JEE Exam > JEE Notes > DPP: Daily Practice Problems for JEE Main & Advanced > DPP for JEE: Daily Practice Problems- Equilibrium (Solutions)
Equilibrium Practice Questions - DPP for JEE
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
Related Exams
About this Document
|
Oct 24, 2024 Last updated |
Document Description: DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) for JEE 2024 is part of DPP: Daily Practice Problems for JEE Main & Advanced preparation.
The notes and questions for DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) have been prepared according to the JEE exam syllabus. Information about DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) covers topics
like and DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DPP for JEE: Daily Practice Problems- Equilibrium (Solutions).
Introduction of DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) in English is available as part of our DPP: Daily Practice Problems for JEE Main & Advanced
for JEE & DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) in Hindi for DPP: Daily Practice Problems for JEE Main & Advanced course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: Equilibrium Practice Questions - DPP for JEE
Description
Full syllabus notes, lecture & questions for Equilibrium Practice Questions - DPP for JEE - JEE | Plus excerises question with solution to help you revise complete syllabus for DPP: Daily Practice Problems for JEE Main & Advanced | Best notes, free PDF download
Information about DPP for JEE: Daily Practice Problems- Equilibrium (Solutions)
In this doc you can find the meaning of DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) defined & explained in the simplest way possible. Besides explaining types of
DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) theory, EduRev gives you an ample number of questions to practice DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
mock tests for examination
,past year papers
,Equilibrium Practice Questions - DPP for JEE
,ppt
,study material
,Sample Paper
,Summary
,Semester Notes
,MCQs
,Viva Questions
,Equilibrium Practice Questions - DPP for JEE
,Objective type Questions
,Free
,Equilibrium Practice Questions - DPP for JEE
,Extra Questions
,Important questions
,video lectures
,practice quizzes
,Exam
,Previous Year Questions with Solutions
,shortcuts and tricks
;
Additional Information about DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) for JEE Preparation
DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) Free PDF Download
The DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) now and kickstart your journey towards success in the JEE exam.
Importance of DPP for JEE: Daily Practice Problems- Equilibrium (Solutions)
The importance of DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) Notes
DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DPP for JEE: Daily Practice Problems- Equilibrium (Solutions).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) Notes on EduRev are your ultimate resource for success.
DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) JEE Questions
The "DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) on the App
Students of JEE can study DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DPP for JEE: Daily Practice Problems- Equilibrium (Solutions),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DPP for JEE: Daily Practice Problems- Equilibrium (Solutions) is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup