Equilibrium Practice Questions - DPP for NEET

 Page 1


1. (b) K
p
 = K
c
 (RT)
?n
 since ?n = 0, K
p
 = K
c
.
2. (c) pH =   pK
a
 + log 
5 = 4 + log         [ Q pK
a
 = – log K
a
]
Given,  K
a 
=  1 × 10
– 4
? pK
a
  =  – log (1× 10
– 4
) =  4
Now from Handerson equation
pH =  pK
a
 + 
Putting the values
5 =4 + 
 =  5 – 4 = 1
Taking antilog
[Salt]/[Acid] = 10 = 10 : 1
3. (a) The reaction given is an exothermic reaction thus according to Le
chatalier’s principle lowering of temperature, addition of F
2
 and
Cl
2 
favour the forward direction and hence the production of ClF
3
.
4. (c) Ag
2
CrO
4
K
sp
 = [Ag
+
]
2 
 = 1.1 × 10
–12
AgCl
K
sp
 = [Ag
+
] [Cl
–
] = 1.8 × 10
–10
Page 2


1. (b) K
p
 = K
c
 (RT)
?n
 since ?n = 0, K
p
 = K
c
.
2. (c) pH =   pK
a
 + log 
5 = 4 + log         [ Q pK
a
 = – log K
a
]
Given,  K
a 
=  1 × 10
– 4
? pK
a
  =  – log (1× 10
– 4
) =  4
Now from Handerson equation
pH =  pK
a
 + 
Putting the values
5 =4 + 
 =  5 – 4 = 1
Taking antilog
[Salt]/[Acid] = 10 = 10 : 1
3. (a) The reaction given is an exothermic reaction thus according to Le
chatalier’s principle lowering of temperature, addition of F
2
 and
Cl
2 
favour the forward direction and hence the production of ClF
3
.
4. (c) Ag
2
CrO
4
K
sp
 = [Ag
+
]
2 
 = 1.1 × 10
–12
AgCl
K
sp
 = [Ag
+
] [Cl
–
] = 1.8 × 10
–10
[Ag
+
] = 
AgBr
K
sp
 = [Ag
+
] [Br
–
] = 5.0 × 10
–13
AgI
K
sp
 = [Ag
+
] [I
–
] = 8.3 × 10
–17
  If we take 
  than maximum [Ag
+
] will be required in case of Ag
2
CrO
4
.
5. (b) MY  M
+
 + Y
–
K
SP 
= s
2
? 
= 6.2 × 10
–13
? s =
 
s = 7.87 × 10
–7
 mol L
–1
NY
3
  N
3+
 + 3Y
–
K
SP
 = s × (3s)
3
 = 27s
4
 = 6.2 × 10
–13
s = 
s = 3.89 × 10
–4
 mol L
–1
? molar solubility of NY
3
 is more than MY in water.
6. (b) ?G°
NO(g)
 = 86.6k J/mol = 86600 J/mol
 = x J/mol
T = 298, K
P
 = 1.6 × 10
12
?G° = – RT ln K
P
Given equation,
2NO(g) + O
2
 (g)  2NO
2
(g)
? 2?G°
NO2
 – 2?G°
NO 
= – R (298) ln (1.6 × 10
12
)
Page 3


1. (b) K
p
 = K
c
 (RT)
?n
 since ?n = 0, K
p
 = K
c
.
2. (c) pH =   pK
a
 + log 
5 = 4 + log         [ Q pK
a
 = – log K
a
]
Given,  K
a 
=  1 × 10
– 4
? pK
a
  =  – log (1× 10
– 4
) =  4
Now from Handerson equation
pH =  pK
a
 + 
Putting the values
5 =4 + 
 =  5 – 4 = 1
Taking antilog
[Salt]/[Acid] = 10 = 10 : 1
3. (a) The reaction given is an exothermic reaction thus according to Le
chatalier’s principle lowering of temperature, addition of F
2
 and
Cl
2 
favour the forward direction and hence the production of ClF
3
.
4. (c) Ag
2
CrO
4
K
sp
 = [Ag
+
]
2 
 = 1.1 × 10
–12
AgCl
K
sp
 = [Ag
+
] [Cl
–
] = 1.8 × 10
–10
[Ag
+
] = 
AgBr
K
sp
 = [Ag
+
] [Br
–
] = 5.0 × 10
–13
AgI
K
sp
 = [Ag
+
] [I
–
] = 8.3 × 10
–17
  If we take 
  than maximum [Ag
+
] will be required in case of Ag
2
CrO
4
.
5. (b) MY  M
+
 + Y
–
K
SP 
= s
2
? 
= 6.2 × 10
–13
? s =
 
s = 7.87 × 10
–7
 mol L
–1
NY
3
  N
3+
 + 3Y
–
K
SP
 = s × (3s)
3
 = 27s
4
 = 6.2 × 10
–13
s = 
s = 3.89 × 10
–4
 mol L
–1
? molar solubility of NY
3
 is more than MY in water.
6. (b) ?G°
NO(g)
 = 86.6k J/mol = 86600 J/mol
 = x J/mol
T = 298, K
P
 = 1.6 × 10
12
?G° = – RT ln K
P
Given equation,
2NO(g) + O
2
 (g)  2NO
2
(g)
? 2?G°
NO2
 – 2?G°
NO 
= – R (298) ln (1.6 × 10
12
)
2?G°
NO2
 – 2 ×  86600 = – R (298) ln (1.6 × 10
12
)
2?G°
NO2
 = 2 × 86600 – R (298) ln (1.6 × 10
12
)
?G°
NO2
 =  [2 × 86600 – R(298) ln (1.6 × 10
12
]
= 0.5 [2 × 86600 – R (298) ln (1.6 × 10
12
)]
7. (a) (i) 
(ii) 
(iii) 
Applying (II + 3 × III – I) we will get
= 
? K = K
2
 ×  / K
1
8. (b) Initially on increasing temperature rate of reaction will increase, so
% yield will also increase with time. But at equilibrium % yield at
high temperature (T
2
) would be less than at T
1
 as reaction is
exothermic so the graph is
9. (c)
Page 4


1. (b) K
p
 = K
c
 (RT)
?n
 since ?n = 0, K
p
 = K
c
.
2. (c) pH =   pK
a
 + log 
5 = 4 + log         [ Q pK
a
 = – log K
a
]
Given,  K
a 
=  1 × 10
– 4
? pK
a
  =  – log (1× 10
– 4
) =  4
Now from Handerson equation
pH =  pK
a
 + 
Putting the values
5 =4 + 
 =  5 – 4 = 1
Taking antilog
[Salt]/[Acid] = 10 = 10 : 1
3. (a) The reaction given is an exothermic reaction thus according to Le
chatalier’s principle lowering of temperature, addition of F
2
 and
Cl
2 
favour the forward direction and hence the production of ClF
3
.
4. (c) Ag
2
CrO
4
K
sp
 = [Ag
+
]
2 
 = 1.1 × 10
–12
AgCl
K
sp
 = [Ag
+
] [Cl
–
] = 1.8 × 10
–10
[Ag
+
] = 
AgBr
K
sp
 = [Ag
+
] [Br
–
] = 5.0 × 10
–13
AgI
K
sp
 = [Ag
+
] [I
–
] = 8.3 × 10
–17
  If we take 
  than maximum [Ag
+
] will be required in case of Ag
2
CrO
4
.
5. (b) MY  M
+
 + Y
–
K
SP 
= s
2
? 
= 6.2 × 10
–13
? s =
 
s = 7.87 × 10
–7
 mol L
–1
NY
3
  N
3+
 + 3Y
–
K
SP
 = s × (3s)
3
 = 27s
4
 = 6.2 × 10
–13
s = 
s = 3.89 × 10
–4
 mol L
–1
? molar solubility of NY
3
 is more than MY in water.
6. (b) ?G°
NO(g)
 = 86.6k J/mol = 86600 J/mol
 = x J/mol
T = 298, K
P
 = 1.6 × 10
12
?G° = – RT ln K
P
Given equation,
2NO(g) + O
2
 (g)  2NO
2
(g)
? 2?G°
NO2
 – 2?G°
NO 
= – R (298) ln (1.6 × 10
12
)
2?G°
NO2
 – 2 ×  86600 = – R (298) ln (1.6 × 10
12
)
2?G°
NO2
 = 2 × 86600 – R (298) ln (1.6 × 10
12
)
?G°
NO2
 =  [2 × 86600 – R(298) ln (1.6 × 10
12
]
= 0.5 [2 × 86600 – R (298) ln (1.6 × 10
12
)]
7. (a) (i) 
(ii) 
(iii) 
Applying (II + 3 × III – I) we will get
= 
? K = K
2
 ×  / K
1
8. (b) Initially on increasing temperature rate of reaction will increase, so
% yield will also increase with time. But at equilibrium % yield at
high temperature (T
2
) would be less than at T
1
 as reaction is
exothermic so the graph is
9. (c)
K
SP
 = [Ag
+
]
2
 [C
2
O
4
? 2–
]
[Ag
+
] = 2.2 × 10
–4
 M
Given that:
? Concentration of C
2
O
4
? 2–
 ions,
? K
SP
 = (2.2 × 10
–4
)
2
 (1.1 × 10
–4
)
= 5.324 × 10
–12
10. (b)    +  
1 –                        
 =  = 
or, =         =  when  = 1
11. (d) Max. pressure of CO
2
 = Pressure of CO
2
 at equilibrium
For reaction,
SrCO
3
(s)  SrO(s) + CO
2
K
p
 = P
CO2
 = 1.6 atm = maximum pressure of CO
2
volume of container at this stage.
V = …(i)
Since container is sealed and reaction was not earlier at equilibrium.
? n = constant.
n = …(ii)
Put equation (ii) in equation (i)
V = 
Page 5


1. (b) K
p
 = K
c
 (RT)
?n
 since ?n = 0, K
p
 = K
c
.
2. (c) pH =   pK
a
 + log 
5 = 4 + log         [ Q pK
a
 = – log K
a
]
Given,  K
a 
=  1 × 10
– 4
? pK
a
  =  – log (1× 10
– 4
) =  4
Now from Handerson equation
pH =  pK
a
 + 
Putting the values
5 =4 + 
 =  5 – 4 = 1
Taking antilog
[Salt]/[Acid] = 10 = 10 : 1
3. (a) The reaction given is an exothermic reaction thus according to Le
chatalier’s principle lowering of temperature, addition of F
2
 and
Cl
2 
favour the forward direction and hence the production of ClF
3
.
4. (c) Ag
2
CrO
4
K
sp
 = [Ag
+
]
2 
 = 1.1 × 10
–12
AgCl
K
sp
 = [Ag
+
] [Cl
–
] = 1.8 × 10
–10
[Ag
+
] = 
AgBr
K
sp
 = [Ag
+
] [Br
–
] = 5.0 × 10
–13
AgI
K
sp
 = [Ag
+
] [I
–
] = 8.3 × 10
–17
  If we take 
  than maximum [Ag
+
] will be required in case of Ag
2
CrO
4
.
5. (b) MY  M
+
 + Y
–
K
SP 
= s
2
? 
= 6.2 × 10
–13
? s =
 
s = 7.87 × 10
–7
 mol L
–1
NY
3
  N
3+
 + 3Y
–
K
SP
 = s × (3s)
3
 = 27s
4
 = 6.2 × 10
–13
s = 
s = 3.89 × 10
–4
 mol L
–1
? molar solubility of NY
3
 is more than MY in water.
6. (b) ?G°
NO(g)
 = 86.6k J/mol = 86600 J/mol
 = x J/mol
T = 298, K
P
 = 1.6 × 10
12
?G° = – RT ln K
P
Given equation,
2NO(g) + O
2
 (g)  2NO
2
(g)
? 2?G°
NO2
 – 2?G°
NO 
= – R (298) ln (1.6 × 10
12
)
2?G°
NO2
 – 2 ×  86600 = – R (298) ln (1.6 × 10
12
)
2?G°
NO2
 = 2 × 86600 – R (298) ln (1.6 × 10
12
)
?G°
NO2
 =  [2 × 86600 – R(298) ln (1.6 × 10
12
]
= 0.5 [2 × 86600 – R (298) ln (1.6 × 10
12
)]
7. (a) (i) 
(ii) 
(iii) 
Applying (II + 3 × III – I) we will get
= 
? K = K
2
 ×  / K
1
8. (b) Initially on increasing temperature rate of reaction will increase, so
% yield will also increase with time. But at equilibrium % yield at
high temperature (T
2
) would be less than at T
1
 as reaction is
exothermic so the graph is
9. (c)
K
SP
 = [Ag
+
]
2
 [C
2
O
4
? 2–
]
[Ag
+
] = 2.2 × 10
–4
 M
Given that:
? Concentration of C
2
O
4
? 2–
 ions,
? K
SP
 = (2.2 × 10
–4
)
2
 (1.1 × 10
–4
)
= 5.324 × 10
–12
10. (b)    +  
1 –                        
 =  = 
or, =         =  when  = 1
11. (d) Max. pressure of CO
2
 = Pressure of CO
2
 at equilibrium
For reaction,
SrCO
3
(s)  SrO(s) + CO
2
K
p
 = P
CO2
 = 1.6 atm = maximum pressure of CO
2
volume of container at this stage.
V = …(i)
Since container is sealed and reaction was not earlier at equilibrium.
? n = constant.
n = …(ii)
Put equation (ii) in equation (i)
V = 
12. (d)
 K
1
 = 1.0 × 10
–5
 = (Given) 
 (Given)
= (1.0 × 10
–5
) × (5 × 10
–10
) = 5 × 10
–15
13. (c) Let s = solubility
K
sp
 = [Ag
+
] [IO
3
? –
] = s × s = s
2
Given K
sp
 = 1 × 10
–8
    = 
=  1.0 × 10
–4
 mol/lit  = 1.0 × 10
–4
 × 283 g/lit
=   = 2.83 × 10
–3
 gm/ 100 ml
14. (a) (HSO
4
)
–
 can accept and donate a proton
(HSO
4
)
– 
+ H
+
  H
2
SO
4 
(acting as base)
(HSO
4
)
–
 – H
+
  SO
4
? 2–
.   (acting as acid)
15. (d) [Cu(H
2
O)
4
]
2+
 + 4NH
3
  [Cu(NH
3
)
4
]
2+
 + 4H
2
O involves lose
and gain of electrons. H
2
O is coordinated to Cu by donating
electrons (LHS). It is then removed by withdrawing electrons.
16. (b) pH of an acidic solution should be less than 7. The reason is that
from H
2
O. [H
+
] = 10
–7
M  which cannot be neglected in comparison
to 10
–8
M. The pH can be calculated as.
from acid, [H
+
]  = 10
–8
M.
from H
2
O, [H
+
] = 10
–7
M
? Total [H
+
] = 10
–8 
+ 10
–7