Page 169 - الكيمياء كتاب الطالب للصف 11 الفصل 1

z yfy^a h 3  £ 6e *a -%ra

d C<4=: ´±¼C59 ’ d98 ± X51 ± ” ¼C59 d b: ± R 8 ± § ¼C59 dgS ± X51 ±
_<
Û
¤d ŠC E ¹C4: ± ÑQ ` R= 4 ± `8: Ò ˜“Áb9 : ±
»b 8 ± Ãb: Э Y „

P n X X P c ¯ ^0 Э D E= b: ±
X T Ãb: Ð­Ò –1.00 E:=7 ±
n
T
Э D E=gS ± Áb51 ±
˜d98 ± X51 ± ÔÒC

E=:9 ´C 920

´„b: d98 ± ¹P4 ± c9 CÕ b 7 ´±¼C5 ± ` Áb9 d `Û=4 ¼C ´„b ¹P :Molar fraction d b: ± R 8 ±
˜Áb9 : ± d ³¹b b: ± C<4=: ´±¼C5 ±

ÎC

˜¼C ^8 d b: ± R 8 ± D ± ¤œ ³b2 ± ¹b= ±Ò `= Ò»P=< ± `= бS „± E= Š± E ¹C4: ± Nhb ˜››
×
1.71 10 –3 : Û& ³»±R E »¹ P; `= Ò»P=< ± P ¹b Ò
H = = 0.0810
×
2
2.112 10 –2
H (g) + I (g) ⇌ 2HI(g)
2 2
×
2.91 10 –3 бS „± P; E C; ±Ò E9 C6 : ± ¹±b: ± ` ^ ´„b ¹P
I = = 0.1378
2 × –2
2.112 10
¤ÔÒC
×
H I = 1.65 10 –2 = 0.7813 H = 1.71 × 10 mol,
í
×
2.112 10 –2 2
I = 2.91 × 10 mol,
í
2
˜¼C ^8 dgS ± X51 ± E:= D ± ¤ ³b2 ± HI = 1.65 × 10 mol
í
H = 0.0810 × 100 = 8.10 kPa
2 (K ) E:= D ± –100 kPa ÔÒC d98 ± X51 ±Ò
p
I = 0.1378 × 100 = 13.78 kPa
2 ˜^ C6 ± ±Q<
HI = 0.7813 × 100 = 78.13 kPa
¤ Û ^ ±
бS „± F C E:= ¤ž ³b2 ±
˜´„b: ± ¹P Ãb: D ± ¤› ³b2 ±
(PHI) 2
K =
p (1.71 × 10 ) + (2.91 × 10 ) + (1.65 × 10 )
í
í
í
(PH 2 × )(PI 2 )
í
×
K = 78.13 78.13 = 54.7 = 2.112 × 10 mol
×
p
8.10 13.78

› ¡