Page 167 - الكيمياء كتاب الطالب للصف 11 الفصل 1
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                                                              E=gS  ± Áb51 ± ÏP     E= C   ´C=9:


                                                                                                            E9  ­


                 `= ÒR =; ±Ò `= Ò»P=< ± Áb9   a Q   ÔQ ± X51 ±       E R4  ΅  `  (K ) ²C   C;;8:  d б ÎC : ± d    ˜£
                                                                                   p
                 ÔÒC    E  C      Û&  ³»±R   E »¹  P;   C= b M±Ò    ˜Áb9 : ± d  ¹b b  ¼C  ^8  dgS  ± X51 ± E:=
                 X51 ± –C< 6  ËÒR3 ± ÑQ  d Ò ˜9.75 × 10  Pa                ¤E= б E ¹C4:9  C Õ 7 Ò _   ÔQ ± ^ C6  ± d
                                                       5
                 X51 ±Ò –3.5 × 10  Pa ÔÒC   `= ÒR =;9  dgS  ±
                                 4
                                                                             2SO (g) + O (g) ⇌ 2SO (g)
                 D  ±  ˜1.1 × 10  Pa  ÔÒC    `= Ò»P=<9   dgS  ±                  2      2          3
                               5
                                                                   E  C  ³»±R  E »¹  P;  E=gS  ± бS „± Áb5h Ðb8
                                        ˜^ C6  ± ±Q<  (K ) E:=
                                                      p                                                 ¤d ŠC
                                                     ¤ Û ^  ±              –O  = 7.0 × 10  Pa – SO  = 1.0 × 10  Pa
                                                                                                          6
                                                                                       6
                                                                             2                  2
                                                                                                          6

                 `   ˜C= b ‡  dgS  ± X51 ± E:=  D  ± ¤› ³b2  ±     ˜^ C6  ± ±Q<  (K ) E:=  D  ± .SO  = 8.0 × 10  Pa
                                                                                                3
                                                                                p
                 Ãb:    ÔÒC    d98 ±  X51 ±  Э  ËR4                                                   ¤ Û ^  ±
                      ¤E= б E ¹C4:9  CÕ7 Ò E=gS  ± Áb51 ±
                                                                   G=  `  ^ C6  ± ±Q<  бS „± E ¹C4  D  ± ¤› ³b2  ±

                          PT= (PN 2) + (PH 2) + (PNH 3)                                ¤E=gS  ± Áb51 ±
                                          5
                                4
                         5
                  9.75 × 10  = (3.5 10 ) + (1.1 × 10 ) + (PNH 3)                         (PSO 3 ) 2
                                                                                   K  =
                  8.3 × 10  Pa = PNH 3  C= b ‡  dgS  ± X51 ±                        p  (PSO 2 ) 2  × (PO 2 )
                       5
                 Áb51 ± G=  `  ^ C6 9  бS „± E ¹C4  D  ± ¤œ ³b2  ±     ¤E ¹C4: ± d  E=gS  ± бS „± Áb5h ÀÛb  ¤œ ³b2  ±

                                                                                  ×
                                                                                     6
                                            ¤E=gS  ±                    K  =   ( 8.0 10 ) 2   = 9.1 × 10 í
                                                                                  2
                                                                               ×
                                                                                       ×
                                                                                 6
                                 (PNH 3 ) 2                              p  ( 1.0 10 ) ×  7.0 10 6
                           K  =
                            p      × )(PH 2 ) 3
                               (PN 2                                         ¤E = 0 ± ½C=7 ± ³P Ò ûh­ ¤ ³b2  ±
                 ¤E ¹C4: ± d  E=gS  ± бS „± Áb5h À Û b  ¤ ³b2  ±                  Pa ×
                                                                                        Pa  =  1  = Pa
                                                                                                   í
                                                                                        ×
                                   (8.3 × 10 )                                    Pa ×  Pa Pa  Pa
                                           5 2
                         K  =
                           p                                                         K  = 9.1 × 10  Pa í
                                                                                               í
                                               5 3
                                     4
                             (3.5 × 10 ) × (1.1 × 10 )                                p
                          K  = 1.48 × 10  í                          E R4  ΅  `  (K ) ²C   C;;8:  d б ÎC : ± d   ˜›š
                           p                                                       p
                           ¤E = 0 ± ½C=7 ± ³P Ò ûh­ ¤ž ³b2  ±      d  ³¹b b  ´±¼C  E …  `  ` ؼC5  dgS  ± X51 ±
                                                                                 ˜d98 ± X51 ± c ¯ E Ch¯ Áb9 : ±
                              Pa ×  Pa    1
                                        =   =  Pa í
                                 ×
                                    ×
                           Pa ×  Pa Pa Pa  Pa 2                    C= b M± ` b8   `= Ò»P=< ± Z  `= ÒR =; ± ^ C6
                                  K  = 1.48 × 10  Pa í                                       ¤E= б E ¹C4:9  C Õ 7 Ò
                                             í
                                   p
                                                                            N (g) + 3H (g) ⇌ 2NH (g)
                                                                              2       2          3
                                                                                                             _<
                                                                                                              Û
                   `8:: ± »P7 C  C<4=:  ÏC »M± C;=7 ­ ±º¯ `8  –E b;4  ÏC »­ E …  c   E=  R   ± ÏC »M± ˺˹ ÎC : ± d  C; P   ± P7
                                     ˜N= j ^8  d  _=7 ± Z: Ù  Э d5 ;  C P;  –´C C   ± «±R ˆ E  C  ± E б d  E P ± `

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