ALP Solutions Chemical Equilibrium Chemistry English JEE JR

8/11/2019 ALP Solutions Chemical Equilibrium Chemistry English JEE JR

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CHEMICAL EQUILIBRIUM - 1

TOPIC : CHEMICAL EQUILIBRIUM

ADVANCED LEVEL PROBLEMS

1. (i) CO(g) + 2H 2 (g) CH 3OH (g)0.15 a0.15 x a 2x x ! x = 0.080.15 x + a 2x + x = 0. 5 PV = n RT

a 2x = 0.35 n = 500082.05.22.8

""

= 0.5

Kc = 2

5.235.0

5.207.0 5.2

08.0

# $ %&

' ( "

= 34320000

= 58.3

Kp = 58.3 (RT) 2 = 2)500082.0(

3.58" = 4141

3.58" = 0.035

(ii) Total pressure will remain 8.2 atm as catalyst reduces only time taken to achieve equilibrium, doesnot affect equilibrium condition / concentrations.

2. 2.0 10 37 =) *

) * ) *933.0482.0x2

2

" ! x = 6.6 10 20

+ [N2] = 0.0482 mol L 1 ; [O 2] = 0.0933 mol L

1 ; [N2O] = 6.6 10 21 mol L 1

3. Ag+ (aq) + Fe 2+ (aq) Fe 3+ (aq) + Ag(s)Millimole before reaction 500 0.150 500 1.09or = 75 = 545 0 0Millimole after reaction (75 x) (545 x) x x! mM = Meq. (both Ag + /Ag and Fe 2+ /Fe 3+ have valency factor unity)

! KC = ]Fe[]Ag[]Fe[2

3

,,

,

+ KC = # $ %&

' ( -#

$ %&

' ( -

1000x545

1000x751000

x

.... (1)

! Conc. = volumeTotalolelimMil

[Ag+] = 1000x75 -

; [Fe 2+] = 1000x545 -

; [Fe 3+] = 1000x

Now 25 mL of mixture requires 30 mL of 0.0832 M or 0.0832 5 N KMnO 4.! Fe 2+ is oxidized by KMnO 4.+ Milliequivalent of Fe 2+ left at equilibrium in 1000 mL

= Milliequivalent of KMnO 4 for 1000 mL

= 25100050832.030 """

= 499.2+ 545 x = 499.2+ x = 545 499.2 = 45.8

Target : JEE (IITs)CHEMISTRY

COURSE : VIJAY (JR)


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Thus, by Eq. (1),

2.4992.2910008.45

10008.45545

10008.4575

10008.45

Kc "".

# $ %&

' ( -"#

$ %&

' ( -

.

KC = 3.1420.

4. For the reaction,

Fe2+

+ 3dipy [Fe(dipy) 3]2+

rforward = K f = [Fe2+] [3 dipy]

= 1.45 10 13 [Fe 2+] [dipy]3

rbackward = Kb [Fe(dipy) 3]2

= 1.22 10 4 [Fe(dipy) 3]2+

At equilibrium r f = r b and K C = bf

KK

= [Fe(dipy) 3]2+ /Fe 2+] [dipy]3

Also, stability constant of complex = Equilibrium constt. of reaction =b

fKK

= 413

1022.1

1045.1-"

" = 1.188 10 17

5. (a) A + 2B 2Ca 2a

a x 2a 2x 2xTotal moles at equilibrium 3a x

Mole fraction of C = xa3x2- = 0.4

2x = 1.2a 0.4x

x =4.2a2.1

x =2a

Now A + 2B 2Ca/2 a a

Total moles =2a5

P A = 8x

2a5

2a

= 58

atm

P B = 2 / a5a

x 8 = 516

atm

P C = 2 / a5 a x 8 = 516 atm

KP =A

2B

2C

P.)P(

)P( = 8

5

KP = 0.625 atm 1

(b) A + 2B 2CMole fraction 0.16 0.32 0.52Partial pressure 0.16 P 0.32 P 0.52 P

KP = 22

)P32.0(xP16.0

)P52.0( = 8

5

P = 5x32.0x32.0x16.08x52.0x52.0

= 26.4 atm


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6. (i) From the graph 0.3 n = 0.6n = 2

(ii) K = (0.6) 2 / 0.3 = 1.2 mol / L

(iii) initial rate of conservation of A = 15.06.0 -

= 0.1 M hr 1

7._ QC = 2

4

104

102

103

103

# $ %&

' ( #

$ %&

' (

# $

%&'

( #

$

%&'

(

=2 10

32243 " = 7.59 10 2 > K c

so, reaction will proceed in backward direction.

8. A(g) B(g) + C(g) K c = 41 2 20.5 1 1 Q c = 2 < Kc

0.5 - x 1 + x 1 + x 4 =) *

x5.0x1 2

-,

x = 0.162 [A] = 0.338 [B] = [C] = 1.162

9. Ag+ + 2CN [Ag(CN) 2]

t = 0 0.1 0.5 0

teq 10 6 0.3 0.1 K 1 = 9

10 10 6

Zn2+ + 4CN [Zn(CN) 4]2

t = 0 0.1 0.5 0

eq. 10 12 0.1 0.1 12 42 10)1.0(1.0

K"

. = 10 15

Substracting two times I st reaction from II nd reaction, we will get the required reaction, so

Keq =12

2

15

109

10

10

"# $ %&

' (

=100

8110 3 " = 810 Ans. 810

10. 2HBr(g) H 2(g) + Br 2(g) Kp = 6.110 5-

10 2p p p2

p210p

## $

%&&'

( - = 6.1

10 5-p = 2.5 10 2

(pH2)eq = (p Br2)eq = 2.5 10 2 bar ; (p HBr)eq = 10.0 bar

11. The equilibrium reaction would involve 2 moles of CO, 1mole of O 2 and 2 moles of CO 2 as the unit of K C islit/mole.So the equilibrium equation is

2CO(g) + O 2 (g) 2CO 2 (g)

The equation, CO(g) +21

O2 (g) CO 2 (g) would have an equilibrium constant with units (lit/mole) 1/2 .

2CO(g) + O 2 (g) 2CO 2(g)

Initial conc.21

= 0.521

= 0.5 0

Conc. at equil. 0.5 x 0.5 2x

x


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+ Kc = ]O[]CO[]CO[

22

22

= 5 10 3 + 5 10 3 =#

$ %

&' ( --

2x

5.0)x5.0(

x

2

2

Since, the value of equilibrium constant is pretty high so we can assume that almost entire CO goes to CO 2.Thus, value of x would be close to 0.5. But concentration of CO, (0.5 x) would not be zero but would be a

small value. Let this value be y. Then the concentration of O 2 at equilibrium would be# $

%&'

( ,2

y25.0 .

+ 5 10 3 =#

$ %&

' ( ,

2y

25.0y

)5.0(

2

2

As value of y is very small,2y

can be easily ignored with respect to 0.25

5 10 3 = 25.0y)5.0(

2

2

"y = 1.4 10 2

[CO] = y = 1.4 10 2 M.

12. CH 3COOH (" ) + C 2H5OH (" ) CH 3COOC 2H5( " ) + H 2O (" ) 1 0.5 1 0.5 0.5 0.5

So, K C = 5.05.05.05.0

""

= 1

Now let a moles of CH 3COOH and b moles of C 2H5OH are taken :

a 3a

b 3a

3a

3a

KC =# $ %&

' ( -"

"

3ab3 / a2

)3 / a()3 / a(or 2 # $

%&' ( -

3a

b = 3a

or 2b = a or ba = 1

2

13. H2 + I2 2HI Kp = 6.26.9)8.30( 2

" = 38

25 x 18 x 2x = 30.8

2HI H 2 + I2 Kp = 6.26.9)8.30( 2

" = 38

1

/ 2

/

2

/38

1

= 2

2

)1(4 /-

/

! / = 0.245

14. PCl 5 PCl 3 + Cl 2a 0 0a(1- / ) (a / ) (a / )Total moles at equilibrium = a(1+ / ) ........ (i)

From n = 5230821.07.11

RTPV

""

. = 0.03959 ........ (ii)

From (i) and (ii) a(1+ / ) = 0.03959

Here a = 5.2085.4

Hence / = 0.8329


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15. 4PH 3(g) P 4(g) + 6H 2(g)1 0.4 0.1 0.6

Meq. = 3.1126.01241.0346.0 ",","

= 30.76

Vapour density = 15.38

16. PCl 5(g) PCl 3 + Cl 2 4.678 110 = n total 0.082 523

5(1 / ) 5/ 5 / n total = 11.998

5(1 / ) + 4 = 12 / = 53

= 0.6

P eq mix. = 12768.4

8 = 3.18

Kp = 22

1P/-

/ = 2

2

6.0118.36.0

-"

= 1.78

17. (a) The given equilibrium is ,CO 2 (g) + C (s) 2 CO(g)

Initial mole 1 0Final mole (1 / ) 2/

KP =2CO

2CO

n)n(

1

nP

123

4567 = ) 1(

)2( 2

//

123

456

/,15

10 = ) 1(20 2

//

or 10 10 / 2 = 20 / 2 .

+ / 2 = 3010

+ / = 31

= 0.577 .

Thus, mole of CO 2 at equilibrium = 1 / = 1 0.577 = 0.423.and mole of CO at equilibrium = 2 / = 2 0.577 = 1.154

Total mole present at equilibrium = 0.423 + 1.154 = 1.577.At equilibrium, PV = nRT

P = 5 atm ; n = 1.577, T = 817 + 273 = 1090.5 V = 1.577 0.0821 1090.

+ V = 28.22 litre.

+ [CO] at equilibrium = 22.28154.1

= 0.041 mol litre 1

.

+ [CO 2 ] at equilibrium = 22.280423

= 0015 mol litre 1 .

(b) For CO 2 (g) + C(s) 2CO (g)Initial mole 1 0Final mole (1 a) 2a

Total mole at equilibrium = 1 a + 2a = 1 + a.

Given a1a 1

, = 1005

+ a = 10095


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KP = )n()n(

2CO

2CO

1

nP

123

456

7

10 =#

$

%&

'

(

# $ %&

' ( "

100

5100

9522

1

100105

P

1111

2

3

4444

5

6

+ P = 0.145 atm.

18. Pentyne-1 Pentyne-2 + Pentadiene-1-2.(A) (B) (C)

At eq. 1.3 95.2 3.5

Keq. = ]A[]C[]B[

= 3.15.32.95 "

= 256.31 .....(i)

Now, for B AA

K1 = ]B[

]A[

then from Eqs. (i) and (ii), K 1 = .eqK]C[

= 31.2565.3

= 0.013.

+ 8 G= 2.303 RT log 10 K = 2.303 8.314 448 log 0.013 = 16178 J = 16.178 kJ.Stability order for A and B is B >A.Similarly for B C

K2 = ]B[]C[

= 2.eq

]B[

]A[K = 2.952.95

3.131.256"

" = 0.037.

+ 8 G = 2.303 R log 10 K = 2.303 8.314 448 log 0.037 = 12282 J = 12.282 kJ.Thus, stability order for B and C is B > C.Given the values of 8G1 and 8G2 , the total stability order is B > C > A.CH 3.CH 2.CH 2.C 9 CH : : : : ; : .alcKOH [CH 3CH 2CH = C = CH 2] :; : [CH3CH 2C 9 C CH 3].

19. 8G = 8H T 8S ; K = e 8G / RTAt 300 K, 8G = ( 41.16 300 0.0424) 1000 = 28440 J/molAt 1200 K, 8G = ( 32.93 1200 0.0296) 1000 = 2590 J/molKP (300 K) = 8.935 10

4

KP (1200 K) = 0.37753Q = 1at 300 K : Q < K , Hence forward direction.at 1200 K : Q > K , Hence backward direction.

20. CO(g) + H 2O(g) CO 2(g) + H 2(g)8G R = ( 394.36) (137.17 228.57) = 28.62 kJ = RT ln k

log K = 303.2248314.862.28

"" = 5

K = 10 5

K =OHCO

HCO

2

22

PP

PP""

CO + H 2O CO 2 + H 21 x OH2P 1 + x 1 + x

10 5 =OH

2

2P)x1()x1(

-,

= 32

1022.1)x1()x1(

-"-,


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122 = )x1()x1( 2

-,

H2O(" ) H2O(g) kp = OH2P (vp)8G rxn. = 28.62 kJkp = 10 5

CO + H 2O CO 2 + H 2

101.325

x 3.16 kp 101.325 kp

x 101.325 kp

x

H2O(" ) H2O(g)8G = 8.56 8.56 = 2.303 RT log kplog kp = 1.5

kp = 3.16 = OH2P

kp = 10 5 = )x325.101()16.3()x325.101()x325.101(

,,,

x = 101.201 kpa ; P CO = 0.124 kPa ; 2COP = 202.65 kPa

21. (a) Because Fe 2O3 is a solid, its "concentration" doesn't change when more Fe 2O3 is added. Therefore,there is no concentration stress, and the original equilibrium is undistrubed.(b) Le Chatelier's principle predicts that the concentration stress of removed CO 2 will be relieved by reactionfrom left to right to replenish the CO 2.(c) Le Chatelier's principle predicts that the concentration stress of removed CO will be relieved by reactionfrom right to left to replenish Q C = [CO 2]

3 / [CO] 3.When the equilibrium is disturbed by reducing [CO], Q C increases, so that Q C > KC. For the system to moveto a new state of equilibrium, Q C must decrease-that is, [CO 2] must decrease and [CO] must increase.Therefore the reaction goes from right to left, as predicted by Le Chatelier's principle.

23. NH4I (s) NH 3 (g) + H I

initial eq. 150 150final eq. 750 30

HI (g) 21

H2 (g) + 21

I2 (g)

eq. 150 2x x x

x2150xx

- = 2 ! x = 60

Final pressure = 750 + 30 + 60 + 60 = 900 mm of Hg.

24. A(g) B (g) + 2C (g)1 x x + 3y 2x y

C (g) 2D (g) + 3B (g)2x y 2y 3y + x

neq = 1 x + x + 3y + 2x y + 2y

i

eq

n

n = 1

y4x21 ,, = 6

13 ! 2x + 4y = 67

1y4x21 ,,

= 613

2x + 4y = 67

12x + 24y = 7 ........(i)

x1yx2

-,

= 94


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18x 9y = 4 4x

38

(22x 9y = 4) ........(ii)

3176

x 24y = 332

........(iii)

(1) + (3)

12x + 3176

x = 332

+ 7

3212

x = 353

x = 21253

x = 0.25 = 41

use in eq. (1)

312 41

+ 24y = 7

24y = 4

y = 61

A(g) B (g) + 2C (g)

1 41 = 4

3 41 + 3 6

1 = 43

42 6

1 = 31

C (g) 2D (g) + 3B (g)

42 6

1 = 31 6

2 = 31

63 + 4

1 = 43

1cK =43

31

43

2# $ %&

' ( "

= 91

= 0.11111

2cK =31

43

31

22#

$ %&

' ( "#

$ %&

' (

= 31

6427

= 0.14

25. 3A2 (g) AA6 (g) KP = 1.6 atm 2

KP = 1.6 = 3A

A

)P(

P

2

2

ALP Solutions Chemical Equilibrium Chemistry English JEE JR

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