Question

n = Products-react (gas only) cuent Kp = kelet Practice exercises: Give the K, expression and the value for the given reactions. 1. NO.(g) = 2 NO. (g) At equilibrium [N.O.]=18.6 torr, (NO )= 4.58 torr. 18. Eberl shin Ke=P(NO₂)² 7600 101024) Ko= 101024)2 To 70.096 P(N2O4) (0.006) 4.58 _60.C 760 2. A673B(g) = 2 C(g)+D At equilibrium P,034 atm, Pc-0.52 atm Ko - (0.32 - Kp = 6.88] 3. 250() + 0x0= 250,(g) At equilibrium (80.)- 14.5 mmHg (0.)- 128 mmHg; [SO]= 0.18 mmHg. SO2 = 0.019 atm kp= 02= 0,168 atm Ko= (0.00024 (503) KP (50,1? ( KPro.019) (0,163) 303=0.00024 atm kp=0.00095 latm = 760 mmHg

Answer 1

11) N₂O4 + 2NO, 'PNo = 4.58 toor . 760 toos = 1atm - 4.58 tors= 1atmx 4.58 toss 760 toor = 6.026x103 atm PNacy = 18.6 toss = 1 atmx 18.6 togs = 0.0245 760 toss atm Kp = (PNoz)² = (6-026x103)² Proizou 0.0245 Kp = 1.48481037

2. Als) + 3B(g) → 20(g) + D(6) Ko= (Pe)² = (0.52 ) = 6.88 TPB) 3 (0.34) 3

Kp = 6.88] 3) 250219 + O2(g) = 250319) Kp of psog) ² (ksoz)? Poz

bson = 14.5 mm Hg : 760 mmHg = katm i. 14.5 mm Hg = 2 month = atm ; x 14.5mming 760 mming 20.01908 ath PO₂ = 128 mm Hg = 1 atm x 128mm Hg 760 mmhg z 0.1684 atm

3) p Soz = 0.18th = mmng latm xo. 18 mm Hg 760 mata -2. 37 xlo atm Kp = (2.37X10-4) 2 6. 01908)² x 0.1684 [Kp = 9.15 0154 Thus, Kp values were calculated