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NEET Chemistry MCQ
Quiz 5
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Q.1
In the reaction N2(g) + O2(g) ↔ 2NO(g) -180.7 kJ, on increasing the temperature, the production of NO [ Chandhigarh CET 2000]
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a) increases
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b) decreases
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c)remains same
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d)cannot be predicted
Explanation
Reaction is exothermic thus reaction moves backward direction, Concentration of NO decreasesAnswer: (b)
Q.2
A 1M solution of glucose reaches dissociation equilibrium according to the equation given below:6HCHO ↔ C6H12O6What is the concentration of HCHO at equilibrium if equilibrium constant is 6×1022 ..[MP PMT 2000]
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a) 1.6×10⁻⁸M
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b)3.2×10⁻⁶M
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c)3.2×10⁻⁴M
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d)1.6×10⁻⁴M
Explanation
Concentration at equilibrium is HCHO=6x and C6H12O6=1-x From formula for constant Answer:(d)
Q.3
Kc for reactionN2(g) +O2(g) ↔ 2NO(g) at 300K is 4.0×10-Kp for above reaction will be (R=2 cal mol- K-1) ..[CBSE PMT 1994]
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b) 4 ×10⁻⁶
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a) 2.4 ×10⁻³
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c) 4 ×10⁻⁶(RT)2
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d) 16 × 10⁻¹²
Explanation
Δng=2-2=0∴ Kp=Kc Answer: (b)
Q.4
The reaction, SO2 + Cl2 → SO2Cl2 is exothermic and reversible. A mixture of SO2(g), Cl2(g) and SO2Cl2(l) is at equilibrium in a closed container. Now a certain quantity of extra SO2 is introduced into the container, the volume remaining the same. Which of the following is are true? [ ISM Dhanbad 1994]
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a)The pressure inside the container will not change
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b) The temperature will not change
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c)The temperature will increase
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d)The temperature will decrease
Explanation
On adding SO2, equilibrium shift forward. As it is exothermic reaction, temperature will increaseAnswer: (c)
Q.5
On the basis of Le Chatellier principle, predict which of the following conditions would be unfavorable for the formation of SO3? Given that 2SO2 + O2 ↔ 2SO3 ; ΔH=-42 kcal .. [ MP PET 1998]
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a) Low pressure
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b) High pressure
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c)high temperature
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d)high concentration of SO2
Explanation
reaction is exothermic thus increase in temperature is not favourableAnswer: (c)
Q.6
4 moles of A are mixed with 4 moles of B when 2 moles of C are formed at equilibrium according to the reactionA + B ↔ C + D The value of the equilibrium constant is ..[ CPMT 1992]
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a) 4
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b) 1
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c)1/2
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d)1/4
Explanation
At equilibrium concentration of A, B, C and D is 2 moles each thus equilibrium constant will be 1 Answer:(b)
Q.7
The rate of forward reaction is two times that of the reverse reaction at a given temperature and identical concentration. K is
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a) 0.5
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b) 1.5
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c) 2.5
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d) 2
Explanation
K=rate of forward/ rate of reverse Answer: (d)
Q.8
Consider the reaction PCl5(g) ↔ PCl3(g) + Cl2(g) in a closed container at equilibrium. At a fixed temperature what will be the effect of adding more PCl5 on the equilibrium concentration of Cl2(g)? [ BHU 1991]
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a)It decreases
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b) It increases
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c)It remains unaffected
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d)It can not be predicted without the value of Kp
Explanation
Answer: (b)
Q.9
Which of the following is not affected by pressure change? [ Pb CET 1999]
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a) 2NO2(g) ↔ N2O4(g)
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b) CO2(g) + H2O (l)↔ CO2 (solution)
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c)H2(g) + I2(g) ↔ 2HI(g)
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d)2O3(g) ↔ 3O2(g)
Explanation
In option 'c' number of moles of gases on reactant and product side is equal thus no change in pressureAnswer: (c)
Q.10
An equilibrium mixture for the reaction2H2S (g) ↔ 2H2(g) + S2(g) had one mole of hydrogen sulphide, 0.2 mole of H2 and 0.8 mole of S2 in a two litre vessel. the value of K in mole/litre is [ IIT 1991]
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a) 0.005
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b) 0.016
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c)0.08
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d)0.160
Explanation
Concentration of hydrogen sulphide=0.5, concentration of H2 is 0.1 and concentration of S2 is 0.4USe formula for equilibrium constant Answer:(b)
Q.11
In the reaction PCl5(g) ↔ PCl3 + Cl2(g) the equilibrium concentration of PCl5 and PCl3 are 0.4 and 0.2 mole/litre respectively. If the value of K is 0.5, what is the concentration of Cl2 mole/litre [ EAMCET 2003]
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a) 2.0
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b) 1.5
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c) 1.0
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d) 0.5
Explanation
For formula Answer: (c)
Q.12
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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c) 5.0 × 1015
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b) 5.0 × 10⁻⁵
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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.13
A weak acid HX has the dissociation constant 1 × 10⁻⁵M. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1M solution of NaX is
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a) 0.0001%
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b) 0.01%
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c) 0.1%
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d) 0.15%
Explanation
For a salt of weak acid (HX) and strong base ( NaOH) We know, Percentage degree of hydrolysis=10-4 × 100=10-2 =0.01% Answer: (b)
Q.14
The equilibrium constant for the reaction H2(g) + I2(g) ↔ 2HI(g) is 32 at given temperature. the equilibrium concentrations of I2 and HI are 0.5×10⁻³ and 8×10⁻³ M respectively. the equilibrium concentration of H2 is ..[Pb CET sample paper 1988]
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c)2 ×10⁻³
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a) 1 ×10⁻³ M
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b) 0.5 ×10⁻³ M
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d)4 ×10⁻³
Explanation
Answer:(d)
Q.15
The decomposition of N2O4 to NO2 is carried out at 280°C in chloroform. When equilibrium is reached 0.2 mol of N2O4 and 2×10⁻³ mol of nO2are present in 2 litre solution. The equilibrium constant for the reaction N2O4 ↔ 2NO2 is ..[AIIMS 1984]
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a) 1 ×10⁻³
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b) 2 ×10⁻³
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c)1 ×10⁻⁵
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d)2 ×10⁻⁵
Explanation
Answer:(c)
Q.16
in a chemical reaction, the rate constant for the backward reaction is 7.5×10⁻⁴ and the equilibrium constant is 1.the rate constant for forward reaction is ..[KCET 1989]
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a) 5 ×10⁻⁴
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b)2 ×10⁻³
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c) 1.125 ×10⁻³
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d) 9 ×10⁻⁴
Explanation
Answer:(c)
Q.17
For the reaction equilibrium N2O4(g) ↔ 2NO2(g)the concentration of N2O4 and NO2 at equilibrium are 4.8×10⁻² and 1.2×10⁻² molL-1 respectively. the value of K for reaction is ..[AIEEE 2003]
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a) 3×10⁻¹ molL-1
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b) 3×10⁻³ molL-1
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c) 3×103 molL-1
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d) 3.3×102 molL-1
Explanation
Use formula for equilibrium constant Answer: (b)
Q.18
The equilibrium constant for reaction, N2(g) + O2 is 4×10⁻⁴ at 2000KIn presence of a catalyst, equilibrium is attended ten times faster. therefore, the equilibrium constant, in presence of the catalyst, at 2000 K is : [ MLNR 1994]
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d)difficult to compute without more data
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a)40 ×10⁻⁴
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b) 4 ×10⁻⁴
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c)4 × 10⁻³
Explanation
Catalyst do not affect equilibrium constantAnswer: (b)
Q.19
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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b) 1 × 108 M
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a) 1 × 10⁻¹³ M
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c) 2 × 10⁻¹³ M
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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.20
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
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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.21
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.22
The dissociation constant of weak 0.1M acid solution is 4.9×10⁻⁸, its percentage ionization is .. [ AFMC 1999]
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a) 0.07%
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b) 0.007%
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c)0.7%
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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.23
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
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b) 1.8×10⁻⁵M
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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.24
Solubility product of PbCl2 at 298 K is 10-At this temperature solubility of PbCl2 in mol/L is .. [ AFMC 2006]
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a)(10-6) 1/2
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b) (10-6) 1/3
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c)(0.25×10⁻⁶) 1/3
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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.25
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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d)none of the above
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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
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.26
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.27
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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d) None of these
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a) 1.08 × 10⁻⁷
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b) 1.08 × 10⁻⁵
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c) 1.08 × 10⁻⁹
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.28
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]
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a)0.2
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b) 0.4
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c)1.6
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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.29
When 0.1 mol of CH3NH2 ( ionization constant,Kb=5 ×10⁻⁴) is mixed with 0.08 mol HCl and the volume is maid up toFind the [H+] of resulting solution. [ IIT 2005]
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a) 8 × 10⁻⁵
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b) 2 × 10⁻¹¹
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c) 1.23 × 10⁻⁴
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d) 8 × 10⁻¹¹
Explanation
Resulting solution contains [salt]=0.08 and [base]=0.1 - 0.08=0.02 Now pKb=-logKb pKb=3.3 pH + pOH=14 pH=140- 3.90=10.1 Applying pH=-log[H+] 10.1=-log [ H+] taking antilog Antilog [ -11 + 0.9]=H+H+=8 ×10⁻¹¹ Answer: (d)
Q.30
In a saturated solution of the sparingly soluble strong electrolyte AgIO3 molecular mass=283, the equilibrium which sets in is AgIO3 ⇔ Ag+ + IO3- If the solubility product constant Ksp of AgIO3 at given temperature is 1.0 ×10⁻⁸, what is the mass of AgIO3 contained in 100 ml of its saturated solution [ AIEEE 2007]
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a)1.0 ×10⁻⁴ g
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b) 28.3 ×10⁻² g
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c)2.83 ×10⁻³ g
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d)1.0 ×10⁻⁷ g
Explanation
AgIO3 ⇔ Ag+ + IO3- s=solubility ∴ Ksp=s2 s2=1.0 ×10⁻⁸ or s=1.0 × 10⁻⁴ mol/L s=1.0 × 10⁻⁴ (283) g/L for 100 ml solution s=2.83 ×10⁻³ g Answer: (c)
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