Question

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4. K for the reaction Br2(g) 2Br(g) is 1.1 x 10 reaction at this temperature. at 1280°C. Calculate the value of K, for this (a) 1.1 x 10 (b) 18 (c) 0.14 (d) 910 (e) 8.3 x 10 5. Which of thg following reactions favors reactants? (a) H2(g)+ F2(g) 2HF(g ) K- 1.2 x 10 (b) 2NO(g)+ Cl:(g) 2NOCI (g): K= 4.6 x 10 cN:(g)+ Olg)= 2NO(g)K 4.7 x 1031 (d) 2NO(g) +O:(g)# 2NO : (g): K 5 x 1012 (e) The reactions in (a), (b), and (c) all favor reactants. 6 At 400 K, bromine (Br2) and chlorine (Cl) react to give BrCl according to the equation. Bra(g) + Cl(g) 2BrCl(g) When equilibrium is reached, the partial pressures of Br2, Cla, and BrCl are 2.5 atm, 2.5 atm and 6.6 atm, respectively. Calculate the equilibrium constant (Kp) for this reaction. (a) Kp 7.0 (b) Kp 0.14 (c) Kp 1.1 (d) K 0.95 7. An HCl solution has a pH of 2.37. Calculate the molarity of HCI in the solution. (a) 2.34x 102 M (b) 2.34x 1012 M (c) 4.27 x 103M (d) 3.75 x 10 M toTCH PL0 174

Answer 1

0 kp = (hege (po gd (poga (B² for any reaction artig, at b Big Costa Pig ..Puan RT = P = ART & P 2 CRT . Кре (ерт) ср ету ке (2) 4 -с- CART) CCBRT) for up = K (RT Dn. reaction Benaigo - 2Br 19, Dn=1 T = 1280°C = 1553K R = 8.20 x 102 Latm in mor kp = Ke (RT) = (1.18 10-3) [18:20 x 102) (1553) 14008 xlors = 0.14 10. For any reaction artig, + bBig clig at a Dig k = [C] [ojd [A] [B] it Kol & [eje [pd ng [R] concentration of product is more and reaction is to reward if KL1 [C] DJd L[A]a CBJb concentration of reactant is more and reaction tends to backward

So, for option (C) only N2(g) + O2(g) - 200 g K = 4.7 g 10r21 as Kroc reactant ride is favoured ② 0 Pon reaction. Brzig, & algo - 2 Brclig kp = (bre² Bre) () as Buce Br = 6.6 atm = 2.5 aton and h = 2.5atm. & Kp = (6.6)² (2,5)(25) = 6.96 7.0 C. He is a strong acid. and it dissociates completely : не (а и* (рлар + сі: (49 and as PH = 2.37 - log [ht] = 2. 3 7 ht] = antilog (2.37 20.oou27 & 4.278103 m as the and dissociates completely, thus Ht concentration is equal to concentration of ha) Ha] = 4.27 x 103 m

16. On aqueous medium always [hot] [oH] = 10m2 as not a 1.99 x 10.5m. a LOH = 1014 m² nou = 0.5025 x 1014 Traxtos x 105 m = 0.5025 X 109M = 5.038 10-10 m Herre, acidic. becans PH = -log (1.99x105) - 5-log(1-99) = 5 - 0.298 = u. 702 E as p4277 , hence medium is acidic 6 for (c) [04] = 5.6*10-9 m. POH = -log (5.64 109) I a- log (5.6 = 9-0.748 = . 252 PH = 14-POH = 14.8.252 = 5.748 fos (oz. t] = [thot] = 1.98 102 PH= og hot] = log (1. ax 10-2) - 2-bog 19 - 2-0278 1.722 | 6) - P1 2 0 , 12 2 = r 86 So option (o) is most acidic.