Question

tant " (16) In question (12) the removal of so, (g) will shift the equilibrium to the (a) right (b) left (c) no shift. (17) In question (12) an increase in pressure will shift the equilibrium to (a) left (b) right (c) middle (d) no shift (18) in question (12) assuming the reaction is endothermic to the right, a decrease in temperature will shift the equilibrium to the (a) left (b) right (c) no shift since it is an equilibrium shift (19) In question (12) a catalyst would shift the equilibrium to the (a) left (b) right (c) no shift (d) middle. (20) What will be Kc for SO, (g) = SO2(g) + 1/20(g). (a) 4.74 (b)0.25 (c) 4.0 (d) 17.0 (e) 0.211 (21) Consider the data below: (22) The order of rxn with respect to l'(aq) is (a) 1 (b) 2 (c) 3 (d) 1.5 (e) none (23) The order with respect to OC (aq) is (a) 1 (b) 2 (c) 1.5 (d) 3 (e) 4 (24) The order with respect to Cl(aq) is (a) 1 (b) 2 (c) 3 (d) Not Applicable (e) 4 (25) The overall order of reaction is (a) 1 (b) 2 (c) 3 (d) 1.5 (e) Need more information

Answer 1

08- t'/a = 50 yes s ? since for tot order decay 0.693 t² = - x = = 0.01386 yr 0.693 t² 0.693 50 0.0144" So, the correct option is (C) +304) = ? Let [AJ. - 100 lie inili al conc" q Radio isotope for 30% decay [Alt : 100 - 30 - 70 for Ist order radioactive decay, we have -xt en fast : en [ASE TAJO -0.014 x tio en 70 100 = Holuing -0.3566 - - 0.014 x t307. 630-4. = 25.69 year Hence the correct option is (b) 010 we know that Arrhenius equation relates rate conetants at two different temperature according to the relation: 1. In kz - Calt k R L T K₂ rate constant @ temp. Tz ( 127 c = 40. ki o rate constant @ temp. Ti (300K) Ea E Activation energy - 5x104 J mole? R = 8.314 Jkt molt k, = 2.4x10 3 Mist 5X10 en de les cas to the red 300 ODJ 8.314 2.103 en kz = 5.011627 2.4 X20-3 = 5.011627 e e . = = 150.1489 2.4x10-3 kg = 0.36035 Myst The closest option is (c) ie 0.150 Mt 27. Qll2n, reactant no products since, it is given that n₂ >n, more no. g moles / molecules are formed An = tue .: randomness increase which, in turn, inculeses the entropy of the system. Hence, the correct option (a) 0124 250, † O F 250, since kp=ke (RT) on on = np. na = 2-3 = -1 T: 827°C - (827 +273 ) K = 1100k R= 0.0821 Latm kt molt кс . kp_ 0.25/ = (RT)^n (0.0821 x 1100 ) -1 0.25 0.0821X100 ki 0.25 x 1100 X 0.082). Kc = 22.57 correct option - (a) 413, 2NO + 2H2 - Ng (9) + 2H₂O (9) Kc = [Na] [H₂O]² (NO)² (H2]² = (5-3x10-2) (2.9x1043) (8.1X10-3) (4.1x1005)? = 44.573 x 108 -I : 4.00 x 1, 6 |l02.1p 4) Y corred option* (b) Qis - 2N2 + O₂ - 2N20 equil. constant ko (N2O)? [N₂] ²102] V= 21 . (1,0) = 2x10-2 met 10-2 M 2 L [02] = = 1.25 X10 M 2. 5x10-5 mol 2 L [Na] = 2.80 x1004 - 1.40x10-4 M 10-4 K = (102M)? (1.25x10-5) 1140x104)2 2.45 X 10 = 4.0816x 10° (NL) - 2x10 ; 102] = 0.00245 ; (1,0) = 0.200M Q = [NO]2 (0.2)2 1NJ102) (2x104)" x 10.00245) = 0.04 - 4.086 X 100 9.8 x 10-1 I since 1Q=K) system will be at equilibrium. 015 since syptem is at equil. , there will be no shift. 016- - Removal of SO₂, will decrease the conch of product. So, according to de- chatlier principle, equilibrium will shift in that direction where this change is nullified so equilibrium will shift in forward diva le to the Right . Due 0177 on = np- na = -1 An increase in pressure, equilibrium will shift in that direction in which decrease in the total number of moles takes place se forward dirn (Right) 0184 250, + Os 250g + oHº Decrease in temperature , will cause the equilibrium to shift in the direction where heat will evolue ie forward dirn. . Rexn will shift to the right Q19- Generally, Generally, addition of catalyst does not disturb the equilibrium. However, it helps in the attainment of equilibreium quickly. so correct option is (C) no shift. Q20+ 50369) - Soz (g) + < 09 ) - kc = [so,] [ OgJh - 1503] 021 7 Data Below Q22 The Data is missing reaction / or data. so e 8+( 22 to 25) requires