −∑ 𝑛 ∆𝐻𝑓(𝑅) ° = ] 2Δ [𝐻𝑓𝐻2 𝑂 ° − [2∆𝐻𝑓𝐻2 ° + ∆𝐻𝑓𝑂2 ° ] Δ 𝐻𝑟 ° = [2(− 242)] − [0]⟹∆𝐻𝑟 ° = − 484 𝐾𝐽/𝑚𝑜𝑙 Δ 𝐺𝑟 ° = ∑ 𝑛 ∆𝐺𝑓(𝑃) ° ∑ 𝑛 ∆𝐺𝑓(𝑅) ° = ] 2Δ [ − ] 0 + 0 [ 𝐺𝑓𝐻2 𝑂 ° T = 25 + 273 T = 298 K Δ 𝐺𝑟 ° = [2(− 228)]⟹∆𝐺𝑟 ° = − 456 𝐾𝐽/𝑚𝑜𝑙 Δ 𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° - 456 = - 484 - ) 298 ( ∆𝑆𝑟 ° ⟹ ∆𝑆𝑟 ° = −28 298 = − 0. 094 𝐾𝐽/𝐾 𝑚𝑜𝑙 ∴ ∆𝑆𝑟 ° = − 0. 094 × 1000⟹ ∆𝑆𝑟 ° = − 94 𝐽/𝐾 𝑚𝑜𝑙 ‫اﻟﺳﺎدس‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬
  • 22.
    1 . Δ𝐺𝑟 ° = Δ𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T,F ( ‫اﻟﺟواب‬ / T 2 . ) A ‫ﯾﺟب‬ ‫اﻟﺗﻣﯾﯾز‬ ‫ﺑﯾن‬ Δ ‫و‬ ‫وﺑﯾن‬ Δ ‫و‬ Δ . 𝐺𝑟 ° ∆𝐺°𝑓 𝐻𝑟 ° 𝐻𝑓 ° ) B ‫ﻗﯾم‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫واﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورة‬ ‫ﺗﺳﺎوي‬ 1 ) C ‫ﻗﯾم‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫واﻟطﺎﻗﺔ‬ ‫اﻟﺣرة‬ ‫ﻟﻠﻌﻧﺎﺻر‬ ‫اﻟﺣرة‬ ‫ﺑﺄﺛﺑت‬ ‫ﺻورة‬ ‫ﺗﺳﺎوي‬ ‫ﺻﻔر‬ (‫)ﺧﯾﺎرات‬ ‫اﻟﺳؤال‬ ‫ﻧوع‬ ‫اﻟﺟواب‬ / A& C 3 . ‫اﻟطﺎﻗﺔ‬ ‫ﻋﻠﻰ‬ ‫ﺗﻌﺗﻣد‬ ‫ﻻ‬ ‫ﻟﻠﺟزﯾﺋﺎت‬ (‫)اﻻﻧﺗروﺑﻲ‬ ‫اﻟﻌﺷواﺋﯾﺔ‬ ‫ﺻﻐﯾرة‬ ‫ﺣرﻛﺎت‬ ‫وھﻲ‬ ،‫اﻟﺣرﻛﯾﺔ‬ ‫اﻟﺣرة‬ ‫واﻟطﺎﻗﺔ‬ ‫اﻻﻧﺛﺎﻟﺑﻲ‬ ‫ﻟﻘﯾم‬ ‫ﺑﺎﻟﻧﺳﺑﺔ‬ ‫ﻟذﻟك‬ ‫ﺗﻛون‬ ‫وﺣدة‬ ΔS ‫ﺻﻐﯾرة‬ ‫ﺑﺎﻟﺟول‬ ) J ( ‫ووﺣدات‬ ΔH ‫و‬ ΔG ‫ﻛﺑﯾرة‬ ‫ﺑﺎﻟـ‬ ) KJ .( ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F ( ‫اﻟﺳﺎﺑﻊ‬ ‫)اﻟدرس‬ ‫اﻟﺣرة‬ ‫ﻛﺑس‬ ‫طﺎﻗﺔ‬ ‫ﺗطﺑﯾﻘﺎت‬
  • 23.
    ‫س‬ 6 / ‫ﺟﺪ‬ ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫اﻟﺘﻲ‬ ‫ﺗﺼﺒﺢ‬ ‫ﻋﻨﺪھﺎ‬ ‫اﻟﺘﻔﺎﻋﻼت‬ ‫اﻟﺘﺎﻟﯿﺔ‬ ،‫ﺗﻠﻘﺎﺋﯿﺔ‬ ‫إذا‬ :‫ﻋﻠﻤﺖ‬ ‫ﻓرع‬ ) A /( ∆𝐻𝑟 = (+ 126)𝐾𝐽/𝑚𝑜𝑙 ∆𝑆𝑟 = (+ 48) 𝐽/𝐾 𝑚𝑜𝑙 /‫اﻟﺣل‬ ΔG ‫ﻧﻔﺘﺮض‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻓﻲ‬ ‫ﺣﺎﻟﺔ‬ ،‫اﺗﺰان‬ ، ‫ﺻﻔﺮ‬ = ∴ ‫وﻧﻮﺣﺪ‬ ‫اﻟﻮﺣﺪات‬ ∆𝐻 ∆𝑆 = 𝑇⟹ 0 = ∆𝐻 − ∆𝑆 ∆𝑆𝑟 = 48 1000 ⟹ ∆𝑆𝑟 = + 0. 048 𝐾𝐽/𝐾 𝑚𝑜𝑙 ∴ 𝑇 = 126 0.048 ⟹ 𝑇 = 2625 𝐾 /‫ﻣﻼﺣظﺔ‬ ‫إذا‬ ‫ﻛﺎﻧﺖ‬ ‫إﺷﺎرة‬ ΔS ‫و‬ ΔH = ‫ﻣﻮﺟﺒﺔ‬ ‫ﻓﯿﺠﺐ‬ ‫ﺗﺴﺨﯿﻦ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻷﻛﺜﺮ‬ ‫ﻣﻦ‬ ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫اﻟﻤﺤﺴﻮﺑﺔ‬ ‫ﻋﻨﺪ‬ ‫اﻻﺗﺰان‬ . ‫وﻋﻠﯿﮫ‬ ‫ﯾﺠﺐ‬ ‫ﺗﺴﺨﯿﻦ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻷﻛﺜﺮ‬ ‫ﻣﻦ‬ ) 2625 K ( ‫ﻟﻜﻲ‬ ‫ﯾﺼﺒﺢ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ .‫ﺗﻠﻘﺎﺋﻲ‬ (B) ‫ﻓرع‬/ ∆𝑆𝑟 = (− 105) 𝐽/𝐾 𝑚𝑜𝑙, ∆𝐻𝑟 = (− 12) 𝐾𝐽/𝑚𝑜𝑙 ‫ﻧﻔﺘﺮض‬ ΔG = ‫ﺻﻔﺮ‬ ‫)ﺣﺎﻟﺔ‬ (‫اﻻﺗﺰان‬ ∴ 𝑇 = ∆𝐻 ∆𝑆 , ∆𝑆 = −105 1000 ⟹ ∆𝑆 = − 0. 105 𝐾𝐽/𝐾 𝑚𝑜𝑙 𝑇 = −12 −0.105 ⟹ 𝑇 = 114. 3 𝐾 ‫وﻋﻠﯿﮫ‬ ‫ﯾﺠﺐ‬ ‫ﺗﺒﺮﯾﺪ‬ ‫إﻧﺎء‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻷﻗﻞ‬ ‫ﻣﻦ‬ ) 114.3 K ( ‫ﻟﯿﺼﺒﺢ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ .‫ﺗﻠﻘﺎﺋﻲ‬ ‫داﺋﻤﺎ‬ ‫ﻓﻲ‬ ‫اﻟﺴﺆال‬ ‫ﻣﺘﺸﺎﺑﮭﺔ‬ ΔS ‫و‬ ΔH ‫ﺳﺘﻌﻄﻰ‬ ‫أﺷﺎرات‬ ‫ﻗﯿﻢ‬ ‫ﻣﻼﺣظﺔ‬/ = ‫ﺳﺎﻟﺒﺔ‬ ‫)ﻧﺒﺮداﻟﺘﻔﺎﻋﻞ‬ ) ΔS ‫و‬ ΔH ‫اذا‬ ‫ﻛﺎﻧﺖ‬
  • 24.
    ‫س‬ 7 : ‫ﺗﺘﻔﻜﻚ‬ ‫ﻛﺎرﺑﻮﻧﺎت‬ ‫اﻟﻜﺎﻟﺴﯿﻮم‬ ‫ﺣﺴﺐ‬ ‫اﻟﻤﻌﺎدﻟﺔ‬ ‫اﻵﺗﯿﺔ‬ : 𝐶𝑎𝐶𝑂3 (𝑠) ∆→ 𝐶𝑎𝑂(𝑠) +𝐶𝑂2 (𝑔) (160 J/K mol) =Δ𝑆°𝑟 ‫ﻓﺈذا‬ ‫ﻋﻠﻤﺖ‬ Δ ‫ﻟﻜﻞ‬ ‫ﻣﻦ‬ ‫و‬ CaO ‫و‬ ‫ھﻲ‬ ‫ﻋﻠﻰ‬ :‫اﻟﺘﻮاﻟﻲ‬ 𝐻𝑓 ° 𝐶𝑂2 𝐶𝑎𝐶𝑂3 -) 1207 / - 635 / - 393.5 ( ‫ﺑﻮﺣﺪات‬ KJ/mol ، :‫ﺟﺪ‬ ‫ﻟﻠﺘﻔﺎﻋﻞ‬ ‫ﺛﻢ‬ ‫أرﺳﻢ‬ ‫ﻣﺨﻄﻂ‬ ‫اﻟﻄﺎﻗﺔ‬ Δ 1) 𝐻𝑟 ° ‫ﻟﻠﺘﻔﺎﻋﻞ‬Δ 2) 𝐺𝑟 ° 3) ‫درﺟﺔ‬ ‫اﻟﺤﺮارة‬ ‫اﻟﺘﻲ‬ ‫ﯾﺼﺒﺢ‬ ‫ﻋﻨﺪھﺎ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﺗﻠﻘﺎﺋﯿﺎ‬ /‫اﻟﺣل‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺎص‬ 1 ( ∆𝐻𝑟 ° = ∑ 𝑛∆ 𝐻𝑓(𝑃) ° − ∑ 𝑛 ∆𝐻𝑓 𝑅 ( ) ° ∆𝐻𝑟 ° = [(− 635) + (− 393. 5)] − [− 1207] ∆𝐻𝑟 ° = (− 1028) + (1207) ∆𝐻𝑟 ° = 178. 5 𝐾𝐽/𝑚𝑜𝑙 ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻏﯿﺮ‬ ‫ﺗﻠﻘﺎﺋﻲ‬ 2 ( T = 25 + 273 = 298 K
  • 25.
    ∆𝑆𝑟 ° = 160 1000 = 0. 16𝐾𝐽/𝐾 𝑚𝑜𝑙 ∴ ∆𝐺𝑟 ° = ∆𝐻𝑟 ° − 𝑇∆𝑆𝑟 ° ∆𝐺𝑟 ° = 178. 5 − (298) (0. 16) ∆𝐺𝑟 ° = 130. 82 𝐾𝐽/𝑚𝑜𝑙 3) (0 = Δ ‫ﻧﻔﺘﺮض‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫ﻣﺘﺰن‬ 𝐺𝑟 ° ) 0 = ∆𝐻𝑟 ° − 𝑇∆𝑆⟹𝑇 = ∆𝐻𝑟 ° ∆𝑆𝑟 ° ⟹ 𝑇 = 178.5 0.16 = 115. 6 𝐾 ‫ﻣﻮﺟﺒﺔ‬Δ ‫و‬ ‫ﺑﻤﺎ‬ ‫اﻧﮫ‬ ‫ﻗﯿﻤﺘﻲ‬ 𝐻𝑟 ° ∆𝑆𝑟 ° ‫ﻟﻜﻲ‬ ‫ﯾﺼﺒﺢ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ .ً‫ﺎ‬‫ﺗﻠﻘﺎﺋﯿ‬ 1115.6 K ً‫ا‬‫إذ‬ ‫ﯾﺠﺐ‬ ‫ﺗﺴﺨﯿﻦ‬ ‫اﻟﺘﻔﺎﻋﻞ‬ ‫إﻟﻰ‬ ‫أﻛﺜﺮ‬ ‫ﻣﻦ‬
  • 26.
    ‫اﻟﺳﺎﺑﻊ‬ ‫اﻟدرس‬ ‫اﺳﺋﻠﺔ‬ 1 . ‫إذا‬ ‫ﻛﺎﻧت‬ ‫إﺷﺎرة‬ ΔS ‫و‬ ΔH = ‫ﺳﺎﻟﺑﺔ‬ ‫ﻓﯾﺟب‬ ‫ﺗﺳﺧﯾن‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻷﻛﺛر‬ ‫ﻣن‬ ‫درﺟﺔ‬ ‫اﻟﺣرارة‬ .‫اﻻﺗزان‬‫ﻋﻧد‬ ‫اﻟﻣﺣﺳوﺑﺔ‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 2 . ‫اذا‬ ‫ﻛﺎﻧت‬ S Δ ‫و‬ ΔH = ‫ﺳﺎﻟﺑﺔ‬ (‫)ﻧﺑرداﻟﺗﻔﺎﻋل‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / F 3 . ‫ﯾﺟب‬ ‫ﺗﺳﺧﯾن‬ ‫اﻟﺗﻔﺎﻋل‬ ‫ﻷﻛﺛر‬ ‫ﻣن‬ ) 2625 K ( ‫ﻟﻛﻲ‬ ‫ﯾﺻﺑﺢ‬ ‫اﻟﺗﻔﺎﻋل‬ .‫ﺗﻠﻘﺎﺋﻲ‬ ‫ﻧوع‬ ‫اﻟﺳؤال‬ ) T, F ( ‫اﻟﺟواب‬ / T