1) The active mass of 96g of O2 contained in a 2 litre vessel is
a) 16 moles/litre
b) 1.5 moles/litre
c) 4 moles/litre
d) 24 moles/litre
2) In a chemical equilibrium te rate constant of the forward reaction is 7.5 10-4 and the equilibrium constant is 1.5. What is the rate constant of the backward reaction ?
3) If the ionisation constant for acetic acid is 1.8 10-5, at what concentration will it be dissociated to 2% ?
b) 0.018 M
c) 0.18 M
d) 0.045 M
4) The molar concentration of 56g of N2 present in a one litre flask is
5) N2 + 3H2 2NH3. Which of the following is the correct statement for this reaction ?
6) When a reversible reaction is at equilibrium, the concentration of one of the products is decreased. The equilibrium state
a) Shifts in the direction of the backward reaction
b) Shifts in the direction of the forward reaction
c) Is not affected
d) Moves in the direction of both reactions
7) In the reaction, 3A + 2B C, the value of Kc will be
8) 1.1 mole of A are mixed with 2.2 moles of B and the mixture is then kept in a 1 litre flask till the equilibrium A+ 2B 2C is attained. At the equilibrium 0.2 moles of C are formed. The equilibrium constant of the reaction is.
9) At 500K, the equilibrium constant for the reaction cis-C2H2Cl2 trans- C2H2Cl2 is 0.6. At the same temperature the equilibrium constant for the reaction, trans-C2H2Cl2 cis-C2H2Cl2is.
10) The equilibrium constant Kp of the reaction N2 + 3H22NH3 changes
a) By catalyst addition
b) By total pressure
c) By temperature
11) For the reaction, PCl5(g) PCl3(g) + Cl2(g)
12) In a chemical equilibrium, the rate constants of the forward and backward reactions are 3.210-4 and 1.210-5 respectively. The equilibrium constant is
13) For the reaction H2(g) + I2(g) 2HI(g) at 72IK, the value of the equilibrium constant Kc is 50. When the equilibrium concentration of both is 0.5 M, the value of Kp under the same conditions will be
14) The decomposition of N2O4 to NO2 is carried out at 280 K in chloroform. When equilibrium has been established 0.2 mole of N2O4 and 2 10-3 mole of NO2 are present in a 2 litre solution. The equilibrium constant for the reaction, N2O4 2NO2 is.
15) @ In the equilibrium, 2SO2(g) + O2(g) 2SO3(g), the partial pressures of SO2,O2 and SO3 are 0.662, 0.101 and 0.331 atm. respectively. What should be the partial pressure of oxygen so that the equilibrium concentration of SO2 and SO3 are equal ?
a) 0.4 atm.
b) 1.0 atm.
c) 0.8 atm.
d) 0.25 atm.
16) One mole of ethyl alcohol was treated with one mole of acetic acid at 25Â°C. 2/3rd of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be
17) The reaction A +2B 2C+D was studied using an initial concentration of B which was 1.5 times that of A. But the equilibrium concentration of A and C were found to be equal. The Kcfor the equilibrium is
18) On adding A to AB (s) A (g) + B (g), the new equilibrium concentration of A becomes double. The equilibrium concentration of B would become
a) 1/2 of its original value
b) 1/4 of its original value
c) 1/8 of its original value
d) Twice of its original value
19) For the reaction, 2A + B C +D, on doubling the concentration of C, the value of Kc will
a) Be doubled
b) Be halved
c) Increase by 2
d) Remain the same
20) When CO2 dissolves in water, the following equilibrium is established.
CO2 + 2H2O H3O+ + HCO–3.
For this the equilibrium constant is 3.8x 10-6 and pH = 6. What would be the ratio of concentration of HCO–3 ion and CO2, i.e. [HCO–3]/[CO2].
21) The effect of increasing the pressure on the equilibrium 2A + 3B 3A + 2B is that the
a) Forward reaction is favoured
b) Backward reaction is favoured
c) effect is nil
d) None of these
22) The equilibrium constant, K for the reaction 2HI (g) H2(g) + I2 (g) at room temperature is 2.85 and that at 698 K is 1.4x 10-2. This implies that
a) HI is exothermic compound
b) HI is very stable at room temperature
d) HI is resonance stabilised
23) The equilibrium constant Kp for a homogeneous gaseous reaction is 10-8. The D GÂ° for the reaction (R = 2 cals K-1 mole-1) at 298 K is
a) + 10.98 k.cal
b) -1.8 k.cal
c) -4.1454 k.cal
d) + 4.1454k. cals
24) In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant will.
a) Also be doubled
b) Be halved
c) Becomes one fourth
d) Remain the same
25) Which of the following will not change the concentration of NH3 at equilibrium ?
N2 (g) + 3H2 (g) 2NH3 (g); D H = – x kJ.
a) Increases of pressure
b) Increase of temperature
c) Decrease of volume
d) Addition of a catalyst
26) Which of the following gaseous reactions will be favoured by low pressure ?
27) When pressure is applied to the equilibrium system, Ice Water, which of the following phenomena will happen ?
a) More ice will be formed
b) Water will evaporate
c) More water will be formed
d) Equilibrium will not be affected
28) For the reaction 2A (g) + B (g) 3C +D(g), two moles each of A and B were taken into a flask. The following must always be true when the system attains equilibrium.
a) [A] = [B]
c) [B] = [C]
d) [A] > [B]
29) In the formation of nitric acid, N2 and O2 are made to combine. Thus N2 + O2 2NO – heat. Which of the following conditions will favour the formation of NO ?
a) Low temperature
b) High temperature
c) Freezing point
d) All are favourable
30) Which of the following factors will favour the reverse reaction in chemical equilibrium ?
a) An increase in the concentration of one of the reactants
b) An increase in the concentration of one of the products
c) Removal of one of the products regularly
31) For the system A (g) + 2B(g) C (g), the equilibrium concentrations are [A] = 0.06,[B]= 0.12 and [C] =0.216 moleL-1. The equilibrium constant is
32) @ 2 moles of HI were heated in a sealed tube at 440Â°C till the equilibrium was reached. HI was found to be 22% dissociated. The equilibrium constant for dissociation is
33) @ 4 moles of A are mixed with 4 moles of B, when 2 moles of C are formed at equilibrium according to the reaction, A + B C +D. The value of the equilibrium constant is
34) If Kp for a reaction A (g) + 2B (g) 3C (g) + D (g) is 0.05 atm. at 1000 K, its Kc in terms of R will be
a) 20000 R
b) 0.02 R
35) For a reaction H2 + I2 2HI at 721 K, the value of the equilibrium constant is 50. If 0.5 moles each of H2 and I2 are added to the system, the value of the equilibrium constant will be
36) The equilibrium constant for a reaction, N2 (g) + O2 (g) 2NO (g) is 4 x 10-4 at 2000 K. In the presence of a catalyst, the equilibrium constant is attained 10 times faster. The equilibrium constant in presence of the catalyst at 2000 K is
37) The factor which changes the equilibrium constant of the reaction,
A2 (g) + B2 (g) ® 2AB (g) is the
a) Total pressure
38) In which of the following reactions, the yield of the products does not increase with an increase in pressure ?
39) At a certain temperature, 50% of HI is dissociated into H2 and I2. The equilibrium constant is
40) @ 1.1 mole of A is mixed with 2.2 mole of B and the mixture is kept in a one-litre flask till the equilibrium is reached. At equilibrium 0.2 mole of C is formed. If the equilibrium reaction is A + 2B 2C , the value of equilibrium constant is
41) The equilibration , cis -2- pentene trans -2- pentene can be brought about by the use of a catalyst. If a more efficient catalyst is used for this reaction, at a given temperature
a) Equilibrium can be achieved faster
b) Equilibrium can be shifted in the forward direction
c) More cis- pentene will be present at equilibrium
d) More trans-pentene will be present at equilibrium.
42) If the equilibrium constant for the reaction, 2HI H2 + I2 is 0.25, then the equilibrium constant for the reaction H2 +I2 2HI is
43) For which of the following reactions is Kp = Kc ?
c) H2(g)+I2(g) 2HI(g)
44) The value of Kc for dissociation of PCl5 as PCl5(g) PCl3(g) + Cl2(g) is 0.0625 moles/litre at 300Â°C. What is Kp for this reaction ?
45) 24 mL of HI are produced from the reaction of 15 mL of H2 and 17.1 mL of I2 vapour at 444Â°C. The equilibrium constant of the reaction H2 + I2 2HI at 444Â°C is
46) The degree of dissociation of HI at 444Â°C was found to be 23.5%. What are the number of moles of HI, H2 and I2 formed at equilibrium if initially the number of moles of HI was 3?
a) 2.295, 0.3525, 0.3525
b) 1.124, 0.2525, 0.2525
c) 0.3535, 0.3535, 0.3535
47) 3.24 moles of iodine and 9.2 moles of hydrogen were heated at 444Â° C until equilibrium was reached. As a result, 6.16 moles of HI were formed. What is the equilibrium constant?
48) If the equilibrium constant for the reaction N2 + O2 2NO is 0.0842 at 3500 K, what is the fraction of the equilibrium mixture of N2 and O2 converted into NO?
49) 3 g moles of PCl5 are heated in a closed 3 litre vessel. At equilibrium PCl5 is 40% dissociated into PCl3 and Cl2. Calculate the equilibrium constant.
50) At 100Â°C the vapour density of N2CO4 is 26.8. What is the percentage dissociation into NO2 molecules at this temperature ?
c) 74.61 %
51) A mixture of hydrogen and iodine vapour in the molecular proportion 2: 1 was heated at 440oC till the equilibrium H2 + I2 2HI was reached. The equilibrium constant K at this temperature was 0.02. What is the percentage of iodine converted into HI ?
52) The equilibrium constant of the reaction H2 + I2 2HI is 50. If the volume of the container is reduced to half of its original value, the equilibrium constant will be
53) The vapour density of completely dissociated NH4Cl would be
54) Which oxide of nitrogen is the most stable ?
a) 2NO2 (g) N2(g)+2O2(g);K=6.7 x 1016 mole litre -1
b) 2NO (g) N2(g)+O2(g);K=2.2 x 1030 mole litre -1
c) 2N2O5 (g) 2N2(g)+5O2(g);K=1.2 x 1034 mole litre -1
d) 2N2O (g) 2N2(g)+O2(g);K=3.5 x 1033 mole litre -1
55) Which will change the value of the equilibrium constant for the reaction between H2 (g) and I2 (g) ?
a) Adding a catalyst
b) Increasing pressure
c) Increasing temperature
d) Increasing concentration of reactants
56) 3.2 moles of HI were heated in a sealed tube at 444Â°C till the equilibrium was reached. The degree of dissociation of HI is 22% at this temperature. The number of moles of HI present at equilibrium are
57) In which of the following reactions will the equilibrium constant have units of concentration ?
d) In none of the above reactions
58) The equilibrium constant K for the reaction 2HI(g) H2(g) + I2(g) at room temperature is 2.85 and that at 698 K is 1.4 x 10-7 This implies that the forward reaction is
d) Neither exothermic nor endothermic
59) The equilibrium constant for the reaction CaSO4.5H2O(s) CaSO4.3H2O(s) + 2H2O(g) is
60) The equilibrium constant for a reaction A + 2B 2C is 40. The equilibrium constant for the reaction
C B + 1/2A is
b) square root of 1/
c) [1/40] X [1/40]
61) The equilibrium constant for a reaction A + B C+D is 1 x 10-2 at 298 K and 2.0 at 373 K. The reaction is expected to be
d) Neither exothermic nor endothermic
62) For a reaction 4A + 5B 4P + 6Q, the equilibrium constant Kc has the units
a) mole / L
c) mol L
63) For a reaction AB2(g) + 1/2B2(g) AB3(g); D H = – xkJ. More AB3 could be produced by
a) Using a catalyst
b) Removing some B
c) Increasing the temperature
d) Increasing the pressure
64) Inert gas has been added to the equilibrium SO2(g) + l/2O2(g) SO3(g). To which direction will the equilibrium shift ?
c) No effect
65) For the reaction 2NH3(g) N2(g) + 3H2(g), the units of Kp will be
66) For the reaction N2O4(g) 2NO2(g), the degree of dissociation at equilibrium is 0.2 at 1 atmospheric pressure. The equilibrium constant Kp will be
67) @ For the equilibrium system 2HX (g) H2(g) + X2(g), the equilibrium constant is 1 x 10 – 5. What is the concentration of HX if the equilibrium concentration of H2 and X2 are 1.2 x 10-3M and 1.2 x 10 -4 respectively?
68) One mole of N2 is mixed with 3 moles of H2 in a one litre container. If 50% of N2 is converted into NH3 by the reaction N2(g) + 3H2(g) 2NH3(g), then the total number of moles of gas at equilibrium is
69) For the reaction A B ; Kc = 1. For reaction B C; Kc = 2, and for reaction C D; Kc = 3. Kc for the reaction A D is.
70) 4 moles of CO2 were heated in 1dm3 vessel under conditions which produced at equilibrium 25% dissociation into CO and O2. The number of moles of CO produced was
71) For a reaction, H2(g) + I2(g) 2HI(g) at a certain temperature, the value of the equilibrium constant is 50. If the volume of the vessel is reduced to half of its original value, the value of the new equilibrium constant will be
72) Kc for the reaction A+B C+D is 10 at 25Â°C. If a container contains 1, 2, 3 and 4 moles/litre of A, B, C and D respectively at 25Â°C, the reaction shall proceed
a) From left to right
b) From right to left
c) Be in equilibrium
73) The system PCl5(g) PCl3(g) + Cl2(g) attains equilibrium. If the equilibrium concentration of PCl3(g) is doubled, the concentration of Cl2(g) would become
a) 1/4 of its original value
b) 1/2 of its original value
c) Twice of its original value
d) Unpredictable since other changes, if any are not specified.
74) The vapour density of N2O4 at a certain temperature is 30. What is the percentage dissociation of N2O4 at this temperature ?
75) XY2 dissociates as XY2(g) XY (g) + Y(g). When the initial pressure of XY2 is 600 mm Hg, the total equilibrium pressure is 800 mm Hg. Calculate K for the reaction assuming that the volume of the system remains unchanged.
76) The vapour density of PCl5 is 104.16, but when heated to 230Â°C, its vapour density is reduced to 62. The degree of dissociation of PCl5 at 230Â°C is
77) The rate constants of certain forward and backward reactions are 8.5 x 10-5 and 2.38 x 10-4 respectively. The equilibrium constant is
78) At 540 K, PCl5 dissociates as PCl5(g) PCl3(g) + Cl2(g). If this reaction is exothermic, which of the following factors would cause the concentration of PCl5 to decrease in the reaction vessel?
a) Adding Cl2 to the reaction mixture
b) Increasing the volume of the reaction vessel
c) Addition of a catalyst
d) Increasing temperature
79) The equilibrium constant for the reaction 3C2H2 C6H6 is 4.0 at T K. If the equilibrium concentration of C2H2 is 0.5 mole/litre, what is the concentration of C6H6?
d) 0.25 M
80) In the system CaF2(s) Ca2+ + 2F–, increasing the concentration of Ca2+ ions 4 times will cause the equilibrium concentration of F– ions to change to
a) 1/4 of the initial value
b) 1/2 of the initial value
c) 2 times of the initial value
d) 4 times of the initial value
81) For a gaseous equilibrium A + 2B C+ 3D, the partial pressures of A, B, C and D are found to be 0.20, 0.10, 0.30 and 0.50 atm. respectively. The equilibrium constant is
82) When H2 is added to an equilibrium mixture 2HI H2 + I2, at constant temperature, the
c) The degree of dissociation of HI decreases
d) Degree of dissociation of HI increases
83) A mixture of chemicals is reacting in a closed vessel in the forward as well as in the backward directions. If we try to increase either the temperature or the pressure of the vessel, then in accordance with Le Chateliers principle the whole system of chemicals reacting in the vessel chooses to react in that direction in which
a) This change is welcomed
b) It opposes this change
c) The whole system collapses
d) The reaction is endothermic
84) Which of the following will not change the concentration of ammonia in the equilibrium
4NH3(g) + 502(g) 4NO(g) + 6H2O(g) ; D H= +ve
a) Increase of pressure
b) Increase of volume
c) Addition of a catalyst
d) Decrease of temperature
85) One mole of nitrogen is mixed with 3 moles of hydrogen in a closed 3 litre vessel. 20% of nitrogen is converted into NH3. Then Kc for the equilibrium 1/2N2 + 3/2H2 NH3 is about
86) @ The equilibrium constant for the reaction A2(g) + B2(g) 2AB(g) is 100 at 25Â°C. What will be the rate constants for the reactions?
(A) 2AB (g) A2 (g) + B2 (g), and
(B) 1/2A2 (g) + 1/2B2 (g) AB (g) at 25Â°C
87) A reaction reaches a state of equilibrium only when
a) The reactants and products stop reacting
b) The concentration of reactants and products become equal
c) The products react together at the same rate at which they are formed
d) All the reactants and products are in the same state of matter
88) The densities of diamond and graphite are 3.5 and 2.3 gm/mL respectively. The increase of pressure on the equilibrium C (diamond) C (graphite) will
a) Favour the backward reaction
b) Favour the forward reaction
c) Have no effect
d) Increases the reaction rate
89) N2 + 3H2 2NH3. For this reaction, in a one litre vessel, after the addition of an equal number of moles of N2 and H2 an equilibrium state is formed. Which of the following is correct?
90) The number of gram molecules of a substance present in unit volume is termed as the
b) Normal solution
c) Molal concentration
d) Active mass
91) For a reaction 2A + B ® C, where the initial concentration of A = 2 moles/litre, B = 1 mole/litre and C = 0, the concentration of B at equilibrium is 0.5 moles per litre. What is the value of the equilibrium constant for the reaction?
92) In the equilibrium 2A(g) + 3B(g) A2B3(g) + Heat, the forward reaction is favoured by
a) High temperature and high pressure
b) High temperature and low pressure
c) Low temperature and high pressure
d) Low temperature and low pressure
93) The reaction N2 (g) + 3H2 (g) 2NH3 ; D H = – 22.4 kcals. The decomposition of ammonia is
d) Isothermal expansion
94) Which of the following statements is wrong?
a) Chemical equilibrium can be attained from either side of the reaction
b) The rate of forward reaction > rate of backward reaction just before reaching the equilibrium
c) After the equilibrium, the reactions do not stop and continue to take place with equal velocities
d) All are correct
95) A + B AB (in gaseous state) is a reversible reaction. It appears in equilibrium that 0.4 mole of AB is formed when one mole each of A and B are taken . What percentage of A changes into AB ?
96) The equilibrium constant K for the reaction AB + C AC + B is 10. The rate of the forward reaction is 104. The rate constant of the backward reaction is
97) For the homogeneous reaction 4NH3 + 5O2 4NO + 6H2O the equilibrium constant Kc has the units of
d) It is dimensionless
98) The equilibrium concentrations of X, Y and YX2 are 4, 2 and 2 respectively for the equilibrium 2X + Y YX2. The value of the equilibrium constant Kc is
99) In the reaction A + 2B 2C +D, the initial concentration of B was 1.5 times that of A, but equilibrium concentrations of A and B are found to be equal. The equilibrium constant for the reaction is
100) Two moles of PCl5 were heated in a closed vessel of 2 litres. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2. The value of the equilibrium constant is
Ans 1) b
Ans Desc 1) No. of moles of O2 = 96/32 = 3. Active mass = 3.0/2 =1.5 moles/litre
Ans 2) a
Ans Desc 2) K = or Kb =
Ans 3) d
Ans Desc 3) Ka = ca2 or 1.8 x 10-5 = c x [0.02]2, since a = 2/100 = 0.02
or c = = 4.5 x 10-2 = 0.045 M
Ans 4) c
Ans Desc 4) No. of moles 56/28 = 2.
The molar concentration or active mass =2/1 =2.
Ans 5) c
Ans Desc 5) . For this reaction D n = 2 – 4= – 2. Thus D H = D E – 2RT.
Ans 6) b
Ans Desc 6) The removal or decrease in the concentration of the product always favours the forward reaction.
Ans 7) b
Ans Desc 7)
Ans 8) a
Ans Desc 8) A + 2B 2C + D
a b 0 0
(a-x) (b-2x) x x
Given 2x = 0.2 and a = 1.1 Thus x = 0. 1 and b = 2.2
Ans 9) b
Ans Desc 9) K1=1/0.6=1.67.
Ans 10) c
Ans Desc 10) In the reaction N2 + 3H22NH3, an increase in temperature always favours the forward reaction.
Ans 11) c
Ans Desc 11) For this reaction, n =2-1=1. Thus Kp = Kc[RT]
Ans 12) b
Ans Desc 12)
Ans 13) c
Ans Desc 13) . Here n = 2 – 2 = 0.
Thus Kp = Kc = 50.
Ans 14) c
Ans Desc 14)
Ans 15) a
Ans Desc 15) : 2.5. PSO3 =0.331 and
PSO3 = PSO2
Ans 16) d
Ans Desc 16) CH3COOH + C2H5OH ® CH3COOC2H5 + H2O
1 1 0 0
(1-2/3) (1-2/3) 2/3 2/3
Ans 17) d
Ans Desc 17) A + 2B 2C +D
a 1.5 0 0
(a-x) (1.5a-2x) 2x x
Now (a — x) = 2x (given)
Thus x = a/3 = 0.33 a.
Ans 18) a
Ans Desc 18) Kc = . Now If [A]=2x[A], [B] should become 1/2 [B] so that Kc remains constant.
Ans 19) d
Ans Desc 19) Kc is independent of the initial concentration of either the reactants or the products.
Ans 20) b
Ans Desc 20)
Ans 21) c
Ans Desc 21) Pressure has no effect on those reactions in which there is no change in volume.
Ans 22) c
Ans Desc 22) An increase in temperature decreases the value of K. This shows that the reaction is endothermic and proceeding in the reverse direction. The backward reaction is thus endothermic and HI is an endothermic compound. The energy of HI is relatively larger than H2 and I2 and so HI is comparatively unstable.
Ans 23) a
Ans Desc 23) D GÂ° = – 2.303 RT log Kp
= – 2.303 x 2 x 298 log 10-8 = 10.98 k.cals.
Ans 24) d
Ans Desc 24) The value of the equilibrium constant is independent of the concentration of reactants.
Ans 25) d
Ans Desc 25) Catalyst does not change the equilibrium concentration.
Catalyst enables the system to reach a state of equilibrium much quickly.
Ans 26) b
Ans Desc 26) As the forward reaction is accompanied by increase in the number of moles (or volume), it is favoured by low pressure.
Ans 27) c
Ans Desc 27) Increase in pressure causes melting of ice. Solid + Heat Liquid
Ans 28) b
Ans Desc 28) Because more of A will be consumed [A] < [B]
Ans 29) b
Ans Desc 29) Endothermic reaction is favoured by high temperature.
Ans 30) b
Ans Desc 30) An increase in the concentration of one of the products will favour the backward reaction according to Le Chatelier’s s principle
Ans 31) a
Ans Desc 31)
Ans 32) c
Ans Desc 32)
Ans 33) b
Ans Desc 33)
Ans 34) d
Ans Desc 34) . Here D n = 4 – 3 = 1. Thus Kp = Kc [RT] or Kc = Kp/RT = 5 x 10-5/R
Ans 35) c
Ans Desc 35)
2a= 7.07- 7.07a
9.07 a = 7.07
a = 0.78.
If 0.5 moles each of H2 and I2 are taken, then equilibrium
conc. of [H2] = 0.5 -0.39 = 0.11,
Conc. of [I2]= 0.5 -039 = 0.11 and
Conc. of [HI] = 0.78.
Ans 36) b
Ans Desc 36) The value of the equilibrium constant does not depend on a catalyst.
Ans 37) c
Ans Desc 37) Equilibrium constant changes with temperature.
Ans 38) a
Ans Desc 38) Pressure has no effect in those reactions for which n = 0.
Ans 39) d
Ans Desc 39) 2HI H2 + I2
1 0 0
1 – 0.5 0.25 0.25
Ans 40) c
Ans Desc 40) A + 2B 2C
1.1 2.2 0
(1.1-0.1) (2.2-0.2) 0.2 at equilibrium.
Ans 41) a
Ans Desc 41) According to Le Chateliers principle, the reaction will shift in the reverse direction, that is, more cis – pentene is formed.
Ans 42) d
Ans Desc 42) Kf = l/Kb or Kb = 1/0.25 = 4.0
Ans 43) d
Ans Desc 43) n = 2 – 2 = 0. Thus Kp = Kc.
Ans 44) b
Ans Desc 44) For this reaction n = 2- 1 = 1. Thus Kp = Kc RT= 0.0625 0.0821 573 = 2.940.
Ans 45) b
Ans Desc 45) According to Avogadros law the number of molecules in a gas under the same conditions of temperature and pressure is proportional to the volume. Hence volumes of gases may be used instead of molar concentrations. Thus
2x = 24 or x=12
Volume of H2 at equilibrium=(a-x)=15- 12 = 3mL.
Vol of I2 at equilibrium = (a – x) = 17.1 -12 = 5.1 mL
Ans 46) a
Ans Desc 46)
|2HI||H2 + I2|
Now a = 3, x = 23.5% = 0.235
Number of moles of HI = a (1 – 0.235) = 3 0.765 = 2.295
Number of moles of H2 = 3 (0.1175) = 0.3525
Number of moles of I2 = 3 (0. 1 175) = 0.3525.
Ans 47) a
Ans Desc 47)
|H2 + I2 2HI|
Now a = 9.2 moles, b = 3.24 moles, 2x = 6.16 or x = 3.08 moles
Here [H2] = 9.2 -3.08 = 6. 12
[I2] =3.24-3.08 = 0.16
[HI] = 2x = 6.16.
Ans 48) a
Ans Desc 48)
|N2 + O2 2NO|
|1-x||1-x||2x at equilibrium|
or x = 0.169 =16.9%.
Ans 49) a
Ans Desc 49) PCl5 PCl3 + Cl2. at equilibrium, the amount of PCl5 is while that of PCl3 and Cl2 is 1 .2 mole each.
Now, V = 3 litre. Thus [PCl5] = 1.8/3 = 0.6;
[PCl3] = 1.2/3 = 0.4 and [Cl2] = 1.2/3 = 0.4
Ans 50) a
Ans Desc 50) In the absence of any dissociation, the vapour density of N2O4 = mol.wt/2 = 92/2 = 46
Now, degree of dissociation,
Ans 51) a
Ans Desc 51)
|H2 + I2 2HI|
3. 98 x2 + 0.06 x- 0.04 = 0 or
x = 0.0025 or 0.25 %
Ans 52) b
Ans Desc 52) Kp or Kc for a given reaction is constant at equilibrium.
Ans 53) b
Ans Desc 53)
For NH4Cl NH3 + HCl. Here x= 1.
Thus Obs mol. wt = Normal mol.wt/2.
Ans 54) a
Ans Desc 54) The smaller the value of K, the smaller will be the tendency to go towards the forward reaction.
Ans 55) c
Ans Desc 55) K changes only with a change in temperature.
Ans 56) a
Ans Desc 56)
|2HI H2 + I2|
Given x = 22/100 and a = 3.2
[Hl] at equilibrium = a(1-x) =3.2 x= 2.496.
Ans 57) d
Ans Desc 57) All these reactions are taking place with no change in the volume or number of moles. Thus the conc. units in the numerator and denominator cancel each other out.
Ans 58) a
Ans Desc 58) The value of K decreases on increasing temperature. This indicates a shift in the reverse direction. Since an increase in temperature shifts the equilibrium towards the endothermic side the forward reaction should be exothermic
Ans 59) c
Ans Desc 59) The concentrations of solids, that is, [CaSO4.5H2O] and [CaSO4.3H2O] are taken as unity.
Thus K = [H2O]2.
Ans 60) b
Ans Desc 60) 2C® A+ 2B;
K= 1/K and
for C ®
Ans 61) b
Ans Desc 61) The K is increasing by increasing the temperature. This shows that the forward reaction is favoured by an increase in temperature. Thus the forward reaction should be endothermic.
Ans 62) a
Ans Desc 62)
Ans 63) d
Ans Desc 63) The given reaction is exothermic and taking place with decrease in no of moles. Thus low temperature and high pressure are favourable conditions.
Ans 64) c
Ans Desc 64) Inert gas at constant volume causes an increase in total pressure but partial pressures of the individual gases remain unchanged. Thus Kp will not change.
Ans 65) d
Ans Desc 65)
Ans 66) c
Ans Desc 66) N2O4 2NO2. Suppose initial moles of N2O4 = 1. At equilibrium the number of moles of N2O4 = 1 – 0.2 = 0.8. The equilibrium moles of NO2 = 2 x 0.2 = 0.4 mole.
Total moles at equilibrium = 0.8 + 0.4 = 1.2 mole. Pressure – 1 atm.
Thus P N 2O4 =
Ans 67) d
Ans Desc 67) K=
Or [HX]2 =
Ans 68) c
Ans Desc 68)
Ans 69) b
Ans Desc 69)
Thus = 1 X 2 X 3 = 6
Ans 70) b
Ans Desc 70)
|CO2 CO + 1/2O2|
|4 0 0|
No. of moles of CO = 4 x 0.25 = 1 .0
Ans 71) b
Ans Desc 71) The reaction is taking place with no change in the number of moles. Thus the expression for K is independent of volume. Hence K will remain the same.
Ans 72) a
Ans Desc 72)
But Kc=10. For increasing the value of K1 to Kc, the forward reaction should occur ie, from left to right
Ans 73) d
Ans Desc 73) Because the degree of dissociation is not known.
Ans 74) a
Ans Desc 74) The mol. wt of N2O4 = 92. Thus vapour density should be 92/2 = 46, but d = 30. Thus
Ans 75) b
Ans Desc 75)
|XY2 XY + Y|
Now 600 – p + p + p = 800
or p = 800 – 600 = 200
Ans 76) b
Ans Desc 76)
Ans 77) a
Ans Desc 77)
Ans 78) b
Ans Desc 78) Change in volume of a gaseous system at equilibrium generally causes a change in pressure.
Ans 79) a
Ans Desc 79)
Ans 80) b
Ans Desc 80)
[F–]=[F–/2] so that Kc remains constant.
Ans 81) d
Ans Desc 81) = 18.75
Ans 82) c
Ans Desc 82) The addition of product causes the reaction to proceed in the reverse direction. Hence the addition of H2 will decrease the dissociation of HI.
Ans 83) b
Ans Desc 83) According to Le Chatelier, When a system is at equilibrium is subjected to a change of pressure, temperature or concentration, the equilibrium gets shifted in such a way as to counteract the effect of that change
Ans 84) c
Ans Desc 84) A catalyst does not change the position of equilibrium.
Ans 85) a
Ans Desc 85)
|1/2N2 + 3/2H2 NH3|
Ans 86) a
Ans Desc 86)
Ans 87) c
Ans Desc 87) An increase in the concentration of any reactant always favours the forward reaction and so the equilibrium gets shifted to the right. Similarly a decrease in the concentration of the products favours the backward reaction.
Ans 88) a
Ans Desc 88) For a given mass, density of diamond > density of graphite. Thus increase in pressure will favour the backward reaction (the volume of diamond is less than that of graphite for a given mass).
Ans 89) b
Ans Desc 89)
|N2 + 3H2 2NH2|
|1 mole||3 moles|
Thus (a-x)> (a-3x)
|N2 + 3H2 2NH3|
(a – x)
(a – 3x)
Ans 90) d
Ans Desc 90)
Ans 91) c
Ans Desc 91) At equilibrium, [B] = 0.5.
No. of moles of B consumed = 1.0 – 0.5 = 0.5.
No. of moles of A consumed = 0.5 x 2 = 1.0.
Now [A] = 2.0 – 1.0, [C] = 0.5 and [B] = 0.5.
Ans 92) c
Ans Desc 92) Exothermic reactions which takes place with decrease in volume also, are favoured by low temperature and high pressure.
Ans 93) a
Ans Desc 93) Formation of ammonia is exothermic. Thus the dissociation of ammonia is endothermic.
Ans 94) d
Ans Desc 94)
Ans 95) b
Ans Desc 95)
|A + B AB|
But x = 0.4
Thus % of A changing to AB = = 40%.
Ans 96) b
Ans Desc 96)
Ans 97) b
Ans Desc 97)
Ans 98) a
Ans Desc 98)
Ans 99) a
Ans Desc 99) [A] = [B] or (1-x) = (1.5 – 2x) or x = 0.5
|A + 2B 2C + D|
Ans 100) a
Ans Desc 100)
|PCl5 PCl3 + Cl2|
V = 2 litres