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NEET Chemistry MCQ
Chemical And Ionic Equilibrium Mcq
Quiz 1
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Q.1
The equilibrium constant at 323°C isWhat would be its value in the presence of catalyst of a catalyst in the forward reaction?A + B ⇔ C + D + 38 kcal
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a) 1000 × conc. of catalyst
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b) 1000
0%
c)1000/ conc. of catalyst
0%
d)can't be predicted
Explanation
Catalyst has no effect on equilibrium constant. It only helps to attend equilibrium fasterAnswer: (b)
Q.2
Find out the solubility of Ni(OH)2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2 is 2 × 10-15.... [NEET 2020]
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a) 1 × 10-13 M
0%
b) 1 × 108 M
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c) 2 × 10–13 M
0%
d) 2 × 10–8 M
Explanation
Ni(OH)2 ⇒ Ni2+ + 2[OH-] KSP = Ni2+ [OH-]2 Concentartion of Ni2+ = s and OH- = 2s On addition of NaOH NaOH ⇒ Na+ + OH- (concentration of OH- = 0.1M New concentration of OH- = 2s + 0.1 Thus New KSP KSP = s × (2s +0.1)2 KSP = s(4s2 +0.01+0.1s) KSP = 4s3 +0.01s + 0.4s2 given KSP =2 × 10-15 We will not consider s3 and s2 as s is very very small ∴ 2 × 10-15 = 0.01s 2×10-15 = 0.01s s = 2× 10-13 M Answer: (c)
Q.3
pH of a saturated solution of Ca(OH)2isThe solubility product (Ksp ) of Ca(OH)2 is: ... [ NEET 2019]
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a) 0.5 × 10–15
0%
b) 0.25 × 10–10
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c)0.125 × 10–15
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d) 0.5 × 10–10
Explanation
pH = 9 thus pOH = 14 -pH = 5 pOH = - log[OH+] ∴ [OH-] = 10-5 Ca(OH)2 ⇒ Ca2+ + 2OH KSP = [Ca2+] [OH-]2 There is 2OH- per Ca2+ Ca2+ = ½ 10-5 KSP= [Ca2+][OH-]2 KSP= ½ 10-15 × (10-5)2 KSP= 0.5×10-15 Answer: (a)
Q.4
Conjugate base for Bronsted acids H2O and HF are : [NEET 2019]
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a) OH- and H2F+, respectively
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b) H3O+ and F-, respectively
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c) OH- and F-, respectively
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d) H3O+ and H2F+, respectively
Explanation
Bronsted base accepts proton Answer:(c)
Q.5
Which will make basic buffer? [NEET 2019]
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a) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH3COOH
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b) 100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH
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c) 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH
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d) 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Explanation
Now according to equation if concentration is doubled reaction will go forward but equilibrium constant will not change. so long as temperature is constant Answer: (c)
Q.6
The solubility of BaSO4 in water is 2·42 × 10–3 gL–1 at 298 K. The value of its solubility product (Ksp) will be ... [NEET 2018] (Given molar mass of BaSO4 = 233 g mol–1)
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a) 1·08 × 10-14 mol2 L–2
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b) 1·08 × 10–12 mol2 L–2
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c) 1·08 × 10–10 mol2 L–2
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d) 1·08 × 10–8 mol2 L–2
Explanation
s = 2.42 ×10-3 BaSo4 ⇒ Ba2+ + SO42- Ksp = (s) (s) KSP = s2 KSP = (1.04 ×10-5)2 KSP = 1.08 × 10-10 mol2 L-2 Answer: (c)
Q.7
For a reaction A + B ↔ 2C2 moles of A and 3 moles of B are allowed to react. If equilibrium constant is 4 at 400°C, then the mole of C at equilibrium is [ AFMC 1997]
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a) 1
0%
b) 2.4
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c)3.6
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d)4
Explanation
Initial concentrations are given and asked to calculate concentration of C at equilibriumAccording to given chemical equation on consumption one mole each of A and B produce 2 mole of CThus if 'x' moles of A and B are consumed then concentration at equilibrium will be concentration of A=2-x and concentration of B=3-x while concentration of C=2x Now according to formula on simplification we getx=1.2 thus answer 2x=2.4Answer: (b)
Q.8
The dissociation constant of weak 0.1M acid solution is 4.9×10-8, its percentage ionization is .. [ AFMC 1999]
0%
a) 0.07%
0%
b) 0.007%
0%
c)0.7%
0%
d)0.0007%
Explanation
According to Ostwald's dilution law, for weak acids,Ka=C α 2Here Ka=4.9×10-8concentration C=0.1M4.9×10-8=0.1(α 2) Thus α=7×10-4=0.07% Answer:(a)
Q.9
Ionization constant of acetic acid is 1.8 ×10-The concentration of H+ ions in 0.1M solution is .. [ AFMC 2000]
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a) 1.8×10-3M
0%
b) 1.8×10-5M
0%
c) 1. 3×10-3M
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d) 1.34×10-3M
Explanation
For weak acid Ka=[H+]2 / C Here Ka=1.8×10-5 C=0.1M on substitution we get 1.8×10-5=[H+]2 / 0.1 On solving we get [H+]2=1.8×10-6Concentration of H+=1.34×10-3MAnswer: (d)
Q.10
Solubility product of PbCl2 at 298 K is 10-At this temperature solubility of PbCl2 in mol/L is .. [ AFMC 2006]
0%
a)(10-6) 1/2
0%
b) (10-6) 1/3
0%
c)(0.25×10-6) 1/3
0%
d)(0.25×10-6) 1/2
Explanation
Let solubility of PbCl2 be x mole / L, then solution will have 'x' moles of Pb-2 and 2x moles of Cl- ionsThus Ksp=[Pb-2][Cl-]210-6=(x)(2x)2on solving above equation we get x=(0.25×10-6) 1/3Answer: (c)
Q.11
If 0.1M of a weak acid is taken, and its percentage of degree of ionization is 1.34%, then its ionization constant will be..[ AFMC 2005]
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a) 0.8×10-5
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b) 1.79×10-5
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c)0.182×10-5
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d)none of the above
Explanation
Formula for weak acid Ka=Cα2 / ( 1 - α) here α=1.34%=0.0134concentration C=0.1MOn substituting values and on solving we getKa=0.182×10-5Answer: (c)
Q.12
The pH of 10-8 M HCl solution is [ AFMC 2008]
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a) 8
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b) more than 8
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c)between 6 and 7
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d)slightly more than 7
Explanation
Here H+ concentration is very less, therefore concentration of H+ of water must be taken in to accountThus Total H+ ions=H+ ions of water + H+ ions of HCl total H+ ions=10-7 + 10-8total H+ ions=10×10-8 + 10-8total H+ ions=11×10-8now pH=- log [H+]pH=-log[10-8]on solving we get pH=6.958 Answer:(c)
Q.13
Solubility product of a salt AB is 10-8M2 in a solution in which the concentration of A+ ions is 10-3M. The salt will precipitate when the concentration of B- ion is kept .. [ AFMC 2008]
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a) between 10-8 to 10-7 M
0%
b) between 10-7 to 10-8 M
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c) > 10-5 M
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d) < 10-8 M
Explanation
Precipitation occurs when ionic product Ip > Ksp [ A +][B+] > Ksp 10-3 × [ B-] > 10-8 On solving we get [ B-] > 10-5 Answer: (c)
Q.14
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.5 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is [ AIEEE 2005]
0%
a)0.30
0%
b) 0.18
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c)0.17
0%
d)0.11
Explanation
NH4HS is solid hence donot contribute to pressure Initial pressure of NH4=0.5atm Increase in pressure due to decomposition=x atm pressure at equilibrium=0.5 + x Initial pressure of H2S=0 Increase in pressure=x br/> pressure at equilibrium=x Total pressure at equilibrium=0.5 + 2x=0.84 atmx=0.17 atm Answer: (d)
Q.15
Phosphorus pentachloride dissociates as follows in closed reaction vessel PCl5 → PCl3 + Cl2 If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is 'x', the particle pressure of PCL3 will be [ AIEEE 2006]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
From given Total number of moles=1-x+x+x=1+x Thus partical pressure of PCl3=mole fraction × total pressure option"a" is correctAnswer: (a)
Q.16
For reaction, H2 + I2 ⇔ 2HI , the equilibrium constant Kp changes with ..[ IIT 1981]
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a) Total pressure
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b) Catalyst
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c)The amount of H2 and I2 present
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d)Temperature
Explanation
Given reaction has equal number of moles on both sides of reaction so equilibrium constant is affected only by temperature and not pressure Answer:(d)
Q.17
The equilibrium constant for the reaction PCl5 ⇔ PCl3 + Cl2 is 16 If the original volume of the container is reduced to half of its original volume, the value of Kp for the reaction at the same temperature will be
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a) 32
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b) 64
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c) 16
0%
d) 4
Explanation
Since volume is reduced thus pressure is increased, so initial concentration is increased but equilibrium constant is independent of initial concentration Answer: (c)
Q.18
In a reaction A2 + 4B2 ⇔ 2AB4 ; ΔH
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a)Low temperature and high pressure
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b) High temperature and low pressure
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c)Low temperature and low pressure
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d)High temperature and high pressure
Explanation
For given reaction ΔH < 0 or it is negative thus reaction is exothermic accompanied with decrease in number of moles. therefore, favoured at low temperature and high pressureAnswer: (a)
Q.19
the equilibrium constant, Kc for reaction N2 + O2 ⇔ 2NO at temperature T is 4 ×10-The value of KC for reaction NO ↔ ½ N2 + ½ O2 at the same temperature is [ AIEEE 2004]
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a) 2.5 × 102
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b) 0.02
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c)4 × 10-4
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d)50
Explanation
If we reverse reaction and divide by two, we getNO ↔ ½ N2 + ½ O2 Answer: (d)
Q.20
In the gas phase reaction C2 H4 + H2 ⇔ C2H6, the equilibrium constant can be expresseed in units
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a) L-1 Mol-1
0%
b) L Mol-1
0%
c)L-2 Mol-2
0%
d)Mol L-1
Explanation
For given reaction formula for equilibrium constant is Answer:(b)
Q.21
The first and second dissociation constant of an acid H2A are 1.0×10-5 and 5.0 ×10-10 respectively. The overall dissociation constant of the acid will be [ AIEEE 2007]
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a) 0.2 × 105
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b) 5.0 × 10-5
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c) 5.0 × 1015
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d) 5.0 × 10-15
Explanation
First dissociation constant H2A → H+ + HA- Second dissociation constant HA- → H+ + A2- For total reaction Note you many use result K=K1 × K2 directly Answer: (d)
Q.22
In a reversible reaction, two substances are in equilibrium. If the concentration of each one is reduced to half, the equilibrium constant will be [ Pajab CET 1997]
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a)Reduced to half of it original value
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b) Doubled
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c)Same
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d)Reduced to one forth of its original value
Explanation
Equilibrium constant does not vary with initial concentrationAnswer: (c)
Q.23
Sulphide ions in alkaline solution react with solid sulphur to form polyvalent sulphide ions. the equilibrium constant for the formation of S22- from S and S32- from S and S2- ions are 1.7 and 5.3 respectively. Equilibrium constant for the formation of S22- from S32- and S is
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a) 1.33
0%
b) 3.11
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c)4.21
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d)1.63
Explanation
2S + S2- ⇔ S32- K2=5.3 .....(1) S + S2- ⇔ S22- K1=1.7 .....(2) ∴ S + S22- ⇔ S32- K=?... (3) Form above equation we can say that eq(1) is two step reaction of eq(2) and eq(3) Thus K2=K × K2 5.3=K × 1.7 K=3.11 Answer: (b)
Q.24
in a system A(l) ⇔ B(l) + 2C(l) doubling the equilibrium concentration of B will cause the equilibrium concentration of C to change to
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a) Two times its original
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b) One half of its original value
0%
c)(1/√2) times the original value
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d)√2 times the original value
Explanation
Equilibrium constant is independent of initial concentration B1 and C1 be the concentrations in first case B2 and C2 be the concentrations in second case Now equilibrium constant of given reaction is K=[B][C]2 Answer:(c)
Q.25
Ag+ + NH3 → [Ag(NH3)+] ; K1=6.8 × 10-3 [Ag(NH3)+] + NH3 → [Ag(NH3)2]2+ K2=1.6 × 10-3 then the formation constant of [Ag(NH3)2]2+ is [ IIT 2006]
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a) 1.08 × 10-7
0%
b) 1.08 × 10-5
0%
c) 1.08 × 10-9
0%
d) None of these
Explanation
Formation of [Ag(NH3)2]2+ takes place in two stapes Now K=K1 × K2 K=6.8 × 10-3 × 1.6 × 10-3 K=1.08 × 10-5Answer: (b)
Q.26
Calculate the partial pressure of carbon monoxide from the following data CO2 + C ↔ 2CO ; Kp=8 × 10-2 CaCo3 ⇔ CaO + CO2 ; Kp=2 [ Orissa JEE 2004]
0%
a)0.2
0%
b) 0.4
0%
c)1.6
0%
d)4.0
Explanation
Adding above two equations we get CaCo3 + C ⇔ CaO + 2CO Since above reaction is two step reaction K'=8 × 10-2 × 2 K'=16 ×10-2 Now since Cao is solid and 2 moles of Co are produced from above equation K'=P2 , Here P is pressure of Co gas Thus P=√K'=√(16×10-2)=0.4Answer: (b)
Q.27
For the reaction, N2 + 3H2 ⇔ 2NH3 at 500°C , the value of Kp is 1.44 × 10-What will be the value of Kp at low pressure where the gases are behaving almost ideally?
0%
a) 1.44 × 10-5
0%
b) (0.082 × 773)2 × 1.44 × 10-5
0%
c)1.44 × 10-5 × (0.082 × 500)2
0%
d)1.44 × 10-5 × (0.082 × 773)3
Explanation
Equilibrium constant has no effect of pressure, it depends on temperature onlyAnswer: (a)
Q.28
The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if [ IIT 1981]
0%
a) Temperature is increased and pressure is kept constant
0%
b)Temperature is reduced and pressure is increased
0%
c)Both temperature and pressure are increased
0%
d)Both temperature and pressure are decreased
Explanation
Given reaction is exothermic, associated with decrease in number of moles. Therefore, the reduction in temperature and increase in pressure is favoured Answer:(b)
Q.29
If equilibrium constant for a reaction is K, then standard free energy change is
0%
a) ΔGo=-RTlogK
0%
b) ΔGo=RTlnK
0%
c) ΔGo=RTlogK
0%
d) ΔGo=-2.303RTlogK
Explanation
Answer: (d)
Q.30
When two reactants A and B are mixed to give products C and D the reaction quotient, Q at the initial stage of reaction ..[ IIT 2000]
0%
a)Decreases with time
0%
b) Is independent of time
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c)Is zero
0%
d)Increases with time
Explanation
Value of reaction quotient, Q increases to equilibrium constant K from initial stage of reaction till equilibriumAnswer: (d)
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