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Equilibrium Practice Questions - DPP for NEET
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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
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FAQs on Equilibrium Practice Questions - DPP for NEET
| 1. What is the concept of equilibrium in chemistry? |  |
Ans. Equilibrium in chemistry refers to a state where the forward and reverse reactions occur at equal rates, resulting in no net change in the concentrations of reactants and products. It is characterized by a dynamic balance between the rates of the chemical reactions involved.
| 2. How is equilibrium achieved in a chemical reaction? | |
Ans. Equilibrium in a chemical reaction is achieved when the rate of the forward reaction is equal to the rate of the reverse reaction. This can be achieved by adjusting the concentration, pressure, or temperature of the system to favor either the forward or reverse reaction, depending on the desired outcome.
| 3. What is Le Chatelier's principle and how does it relate to equilibrium? | |
Ans. Le Chatelier's principle states that when a system at equilibrium is subjected to a stress, it will respond by shifting in a direction that reduces the effect of the stress. This principle helps predict how changes in concentration, pressure, or temperature will affect the equilibrium position of a chemical reaction.
| 4. How does temperature affect the equilibrium in a chemical reaction? | |
Ans. Temperature affects the equilibrium in a chemical reaction by altering the rate of the forward and reverse reactions. An increase in temperature generally favors the endothermic reaction, while a decrease in temperature favors the exothermic reaction. This shift in equilibrium position is in accordance with Le Chatelier's principle.
| 5. What is the significance of equilibrium constants in chemical reactions? | |
Ans. Equilibrium constants, represented by Kc or Kp, provide quantitative information about the position of equilibrium in a chemical reaction. They are calculated by dividing the product of the concentrations or pressures of the products by the product of the concentrations or pressures of the reactants, each raised to their stoichiometric coefficients. The value of the equilibrium constant indicates the extent to which the reaction proceeds towards the products or the reactants at equilibrium.
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