Equilibrium slideshow neet 2024 online class

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1. 4

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NEET 2024 m)

Cou IBRIUN

DIKSHA To

2. Introduction

e Equilibrium is a state in which there are no observable changes as
time goes by.

e When a chemical reaction has reached the equilibrium state, the
concentrations of reactants and products remain constant over time

however, there is much activity at the molecular level

3. 5 | v
0 ange incthe val.

4. Capote
| h+B— ai
Alen e neg” ya b e

Conca cowtant

tow =a Same ( (rra)

5. A+B = Crp

Rate q {un ward Run (RP)

= Rate e) hackuward Pe
(P+ R)

6. At ey" Conc" = Comtt
N) may | meg net same.

Rale= Same
S gmatant

7. (ey :

+ Fa
(a, (“y mt) J (b)

8. + q" + m
(a) not hue Ww Y

9. Types of chemical reactions

Irreversible reaction Reversible reaction

Reaction which proceeds in forward | Reaction which proceed in forward
direction only as well as backward direction
PLU,

Attain the equilibrium state and

DS not attain equilibrium state never go to completion

Generally thermal dissociation are Generally thermal dissociation are
held in open vessel held in closed vessel
2KCIO, (s) — 2KCI (s) + 30, (9) PCI, (g) = PCI, (g) + Cl, (9)

HCl(aq) + NaOH(aq) > NaCl(s) + H,O N,(g) + 0,(g) + 2NO(g)

10. Types of equilibria on the basis of process

Types of equilibria on the basis of process

| |

Physical equilibrium Chemical equilibrium

Equilibrium in physical process is Equilibrium in chemical
called physical equilibrium. process is called chemical
For example equilibrium.
Phase changes like : For example
H,0(1) + H,0(g) H,(g) + CI,(g) + 2HCI(g)
Solvation :

NaCl(s) _H,0. NaCl(aq)

excess

11. Types of equilibria on the basis of physical state

Types of equilibria on the basis of physical state

MC y
Homogeneous equilibrium Heteroyeneous equilibrium
When all reactants and products are When more than one phase are
in same phase present
H,(g) + Cl,(g) + 2HCI(g) 3Fe(s) + 4H,0(g) + Fe,0,(s) + 4H,(9)

SO,(g) + NO,(g) + SO,(g) + NO(g) 2Na,0,(s) + 2H,0 + 4NaOH + 0,(g)

12. Chemical Equilibrium
R,+R, = P,+P,

Reactants Products

A Reversible reaction

e As soon as P, and P, are formed, they start the backward reaction. As
concentrations of R, and R, decrease rate of forward reaction decreases
and rate of backward reaction increases.

At equilibrium :
e Rate of forward reaction (r,) = rate of backward reaction (r,)
e All measurable parameters become constant with respect to time.

13. ah+bB = cc+dD

nok, Caley | #4 i
Ap= Ep UND) se ap ld ul
Re Be cm?
ep

Ke= ne

MOT

14. Characteristics of chemical Equilibrium

e Itisald ic equilibrium)- at this stage, reaction take place in both
the direction with same speed so., there is no net change

ya equilibrium the reaction proceeds both the side , equally.
X= .
Equilibrium can be obtained from either side of the reaction.

YL catalyst an alter the rate of approach of equilibrium but does not
change the state of equilibrium

SN equilibrium, free energy change(AG =0 ) OS=0 , Aw +0

Equilibrium state can be affected by altering factors like pressure,

volume, concentration and temperature etc.(Le chatelier's Principle)

15. Law of mass action

[By Guldberg and Waage]

e Active Mass = Molar concentration i.e. Moles/ Litres

Active Mass =

It is represented in square brackets i.e. [] e.g. [A], [N,] etc.

#ote : Active masses ard dimensionless quantities but for our purposes we
generally take them with dimensions of molarity, partial pressure,\etc.

17. K, (Equilibrium constant in terms of partial pressure)

e Relation between Kp and Kc :

An = sum of the number of moles of

gaseous products - sum of the number of
moles of gaseous reactants

e Unit of k, = (mol/L) 2%

e Unit of k, = (atm)*™

18. Given the reaction between 2 gases represented by A, and B, to give
the compound AB(¿)" ~~

Aa) *Ba(g) © 2AB (y)

At equilibrium, the concentration of

A, = 3.0 x 10? M, of B, = 4.2 x 107? M, of AB = 2.8 x 107? M

If the reaction takes place in a sealed vessel at 527°C, then the value of

NEET 2012

19. Given the reaction between 2 gases represented by A, and B, to give
the compound AB(¿)"

Aa) *Ba(g) © 2AB (y)

At equilibrium, the concentration of

A, = 3.0 x 107? M, of B, = 4.2 x 107? M, of AB = 2.8 x 107? M

If the reaction takes place in a sealed vessel at 527°C, then the value of

NEET 2012

20. Given the reaction between 2 gases represented by A, and B, to give
the compound AB, On

A(g) * B2(9) = 2AB(g)
At equilibrium, the concentration of

A, = 3.0 x 107? M, of B, = 4.2 x 107? M, of AB = 2.8 x 10? M
If the reaction takes place in a sealed vessel at 527°C, then the value of

K. will be NEET 2012

Ay) + Bag) = 2AB(y)
_ AB}?
Ko [42][B2]

(2.8103)? O
7 (30x10 *)(42x10 3) 3.0x4.2

= 0.62

21. A 20 litre container at 200 K contains CO (y) at pressure 0.4 atm
and an excess of SrO (neglect the volume of solid SrO). The
volume of the container is now decreased by moving the movable
piston fitted in the container. The maximum volume of the
container, when pressure of CO, attains its maximum value, will be
(Given that : SCO, = SrOf,) + CO, (o), K, = 1.6 atm) NEET 2017

© [10 litre
© | 4litre
O | 2 litre
Lo (E litre

23. A 20 litre container at 200 K contains CO (y) at pressure 0.4 atm
and an excess of SrO (neglect the volume of solid SrO). The
volume of the container is now decreased by moving the movable
piston fitted in the container. The maximum volume of the
container, when pressure of CO, attains its maximum value, will be
(Given that : STCO, = SrO() + CO.) K, = 1.6 atm) NEET 2017

© /10 litre |
© ‘Altre ]
O [2 litre |
Cosine DO )

24. A 20 litre container at 200 K contains CO, (,) at pressure 0.4 atm
and an excess of SrO (neglect the volume of solid SrO). The
volume of the container is now decreased by moving the movable

piston fitted in the container. The maximum volume of the
container, when pressure of CO, attains its maximum value, will be
NEET 2017

SrCO yo) = SrO(,) + CO.
A pco, X Psro
Pr Psrco,

2(9) À K, =1.6 atm

16 = Peo, (+ Psro = Psrco, = 1)

Let the maximum volume of the container when
pressure of CO, is 1.6 atm be VL

During the process, PV = constant

-.0,4 x20=1.6 x V

_ 0.4 x 20

>V =5L

1.6 a

25. In which of the following equilibrium K, and K are not equal?

—

O0: 0" 00 ed

at
© (so... + NO) = 50 „+ NO.) 2-250
O0 "70 2-2=0 |

| u! Oy@)* 260) 2-14) |
%

NEET 2010

26. In which of the following equilibrium K, and K are not equal?

NEET 2010

© (2NO = No + 29 |
O {SO * NO) © SOsg * NO |
O0 "70 |
®© 2C (+ 0, = 260.4)

27. In which of the following equilibrium K, and K are not equal?

NEET 2010

K_ and K, are related by the equation,
Pp e
K, = K (RT) %%g
When An, = difference in the no. of moles of products
and reactants in the gaseous state.

for 2C,.) + O,,_) = 2CO.
An, = vit ay? » #0 F9)

28. Factors affecting equilibrium constant

+ Kdepends on the stoichiometry of the reaction.
If two chemical reactions at equilibrium having equilibrium constants K,
and K, are added then the resulting equation has equilibrium constant
K=K, x K, on adding

Equation Equilibrium constant

CE mom

29. k= D»
rk,

(4) muh
ply with nm”

ne = nB
- 3t= 2% | el
E 2° >
( — B,k
k, = |
kl"

31. ®

h=B %

CE PF } k,
E, OT
(Add) (sub)
A+c= BM A-c= RD

AZ BEC
et

2

32. Factors affecting equilibrium constant
If the reaction having equilibrium constant K, is reversed then resulting
equation has equilibrium constant 1/K,

A(g) + Big)= C(g)+D(g) |, ky
On reversing, C(g) + D(g) = Alg) + B(g) K = 1/k,
If a chemical reaction having equilibrium constant K, is multiplied by a
factor n then the resulting equation has equilibrium constant K = (K)", n

can be fraction

Equation Equilibrium constant

D,(g) = 2A(g) K,

(rai 1/2 D, (g) = A(g) KEK = JR

33. R=)”

If the equilibrium constant for N

= 2N0 q) i is K, the

2(g) + 26)

equilibrium constant for — 3 Vaio) + 3040 = = NO,\will be
NEET 2015

= | u N
K |

Y
'-()
O: '
\_
CE Belk

Vedantu

34. If the equilibrium constant for N,(,) + O,

'a(g) = 2NO „is K, the

1 1
equilibrium constant for 3 Mao) + 3 Oxo) = NO,,,will be
NEET 2015

35. If the equilibrium constant for N,¡¿, + O,(,) = 2NO;,) is K, the

2(9)

1 1
equilibrium constant for 320) + 3 Oxo) = NO,,,will be
NEET 2015

If the reaction is multiplied by 1/2, then new equilibrium
constant, K' = K'2,

36. Factors affecting equilibrium constant

e Equilibrium constant is dependent only on the temperature.
It means Ko and K, will remain constant at constant temperature no matter
how much changes are made in pressure, concentration, volume or

catalyst TO E,
However if temperature is changed,
ÚS

CS
BEE)
5% 7 2303R|T, 7%
I -IZRAZIS

If T, > T, then K, > K, provided AH = +ve (endothermic reaction)

not,

¡Van't hoff equation
—_

K, <K, if AH = -ve (exothermic reaction)

37. (x=) Exo Endo (1H =+:e)

ER | ED

— TT,
AT TY, 2:
Tr / Lor CET

\ \

38. For a given exothermic reaction, K and K’, are the equilibrium
constants at temperatures T, and T,, respectively. Assuming that
heat of reaction is constant in temperature range between T, and
T, ‚it is readily observed that 1 “TL 7 NEET 2014

39. For a given exothermic reaction, K and K’, are the equilibrium
constants at temperatures T, and T,, respectively. Assuming that
heat of reaction is constant in temperature range between T, and
T, , itis readily observed that NEET 2014

40. For a given exothermic reaction, K and K’, are the equilibrium
constants at temperatures T, and T,, respectively. Assuming that

heat of reaction is constant in temperature range between T, and

T, , itis readily observed that NEET 2014
Ky _ __AH [1 1

log = — 303% E ii

For exothemic reaction, AH = —ve i. e. ‚Heat is evolved.

The temperature T, is higher than T, .

1 1 \: A
Thus, ( _ 4)is negative.
So, log K', — log K, > log K’,

or Ky > K’,

41. Given that the equilibrium constant for the reaction,
280,(4) + 9,0) = 280.4) Ic
Has a value of 278 at gYarticular temperature. What is the value

of the equilibrium constant for the following reaction at the same
1
temperature? SO, = SOx.) + 320)

NEET 2012
© [18 x 10° 0) Revered
Le (26.10 E) divided ys

si _ L

Os Fe
| 6.0 x 10 +) la
Os. | we

42. Given that the equilibrium constant for the reaction,

280,(4) + 9,0) = 280.4)

Has a value of 278 at a particular temperature. What is the value

of the equilibrium constant for the following reaction at the same
1

temperature? SO, = SOx.) + 30%)

NEET 2012

43. Given that the equilibrium constant for the reaction,
25029) + Oz) = 250 (9)
Has a value of 278 at a particular temperature. What is the value

of the equilibrium constant for the following reaction at the same
temperature? SO, = SO.) + 50x à

NEET 2012
2SOn9) + Orig) = 280g, K = 278 ...(i)

By reversing the equation (i), we get

250y(g, = 250219) + O219) (ii)
Equilibrium constant for this reaction is,

NEE
==.
By dividing the equation (ii)by2, we get desired equation,
Sosiy) SOz¡g) + $029) -- (ii)

Equilibrium constant for this reactin,

K" = VK’ = [4 = [45 = 0.0599 = 0.06 or 6 x 10? Nedantic

2

45. The equilibrium constants of the following are
N, + 3H, = 2NH,; K,
N, +0, = 2NO; K,

H, +14 0,= H,O; K,
The equilibrium constant (K) of the reaction:
2N Hz + e 2N 0 + 3H,0 will be NEET 2017, 2007, 2003

© K,K,°/K,

© | Kakzl K, ]
O | K,°k,/ K, |
O xx. | an

46. From the given equations,

1
2NH, = N, + 3H;
K,

N, +O, = 2NO; K,

3H, + o, = 3H,0; K}

By adding equations (i), (ii) and (iii), we get
KK}

2NH, +20, = 2N0+3H,0, K=
1

47. 1. Predicts the extent of a reaction on the basis of its magnitude

2. Predicts the direction of the reaction

3. Calculate equilibrium concentrations

48. Predict the extent of reaction
- _[Product] Ng
Ko, Reactant de
rw) | (Bagesad)

7

Case - | Case - Il Case - Ill
K >103 10% <K,, < 10° Ko < 107
eq Appreciable “

[Product] > [Reactant] concentrations of both [Product] < [Reactant]
In this case, the reactants and products In this case, the
reaction is are present reaction is
( product favourable’) Po > reactant favourable

49. Predict the extent of reaction

Negligible Extremely

(very low) large

Reaction 10° L 10° Reaction

hardly, Both reactants and press

proceeds products are — products are almost to
resent at

completion
equilibrium
—

50. Predict the direction of reaction

At each point in a reaction, we can write a ratio of concentration terms
having the expression same as the equilibrium constant expression.

This ratio is called the reaction quotient denoted by the symbol Q.
Reaction quotient (Q) helps in predicting the direction of a reaction.

The reaction quotient is a variable quantity with time.

51. — —
1
ke c 8-

&>te
Backwoxd

Q= Ke
Equilihnium.

Q< ke
Food Rin

52. Predict the direction of reaction

aA + bB = cC +dD

CD;

NT

Q > At any time during reaction

The concentration [C] , [D], [A], [B]
are not necessarily at equilibrium

© Reaction quotient

Equilibrium
reached

Time

53. Predict the direction of reaction

If Q, >K
e When Qcis higher, [products] are higher.
e Reactants + Products (backward)

e Hence more reactants are formed till equilibrium
is reached

54. Predict the direction of reaction

If Q, <K,
e When Qc is Lower, [products] are Lower.
e Reactants — Products (Forward)

e Hence more products are formed till equilibrium
is reached

55. Predict the direction of reaction
FA =K,

e Reaction is at equilibrium, thus doesn't move in any direction.

56. Calculation of equilibrium concentration

1. Write the balanced equation for the reaction
2. Under each reactant and product write the details:

Initial concentration
Change in concentration before equilibrium
Concentration at equilibrium

3. Substitute the value in the equilibrium equation

57. The reaction, 2A() + Bog) = 3C (5) + Do) is begun with the
concentrations of A and B both at an initial value of 1.00 M. When

equilibrium is reached, the concentration of D is measured and

found to be 0.25 M. The value for the equilibrium constant for this
reaction is given by the expression NEET 2010

O [ [(0.75)*(0.25)] « [(1.00)2(.00)) |

Born) :[(0.50)2(0.75)] |

© [(0.75)*(0.25)] : [(0.50)*(0.25)] |
O | [(0.75)°(0.25)] + [(0.75)?(0.25)] |

58. 24 +B 230 +D
i | 0 0

YN
eg lam be 3x x
0-50 015 0% 0.5 (1=0.)

k= (dr) ‚k.= 0%) (0-25)
(A (3) (0 st) (o 45)

59. The reaction, 2A() + Bog) = 3C (5) + Do) is begun with the
concentrations of A and B both at an initial value of 1.00 M. When

equilibrium is reached, the concentration of D is measured and

found to be 0.25 M. The value for the equilibrium constant for this
reaction is given by the expression NEET 2010

O | [(0.75)*(0.25)] : [(.00)*(.00)) |

R J [(0.75)*(0.25)] : [(0.50)2(0.75)]

© [(0.75)°(0.25)] : [(0.50)?(0.25)] |
O | [(0.75)°(0.25)] + [(0.75)?(0.25)] |

60. The reaction, 2A() + Bog) = 3C (5) + Do) is begun with the
concentrations of A and B both at an initial value of 1.00 M. When
equilibrium is reached, the concentration of D is measured and

found to be 0.25 M. The value for the equilibrium constant for this
reaction is given by the expression NEET 2010

2A + Big) = 3c + D
Initial moles: 1 1 0° d”
Moles ateq.:1-(2 x 0.25) 1-0.25 3x0.25 0.25

=0.5 = 0.75 = 0.75 = 0.25
aja _ lero]
Equilibrium constant, K = aria]
_ (0.75)°(0.25)

(0.5)? (0.75)

61. The reaction quotient (Q) for the reaction
N,(9) + 3H,(9) = 2NH,(g)
Is given by q= INGE . The reaction will process from right

: TA]
to left if NEET 2003

O + |
lc JONA |
O ]

62. The reaction quotient (Q) for the reaction
N,(g) + 3H,(g) = 2NH,(g)

Is given by q= Her . The reaction will process from right
2

to left if NEET 2003

63. The reaction quotient (Q) for the reaction
N,(g) + 3H,(g) = 2NH,(g)

Is given by g= INGE . The reaction will process from right

INR]
to left if

NEET 2003

For reaction to proceed from right to left,

Q > e
(backward (forward

rate) rate) ¡.e. the reaction will be fast in backward
direction i.e r, > r,

64. Relation between Equilibrium constant & gibbs free
energy
e From thermodynamics
AG = AG? + RTInQ
AG = Gibbs energy change

R = Gas constant
T = temperature in Kelvin

= a i m
Q = Reaction quotient At 1 A =0

At equilibrium, when AG = 0 and Q = Ka then - AG + 0

AG° =-RTInK, _
AGO = - 2.303 RT log K, & - E

66. Relation between Equilibrium constant & gibbs free
energy

AG*= -RT InK,,

AG? = -2.303 RT log K,,

o Spontaneous Reaction in forward
AG°<0
reaction direction

at AG°>0 Non spontaneous | Reaction in backward
reaction direction

67. M", AG? = -2:303RTlog, Ke

68. Hydrolysis of sucrose is given by the following reaction:
Sucrose + H,0 = Glucose + Fructose. If the equilibrium constant

(K,) is 2 x 10% at 300 K, the value of A.G° at the same temperature

will be NEET 2020

SI Jmol'K!x 300 K x In(2x 10%) |

vo y
© | 8.314 J mol'K" x 300 K x In(2 x 1013) | AU=-RT In (sx )
O | 8.314 J mol"'k™ x 300 K x In(3 x 10) |

© | -8.314 J mot'k" x 300 K x In(4 x 10") |

69. Hydrolysis of sucrose is given by the following reaction:
Sucrose + H,0 = Glucose + Fructose. If the equilibrium constant

(K,) is 2 x 10% at 300 K, the value of A.G° at the same temperature
will be NEET 2020

© -8.314 J mol'K™ x 300 K x In(2 x 10")

© | 8.314 J molk? x 300 K x In(2x 10%) |
O | 8.314 J molK” x 300 K x In(3 x 10%) |

© | -8.314 J mol'k" x 300 K x In(4 x 101%) ]

70. Hydrolysis of sucrose is given by the following reaction:
Sucrose + H,0 = Glucose + Fructose. If the equilibrium constant

(K,) is 2 x 10% at 300 K, the value of A.G° at the same temperature
will be NEET 2020

AG = AG? + RT InQ

At yop AG =OandQ=K,
*.0=AG°+RTINK,

> AG? =-RTIn Ke

= -8.314 JK mol? x 300 K x In (2 x 10")

71. The equilibrium concentrations of the species in the reaction A + B

=C + Dare 2, 3, 10 and 6 mol L", respectively at 300 K. AG? for the
reaction is (R = 2 cal/mol K)

© | 1372.60 cal |
© | -137.26 cal
© | -1381.80 cal |
© | 13.73 cal |

72. A+BSec+d

Cy o
k= (le)? AG = -2:303 RT log, lo
QU) o
An =(2-303) (2) (309)

k=10 We

73. The equilibrium concentrations of the species in the reaction A + B
=C + Dare 2, 3, 10 and 6 mol L", respectively at 300 K. AG? for the

NEET 2023

reaction is (R = 2 cal/mol K)

© | 1372.60 cal |

© | -137.26 cal |
© | -13.73 cal |

74. The equilibrium concentrations of the species in the reaction A + B
=C + Dare 2, 3, 10 and 6 mol L”, respectively at 300 K. AG? for the

reaction is (R = 2 cal/mol K)

NEET 2023

AG = -2.303Rt log K
Here, RB = 2cal do = 300K

On solving we will get,
AG = —1381.8 Cal

75. Le chatelier's Principle

e If a change is applied to the system at equilibrium, then equilibrium

will be shifted in that direction in which it can minimise the effect of

change applied and the equilibrium is established again under new
conditions.

76. Le chatelier's Principle

Ba erect of concentration |
BE erectortemperature |
_ EZ

77. Le chatelier's Principle
e Effect of concentration :

N, (9) + 3H,(9)+ 2NH, (9)
Concentration Direction of equilibrium shift
Addition of reactants Forward
Removal of reactants | Backward
Addition of Products | Backward

ye Removal of products | Forward
eYNote : The Édition of any solid component Hoes not affect the
equilibrium.
o.

78. R —
e?
RT, prund US
t (9) Go
ockwoMd. N
Al jr wooffect

PT

| Battu

Dy. Am Gry or Dec conc So!
pp od ham

79. Le chatelier’s Principle Dry = MM

Effect of pressure :

On increasing pressure, equilibrium will shift in the direction in which
pressure decreases (no. of moles in the reaction decreases) and vice

versa.
Moles (A n) Pressure Direction of shift

Increases

Backward direction

Decreases Forward direction

Increases

Forward direction

Decreases

Backward direction

Increases or
decreases

No shift

80. Pressure. > PT, mo: mes ave RSS
mary?

PL | moles ate more

Volume > VP ¡md moler mae,
dai Vi y No SP males leg

81. Le chatelier's Principle

e Effect of volume :

1

va P
Direction of
(A n,) Volume pressure equilibrium
An>0 "> Decreases Increase pe
An>0M Vie increases Decreases Forward direction
An<0Npe Ta Decreases Increase Forward direction
An<0 We h R Increases Decreases di cay
An=0 Np = Increases or Increases or No shift
decreases decreases

82. Le chatelier’s Principle
e Effect of inert gas addition :

Inert gas addition has no
effect at constant volume

Np?
An, > 0, reaction will shift in forward
direction :
An, < 0, reaction shift will in [ Np< "a:
backward direction

| b
An, =0, no effect : (mo Me it)

83. > cmt vo L= meet

ar a comite Preuve

(React 0 Mee MO q mola)

84. Le chatelier's Principle
e Effect of temperature :

To L
k
TU) EEK

logKz — logK, = -ve

T 1, K |, Equilibrium shift
in backward direction

He
logK; — logK, = +ve logKz — logK, = 0
k>k logK, = logk ah
T 1, K 7, Equilibrium shift in No effect)pf temperature on
rward direction, me Wad lecken,

‘onc. of product

conc. of reactant f

85. und > Pressu
opp x. the change’in = ? volume >
Fr, ng] males les PAL

86. "p= |, m=? | MR

Which one of the following conditions will favour maximum

formation of the product in the reaction
Ata) + Bag) = Xzray AH = X kd?

NEET 2018
Low temperature and high pressure
uw
pistes
© [ Low temperature and low pressure | T a

© | High temperature and high pressure | TT
] k,? b,

© | High temperature and low pressure

87. Which one of the following conditions will favour maximum

formation of the product in the reaction

Aig) + Bolg) = Xo(gy AH = X ks?

© Low temperature and high pressure

© | Low temperature and low pressure

© | High temperature and high pressure )

© High temperature and low pressure )

NEET 2018

88. Which one of the following conditions will favour maximum
formation of the product in the reaction
AH = -X kJ?

ag) * Bag) = X2(g)

NEET 2018

On increasing the pressure and decreasing the
temperature, equilibrium will shift in forward direction.

89. For the reversible reaction,
+ 3Ha(g) = 2NH3g) + heat {CAD

N
2(g)
The equilibrium shifts in forward direction

29)
NEET 2014

O By increasing the concentration of NH, — Back 1
O By decreasing the pressure PL 1 Bele |

© [ By decreasing the concentrations of Nya) and Haq Back

By increasing pressure and decreasing |
temperature.

90. For the reversible reaction,

No) + 3H2(g) = 2NH,(.) + heat
The equilibrium shifts in forward direction NEET 2014
© | By increasing the concentration of NH,(4) 1

© [ By decreasing the pressure ]

© [ By decreasing the concentrations of Nya) and Hao)
© By increasing pressure and decreasing
temperature.

91. For the reversible reaction,
Nov) + 3H

2(9) = 2NH, (4) + heat
The equilibrium shifts in forward direction

NEET 2014

As the forward reaction is exothermic and leads to
lowering of pressure (Produces lesser number of
gaseous moles) hence, according to Le Chatelier's
principle, at high pressure and low temperature, the
given reversible reaction will shift in forward direction
to form more product.

Vedantu

92. NEET 2024 CRASH COURSE 4.0

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+ 15 Full length mock tests and 20 practice tests

+ Comprehensive Test Analysis and Weakness
Identification

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. STE yasTER TEACHE

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94. Acids , base theories

Acids Dissociates in water to give hydrogen ions ( H* )
[std campo: A ER
HX (aq) x

Bases Dissociates in water to give hydroxyl ions (OH)
Example : Ca(OH),
+ S
NaOH (aq) — Na + OH

Acids are proton donors and bases are proton acceptor)
Acid: (Fe(H,O)¿P*, HCI , Base: NH,
NH, (aq) + H,O (I) = NH,' (aq) + OH (aq)
Base Acid

95. + Modified. Arhenius concept =

HA+HD <= Hp +A

+ r
H +0 > HD

(=)
x dvonium
vo yo Y aes

96. ae Gov ugaie Ban x ñuda
HS E HU HS
(A) (Ce)

SA — w cb
WA>SCB

97. a) Hy80y & Sy (ar) X
b) H so, % #80, (Mt) Y
Rad 4 ik CB fas att
Only me ÿ

98. N + yw —> Nt,
(2) (cA)

>>
ait one H*

99. Acids , base theories

Acids accepts electron pair and bases donates
electron pair Acid : AICI,, BF, , Base: NH, PH,
BF, + :NH,— BF,:NH,

€ de] ene Acid Base

€hich

100. 1

Conjugate base for Bronsted acids lo and HF are

NEET 2019

O | H¿O* and H,F*, respectively |
© | OH’ and H,F*, respectively |

© | H,O* and F, respectively

oO | OH and F-, respectively

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