Jee 2014 Booklet3 Hwt Solutions Chemical Equilibrium

Vidyamandir Classes

Solutions to Home Work Test/Chemistry Chemical Equilibrium 1.(A)

HWT - 1

 2CO(g) CO 2(g)  C (s)  0 2p

0.5 0.5  p

2  2p   0.62 PCO  PCO2 0.5  p 0.2 2

Kp 

= 1.8 0.5  P  2P  0.8

Now



P = 0.3

 PCl3(g)  Cl2(g) PCl5(g) 

2.(A) t=0 t = teq

4 2

0 2 2 2  [PCl3 ] [Cl2 ] Kc   4 4 2 [PCl5 ] 4

0 2 = 0.5

6.(A)

K p  PNH3 PH 2S  50  50  2500 atm 2

7.(C)

3A(g) t 0 3 t  t eq 3  3x Now

   4B(g)   2C(g)  3D(g) 4

4  4x

2x

3x

2x = 1 1 x 2



3 1 2x . 3x 2 1 Kc   3  3  3x   4  4x  2 2 2

 C  D A  B 

9.(D) t0

a

a

t  t eq a  x a  x

0

0

x

x

x  3 a  x 

4x = 3a x

3 a. 4

3 3 a . a 4 4 9 Kc  a a . 4 4

VMC/Chemical Equilibrium

1

HWT-Solutions/Chemistry

Vidyamandir Classes

Chemical Equilibrium SO 2 

1.(D) t=0

5

t = teq

2

HWT - 2

1  SO3 O 2  2 5 0 5

3 2

3

total moles at equilibrium = 2 

7 17 3 2 2

 2 7 PO2  0.4 atm.     17 2 

2.(A)

Q

[PCl3 ] [Cl2 ]  1  K eq [PCl5 ]

Hence reaction will go backward. 5.(C)

Eqn III

= Hence

6.(A)

 P2  KP  y  1 Px 

   

Eqn I + K3 = K1 K2. K P2 

   x   2y  t0 1 0  at t  t eq 1  x 2x  

Eqn II

PP PQ PZ 2

Given: 7.(A)



Let total pressure P1

 P  Q  Z    1 0 0  Let total pressure P2 1 x x x 

 2x  2   P1 1 x   K P1  1  x    P1 1  x 



 x   x    P2   P2 1  x 1 x    K P2  1 x    P2 1  x 



P 1 3

K P1 : K P2  1 : 9

 Y(g)  Z(g) X (g) 

a

0

a 2

0

a 2

a 2 P P . P PX  and K P  3 3  1 P 3 3

and P  3

Hence

PX = 1.

 PCl3(g)  Cl2(g) PCl5(g) 

8.(B) t=0 t = teq

5 3

0 0 2 2 [PCl3 ] [Cl2 ] 44 Kc    2.66 [PCl5 ] 6

 PCl3  Cl2 PCl5 

10.(D) t=0 t = teq

3 1.5

3 4.5

VMC/Chemical Equilibrium

2 3.5

Hence n PCl3  4.5

2

HWT-Solutions/Chemistry

Vidyamandir Classes

Chemical Equilibrium

2.(C)

3.(B)

HWT - 3

0.4 0.4  [PCl3 ] [Cl 2 ] 4 4 Kc   1.2 [PCl5 ] 4 =

0.42 0.4  4  1.2 12

=

4  0.033 10  12

 2NH3(g)  CO 2(g) NH 2COONH 4(s)  2P 3

KP 

P 3

4P3  2.9  105 27

27  2.9  105  195  106 4 Hence P = 0.0582 P3 

t=0 t = teq 8.(B)

1  SO3(g) O 2(g)  2 4 0 3 2

SO 2(g) 

7.(B)

4 2

 6HCHO C6H12O6  1 1 x

Kc  

0 6x

 6x 6  1 1  x  6  1022  1 6x    6  1022 

1/ 6

   



Calculate 6x.

Chemical Equilibrium 1.(B)

Use K p  K c  RT 

6.(D)

Final Keq = K1 × K2

10.(C)

HWT - 4

ng

  N 2(g)  3H 2(g)   2NH 3(g) at t = 0 2 at t = teq 2  x

4 4  3x 2x 

VMC/Chemical Equilibrium

0 2x

34 (Given) 17 x=1

3

HWT-Solutions/Chemistry

Vidyamandir Classes

Chemical Equilibrium KP

HWT - 5

2.(B)

KP 

3.(D)

Value of equilibrium constant depends upon temperature only. Changing volume will change  only.

 RT 2

Since n g  2 .

  NH 4 HS(s)   NH 3(g)

6.(A)

 H 2S(g)

t=0 0.5 0 t = teq 0.5 + P P 0.5 + 2P = 0.84 2P = 0.34 P = 0.17 K P  PNH 3 PH 2S  0.67  0.17  0.11

  A(g)  3B(g)   4C(g)

10.(B) t=0 t = teq

a ax

a a  3x

a  x  4x

KC 

[C]4 [A] [B]3



0 4x 

x

 4a / 54  4a   2a   5  5     

3

a 5

44



4  23

 8.

Chemical Equilibrium   PCl5(g)   PCl3(g)  Cl2(g)

1.(B) t=0 t = teq

2 1.2

KC 

4.(A)

HWT - 6

If KP > KC

0 0.8 0.8

2  1.2 2

0.8

0 0.8

2  0.4  0.4  0.266

0.6

n g  0



Hence, forward reaction is favored at low pressure.

  A  B  CD

5.(C) at

t=0 t  t eq KC 

6.(B)

a

a

0

0

2a 3

2a 3

a 3

a 3

1  0.25 4

[Products] > [Reactants] K must be greater than 1. 

VMC/Chemical Equilibrium

4

HWT-Solutions/Chemistry

Vidyamandir Classes   N 2(g)  3H 2(g)   2NH 3(g)

9.(C) t=0 t  t eq

1 1 x

3 3  3x

0 2x

1  x  0.6



x = 0.4

Total moles = 4  2x  3.2 10.(A)

Addition of inert gas at constant pressure drives the reaction to the side with higher number of gaseous moles.

Chemical Equilibrium 8.(B)

HWT - 7

    NH 4 Cl(s)   NH 4(aq.)  Cl(aq.) H   ve

Increasing temperature will shift the reaction to the right.

Chemical Equilibrium

HWT - 8

6.(D)

Addition of Cl2 at constant volume will take the reaction to the right.

10.(A)

2A  B  Pr oducts

Rate = k [A]2 [B]

Chemical Equilibrium

HWT - 9

  X  Y  Z

2.(A) t=0 t = teq

5 3

3 1

2

2 k eq  3

  A  2B   2C

9.(D) t=0 t = teq

kc

2 2.5 1 2   2 2

3 4

2 1

2

 4   2.5  2  2     



1 4 16 2.5  4 2

VMC/Chemical Equilibrium

 0.05

5

HWT-Solutions/Chemistry