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NEET Chemistry 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
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c)1000/ conc. of catalyst
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d)can't be predicted
Explanation
Catalyst has no effect on equilibrium constant. It only helps to attend equilibrium fasterAnswer: (b)
Q.2
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.3
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.4
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
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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.5
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
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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.6
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]
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a)0.30
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b) 0.18
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c)0.17
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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.7
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)
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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.8
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.9
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
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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.10
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.11
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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d)50
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c)4 × 10⁻⁴
Explanation
If we reverse reaction and divide by two, we getNO ↔ ½ N2 + ½ O2 Answer: (d)
Q.12
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
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b) L Mol-1
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c)L-2 Mol-2
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d)Mol L-1
Explanation
For given reaction formula for equilibrium constant is Answer:(b)
Q.13
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.14
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
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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.15
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
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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.16
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?
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a) 1.44 × 10⁻⁵
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b) (0.082 × 773)2 × 1.44 × 10⁻⁵
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c)1.44 × 10⁻⁵ × (0.082 × 500)2
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d)1.44 × 10⁻⁵ × (0.082 × 773)3
Explanation
Equilibrium constant has no effect of pressure, it depends on temperature onlyAnswer: (a)
Q.17
The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if [ IIT 1981]
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a) Temperature is increased and pressure is kept constant
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b)Temperature is reduced and pressure is increased
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c)Both temperature and pressure are increased
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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.18
If equilibrium constant for a reaction is K, then standard free energy change is
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a) ΔGo=-RTlogK
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b) ΔGo=RTlnK
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c) ΔGo=RTlogK
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d) ΔGo=-2.303RTlogK
Explanation
Answer: (d)
Q.19
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]
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a)Decreases with time
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b) Is independent of time
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c)Is zero
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d)Increases with time
Explanation
Value of reaction quotient, Q increases to equilibrium constant K from initial stage of reaction till equilibriumAnswer: (d)
Q.20
The equilibrium constant for reversible reaction N2 + 3H2 ⇔ 2NH3 is K and for the reaction ½ N2 + (3/2)H2 ⇔ NH3 The equilibrium constant is K'. K and K' will be related as [ MP CET 1998]
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a) K=K'
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b) K'=√(K)
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c)K=√(K')
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d)K × K'=1
Explanation
Equilibrium constant is independent of the initial concentration. It varies with temperature only If equilibrium constant is K for reaction A + B ⇋ C + D If equilibrium constant is K' for reaction n(A+B) ⇋ n(C+D) Then K' = Kn ∴ K' = K1/2 Answer: (b)
Q.21
Which oxide of nitrogen is most stable at 273 K ?
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a)
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b)
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c)
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d)
Explanation
Value of K gives an idea about the extent of reaction. Larger the value of K, more is the extent of reaction. NO2. with the least value of K is dissociated to least extent among the given oxides Answer:(d)
Q.22
The equilibrium constant for reaction SO3 → SO2 + ½O2 is Kc=4.9 × 10-the value of Kc for the reaction 2SO2 + O2 → 2SO3 will be [ AIEEE 2006]
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a) 416
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b) 2.4 × 10⁻³
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c) 9.8 × 10⁻²
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d) 4.9× 10⁻²
Explanation
For reaction SO3 → SO2 + ½O2 For reactionSO2 + ½O2 → SO3 For reaction 2SO2 + O2=2SO3 Answer: (a)
Q.23
the equilibrium constant Kp for the reaction PCl5 ↔ PCl3 + Cl2 is 16 If the 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
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d)4
Explanation
Equilibrium constant is independent of initial concentration. It varies with temperature onlyAnswer: (c)
Q.24
For a reaction N2 + 3 H2 → 2NH3; °H=-22 kcal. If energy of activation for this reaction is 70 kcal, the energy of activation for backward reaction is
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a) 70 kcal
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b) 92 kcal
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c)48 kcal
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d)None of these
Explanation
Backward reaction is endothermic Δ H=Ea(forward) - Ea(backward) Ea(backward)=70 +2 0=92 kcalAnswer: (b)
Q.25
the value of reaction quotient before any reaction occurs is
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a) ∞
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b) -1
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c)0
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d)1
Explanation
Before occurrence of reaction only reactants are present in the reaction mixture. Concentration of product is zero Answer:(c)
Q.26
Consider the following equilibrium in closed container N2O4 ⇔ 2NO2 . At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statement holds true regarding the equilibrium constant (Kp) and degree of dissociation ( α) [ IIT 2002]
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a) Neither Kp nor α changes
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b) Both Kp nor α changes
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c) Kp changes but α does not change
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d) Kp does not change but α changes
Explanation
At constant temperature, equilibrium constant Kp does not change. By decreasing volume at constant temperature, pressure is increased which shifts the equilibrium in the direction in which number of moles decreases. i.e. backward direction. Since direction of equilibrium is changed, degree of dissociation α would also change Answer: (d)
Q.27
For reaction Co + K2O ⇔ Co2 + H2 at given temperature, the equilibrium amount of CO2 can be increased by [ IIT 1998]
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a)Adding a suitable catalyst
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b) Adding an inert gas
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c)Decreasing the volume of the container
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d)Increasing the amount of CO
Explanation
Increase in concentration of one of the reactants shifts the equilibrium in direction in which that reactant is consumed. i.e in forward direction to produce more of productsAnswer: (d)
Q.28
The yields of NH3 in the reaction N2 + 3H2 ↔ 2NH3 ; ΔH=-22.08 kcal is affected by [ Pb CET 1997]
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a) Change in pressure and temperature
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b) Change in temperature and concentration of H2
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c)Change in pressure and concentration of N2
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d)Change in pressure, temperature and concentration of N2
Explanation
option 'd' by Le Chatelier principleAnswer: (d)
Q.29
If 1.0 mole of I2 is introduced into 1.0 litre flask at 1000K which leads to dissociation, at equilibrium, ( Kc=10-6), which one is correct?
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a) [ I2(g)] > I-]
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b) [ I2(g)] < I-]
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c) [ I2(g)]=I-]
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d) [ I2(g)]=½ I-]
Explanation
Equilibrium constant has no effect of pressure, it depends on temperature only The reaction takes place as follows I2( g) ⇋ 2I (g), Kc=10-6 Clearly, the equilibrium constant favours I2( g) Answer:(a)
Q.30
At constant temperature, the equilibrium constant ( Kp) for the decomposition reaction N2O4 ⇔ is expressed by Kp=4x2P/(1-x2) where P is pressure, x is extent of decomposition. Which of the following statement is true [ IIT 2001]
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a) Kp increases with increase with increase of P
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b) Kp increases with increase with increase of x
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c) Kp increases with decrease with increase of P
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d) Kp remains constant with change in P or x
Explanation
Kp changes only with temperature Answer: (d)
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