Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The enthalpy and entropy change for the reaction:

Br2 (l) + Cl2 (g)→ 2BrCl (g)

are 30 kJ mol-1 and 105 JK-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is:

  • 285.7 K 

  • 273 K

  • 450 K

  • 450 K


A.

285.7 K 

At equilibrium, Gibbs free energy change (ΔGo) is equal to zero. The following thermodynamic relation is used to show the relation of ΔGo with the enthalpy change (ΔHo) and entropy change (ΔSo)

ΔGo  = ΔHo - ΔSo
0 = 30 x 103 (J mol-1) - T x 105 (J K-1) mol-1

Therefore comma
straight T space equals space fraction numerator 30 space straight x space 10 cubed over denominator 105 end fraction space straight K thin space equals space space 285.71 space straight K

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2.

The orientation of an atomic orbital is governed by:

  • azimuthal quantum number

  • spin quantum number

  • magnetic quantum number

  • magnetic quantum number


C.

magnetic quantum number

The orientation of an atomic orbital is governed by magnetic quantum number.

799 Views

3.

In which of the following molecules are all the bonds not equal? 

  • ClF3

  • BF3

  • AlF3

  • AlF3


A.

ClF3

In ClF3 all bonds are not equal due to trigonal -bipyramidal (sp3d-hybridisation) geometry of ClF3 molecule.


BF3 and AlF3 show trigonal symmetric structure due to sp2-hybridisation.


NF3 shows pyramidal geometry due to sp3 hybridization.

1107 Views

4.

The enthalpy of combustion of H2, cyclohexene (C6H10) and cyclohexene (C6H12) are -241, -3800 and -3920 kJ per mol respectively.The heat of hydrogenation of cyclohexane is:

  • -212 kJ mol

  • +121 kJ mol

  • +242 kJ per mol

  • +242 kJ per mol


A.

-212 kJ mol



ΔH= [ΔH of combustion of cyclohexane -(ΔH of combustion of cyclohexene +ΔH of combustion of H2)]

= -[-3920 -(3800-24)] kJ
= - [3920 + 4041] kJ
=-[121] kJ
=--121 kJ

1278 Views

5.

Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE?

  • H2 (g) + Br2 (g) →2HBr (g)

  • C (s) + 2 H2O (g) → 2 H2 (g) + CO2 (g)

  • PCl5 (g) →PCl3 (g) + Cl2 (g) 

  • PCl5 (g) →PCl3 (g) + Cl2 (g) 


A.

H2 (g) + Br2 (g) →2HBr (g)

As we know that
 ΔH = ΔE + PΔV

ΔH = ΔE +ΔnRT ..(1)
where ΔH → change in enthalpy of the system (standard heat at constant pressure)

Δ E → change in internal energy of system (Standard heat at constant volume)
Δn → no. of gaseous moles of product - no. of gaseous moles of reactant

R → gas constant
T → absolute temperature

If Δ n = 0 for reactions which is carried out in an open container, therefore, Δn = 0 for reactions which are carried out in an open container, therefore, ΔH  =ΔE
so for reaction (1) Δn = 2-2 = 0
Hence, for reaction (1) , ΔH =ΔE 

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6.

Which of the following pairs consitutes a buffer?

  • HNO2 and NaNO2

  • NaOH and NaCl

  • HNO3 and NH4NO3

  • HNO3 and NH4NO3


A.

HNO2 and NaNO2

A pair constituent with an HNO2 and NaNO2 because HNO2 is weak acid and NaNO2 is a salt of the weak acid (HNO2) with a strong base (NaOH). Hence, it is an example of acidic buffer solution.

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7.

The hydrogen ion concentration of a 10-8 M HCL aqueous solution at 298 K (Kw = 10-14) is:

  • 1.0 x 10-6

  • 1.0525 x 10-7 M

  • 9.525 x 10-8 M

  • 9.525 x 10-8 M


B.

1.0525 x 10-7 M

In aqueous solution of 10-8 M HCl, [H+] is based upon the concentration of H+ ion of 10-8 M HCl and concentration of H+ ion of water Kw of H2O = 10-14 = [H+][OH-] or [H+] = 10-7 M (due to neutral behaviour) os, in aqueous solution of 10-8 M HCl,
[H+] = [H+] of HCl +[H+] of water

= 10-8 + 10-7 = 11 x 10-8 M
1.10 x 10-7 M

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8.

For the reaction,

CH4  (g) + 2 O2 (g) ⇌ CO2 (g) + 2H2O (l), 

ΔrH = - 170. 8 kJ mol-1

Which of the following statements is not true? 

  • At equilibrium, the concentrations of CO2 (g) and H2O (l) are not equal

  • The equilibrium constant for the reaction is given by Kpfraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O subscript 2 right square bracket end fraction

  • Addition of CH4 (g) or O2 (g) at equilibrium will cause a shift to the right

  • Addition of CH4 (g) or O2 (g) at equilibrium will cause a shift to the right


B.

The equilibrium constant for the reaction is given by Kpfraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O subscript 2 right square bracket end fraction

For the reaction,

CH4  (g) + 2 O2 (g) ⇌ CO2 (g) + 2H2O (l), 

ΔrH = - 170. 8 kJ mol-1

This equilibrium is an example of heterogeneous chemical equilibrium.Hence, for it

straight K subscript straight c space equals space fraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O right square bracket squared end fraction space... space left parenthesis straight i right parenthesis
left parenthesis equilibrium space constant space on space the space basis space of space conc. right parenthesis
and space straight K subscript straight p space equals space fraction numerator straight P subscript Co subscript 2 end subscript over denominator straight P subscript CH subscript 4 end subscript space straight x space straight P subscript straight O subscript 2 end subscript squared space end fraction space space space space.... space left parenthesis ii right parenthesis
left parenthesis Equilibrium space constant space according space to space partial space pressure right parenthesis
Thus space in space it space concentration space of space CO subscript 2 space left parenthesis straight g right parenthesis space and space straight H subscript 2 straight O space left parenthesis straight l right parenthesis space are space not space equilibrium
The space equilibrium space constant space left parenthesis straight K subscript straight p right parenthesis space equals space fraction numerator left square bracket CO subscript 2 right square bracket over denominator left square bracket CH subscript 4 right square bracket left square bracket straight O subscript 2 right square bracket space end fraction
not correct expression.
In addition of CH4 (g) or O2 (g) at equilibrium Kc = will be decreased according to expression (i) but Kc remains constant at constant  at constant temperature fro a reaction, so for maintaining the constant value of Kc, the concentration of CO2 will increase in same order. Hence, on the addition of CH4 or O2 equilibrium will cause to the right.

This reaction is an example of an exothermic reaction.

609 Views

9.

Given: The mass of electron is 9.11 x 10-31 kg
Planck constant is 6.626 x 10-34 Js,
the uncertainty involved in the measurement of velocity within a distance of 0.1 A is:

  • 5.79 x 106 ms-1

  • 5.79 x 107 ms-1

  • 5.79 x 108 ms-1

  • 5.79 x 108 ms-1


A.

5.79 x 106 ms-1

By Heisenberg's uncertainty principle

increment straight p space. increment straight x space greater or equal than space fraction numerator straight h over denominator 2 straight pi end fraction
or space increment straight v. increment straight x space greater or equal than fraction numerator straight h over denominator 4 πm end fraction
increment straight p space rightwards arrow space uncertainty space in space momentum
increment straight x space rightwards arrow space uncertainty space in space position
increment straight v space rightwards arrow space uncertainty space in space velocity
straight m rightwards arrow space mass space of space particle
Given comma
increment straight x space equals space 0.1 space straight A space equals space 0.1 space straight x space 10 to the power of negative 10 end exponent space straight m
straight m equals space 9.11 space straight x space 10 to the power of negative 31 end exponent space kg
straight h space equals space planck space constant space equals space 6.626 space straight x space 10 to the power of negative 34 end exponent space Js
straight pi space equals space 3.14
In space uncertain space position space increment straight v. increment straight x space equals space fraction numerator straight h over denominator 4 πm end fraction
increment straight v space straight x space 0.1 space straight x space 10 to the power of negative 10 end exponent space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent over denominator 4 space straight x space 3.14 space straight x space 9.11 space straight x space 10 to the power of negative 31 end exponent end fraction
increment straight v space equals space fraction numerator 6.626 space straight x space 10 to the power of negative 34 end exponent over denominator 4 space straight x space 3.14 space straight x space 911 space straight x space 10 to the power of negative 31 end exponent straight x space 0.1 space straight x space 10 to the power of negative 1 end exponent end fraction ms to the power of negative 1 end exponent
equals space 5.785 space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent
5.79 space straight x space 10 to the power of 6 space ms to the power of negative 1 end exponent

784 Views

10.

Identify the correct statement for the change of Gibbs energy for a system (ΔGsystem) at constant temperature and pressure

  • If ΔGsystem > 0, the process is spontaneous

  • If ΔGsystem =0, the system has attained equilibrium

  • If ΔGsystem < 0, the system is still moving in a particular direction

  • If ΔGsystem < 0, the system is still moving in a particular direction


B.

If ΔGsystem =0, the system has attained equilibrium

If the Gibbs free energy for a system (ΔGsystem) is equal to zero, then system is present in equilibrium at a constant temperature and pressure.

3284 Views