UrbanPro

Learn Equilibrium with Top Tutors

What is your location?

Please enter your locality

Are you outside India?

Back

Equilibrium

Equilibrium relates to CBSE/Class 11/Science/Chemistry

Top Tutors who teach Equilibrium

1
Satara, Satara
8 yrs of Exp
800per hour
Classes: Class 11 Tuition, Class 12 Tuition

I am a teacher. My total teaching experience is 8 years. I teach for classes 11th and 12th as well as NEET / JEE Main. I am working in a reputed institute...

2
Vishesh Pal Class 11 Tuition trainer in Delhi Platinum
Vasant Kunj, Delhi
Verified
12 yrs of Exp
1200per hour
Classes: Class 11 Tuition, NEET-UG Coaching and more.

I am a biology teacher for NEET. I have prior experience of teaching in India's renounced institutes and online platforms like Aakash, Unacademy,...

3
Mandawali, Delhi
11 yrs of Exp
Classes: Class 11 Tuition, BSc Tuition and more.

I have handled a group of 20+ students, but now I focus most on the quality and the result of my students, 11th class is very important and personal...

Do you need help in finding the best teacher matching your requirements?

Post your requirement now
4
Anuj Jain Class 11 Tuition trainer in Kota Featured
Electricity Board Area Kota Thermal Power Station Colony, Kota
8 yrs of Exp
400per hour
Classes: Class 11 Tuition, Class 10 Tuition and more.

I have experience to teach upto class 12th student chemistry. I have specialization in physical , inorganic chemistry. I teach JEE, NEET, & BOARD...

5
Mayank Luniwal Class 11 Tuition trainer in Gandhinagar Featured
Basan, Gandhinagar
1 yrs of Exp
300per hour
Classes: Class 11 Tuition, Engineering Entrance Coaching and more.

Hi! I have a Master's degree in Chemistry from IIT Gandhinagar, and I’ve been teaching students for the past two years as a home tutor. I love making...

6
Baguihati, Kolkata
Verified
5 yrs of Exp
650per hour
Classes: Class 11 Tuition, Class 12 Tuition

I have completed my undergraduate studies at IIEST Shibpur in 2017 earning B.E degree in Mechanical Engineering. I had a fondness for chemistry since...

7
Sundara Rao Ganti Class 11 Tuition trainer in Hisar Featured
Prem Nagar, Hisar
Super Tutor
20 yrs of Exp
600per hour
Classes: Class 11 Tuition, Class 12 Tuition

I am very expert in teaching the basics with simple examples and make to understand the concepts very simple way. It will helpful for the students...

8
Lav Kumar Soni Class 11 Tuition trainer in Arrah Featured
Arrah Chowk, Arrah
Super Tutor
14 yrs of Exp
400per hour
Classes: Class 11 Tuition, Medical Entrance Coaching and more.

I've been passionately teaching chemistry for class 11th, 12th, JEE Mains, and NEET UG for the past 10 years. Armed with an MSc in Chemistry, my mission...

9
Hrithik C. Class 11 Tuition trainer in Bangalore Featured
Ashok Nagar D' Souza Layout, Bangalore
Super Tutor
8 yrs of Exp
300per hour
Classes: Class 11 Tuition, Class 12 Tuition and more.

I have been teaching students since past 8 years and in this journey many of them were able to crack their examination.My way of teaching makes approach...

10
Anamika Class 11 Tuition trainer in Faridabad Featured
Sector 39 Charmwood Village, Faridabad
Super Tutor
6 yrs of Exp
500per hour
Classes: Class 11 Tuition, Class 10 Tuition and more.

In my experience providing tuition for Class 11 students, I aimed to create a supportive and engaging learning environment tailored to their unique...

Reviews for top Class 11 Tuition

Average Rating
(4.8)
  • S
    review star review star review star review star review star
    15 Mar, 2013

    Raghav attended Class 11 Tuition

    "Extremely helpful. Has been teaching me for 4 years and I am majoring in Computer..."

    V
    review star review star review star review star review star
    20 Mar, 2013

    Hemagowri attended Class 11 Tuition

    "My son really liked his teaching . As a result, he has done his exams well. I would..."

    A
    review star review star review star review star review star
    29 Mar, 2013

    Maya attended Class 11 Tuition

    "Amit's method of teaching is very good. He ensures the student understands the subject..."

    S
    review star review star review star review star review star
    16 Apr, 2013

    Saransh attended Class 11 Tuition

    "The time I was preparing for Medical and Engineering Entrance, he supported me in..."

  • C
    review star review star review star review star review star
    16 Apr, 2013

    Nitish attended Class 11 Tuition

    "He is an ideal Teacher / Lecturer. He has excellent command on his subject."

    S
    review star review star review star review star review star
    19 Apr, 2013

    Debanshu attended Class 11 Tuition

    "Extremely good teacher. Very helpful and responsible. Comes on time most of the days..."

    R
    review star review star review star review star review star
    02 May, 2013

    Rohan attended Class 11 Tuition

    " I was very weak in maths but not now, scored 78. best thing abt sir is he knows..."

    N
    review star review star review star review star review star
    08 May, 2013

    Piyush attended Class 11 Tuition

    "I am writing this to support Narayan Jalan for being the pioneer in teaching students...."

Get connected

Equilibrium Questions

Ask a Question

Post a Lesson

Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

To predict the effect of temperature on the equilibrium constant (KK) of the given reaction, we can use the relationship between the standard Gibbs free energy change (ΔG∘ΔG∘) and the equilibrium constant: ΔG∘=−RTln⁡KΔG∘=−RTlnK where: ΔG∘ΔG∘ is the... read more

To predict the effect of temperature on the equilibrium constant (KK) of the given reaction, we can use the relationship between the standard Gibbs free energy change (ΔG∘ΔG) and the equilibrium constant:

ΔG∘=−RTln⁡KΔG=−RTlnK

where:

  • ΔG∘ΔG is the standard Gibbs free energy change,
  • RR is the gas constant (8.314 J/mol·K),
  • TT is the temperature in Kelvin,
  • KK is the equilibrium constant.

At equilibrium, ΔG∘ΔG is zero. Therefore, we can rearrange the equation to solve for KK:

K=e−ΔG∘RTK=eRTΔG

Given that the reaction is:

CaCO3(s)→CaO(s)+CO2(g)CaCO3(s)→CaO(s)+CO2(g)

with the given standard enthalpy changes (ΔH∘ΔH):

ΔHCaCO3∘=−1206.9 kJ/molΔHCaCO3=−1206.9kJ/mol ΔHCaO∘=−635.1 kJ/molΔHCaO=−635.1kJ/mol ΔHCO2∘=−393.5 kJ/molΔHCO2=−393.5kJ/mol

we can use these values to calculate the change in standard Gibbs free energy (ΔG∘ΔG) using the equation:

ΔG∘=ΔHproducts∘−ΔHreactants∘ΔGHproducts−ΔHreactants

ΔG∘=(−635.1+(−393.5))−(−1206.9) kJ/molΔG=(−635.1+(−393.5))−(−1206.9)kJ/mol ΔG∘=118.3 kJ/molΔG=118.3kJ/mol

Now, we can calculate KK at two different temperatures and compare their values:

  1. At T1=298 KT1=298K: K1=e−118.3×1038.314×298K1=e8.314×298118.3×103

  2. At T2=350 KT2=350K: K2=e−118.3×1038.314×350K2=e8.314×350118.3×103

To determine the effect of temperature on KK, we can compare the values of K1K1 and K2K2. If K2>K1K2>K1, then increasing the temperature increases the equilibrium constant KK, indicating that the reaction shifts towards the products at higher temperatures. Conversely, if K2<K1K2<K1, then increasing the temperature decreases the equilibrium constant KK, indicating that the reaction shifts towards the reactants at higher temperatures.

By calculating K1K1 and K2K2 using the above equations, we can determine the effect of temperature on the equilibrium constant KK of the given reaction.

 
 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

In the given reaction: NH3+BF3→H3N:BF3NH3+BF3→H3N:BF3 Ammonia (NH3NH3) acts as the base, while boron trifluoride (BF3BF3) acts as the acid. This reaction involves the formation of a coordinate covalent bond between the lone pair of electrons on the nitrogen atom in ammonia and the empty... read more

In the given reaction:

NH3+BF3→H3N:BF3NH3+BF3→H3N:BF3

Ammonia (NH3NH3) acts as the base, while boron trifluoride (BF3BF3) acts as the acid. This reaction involves the formation of a coordinate covalent bond between the lone pair of electrons on the nitrogen atom in ammonia and the empty orbital on the boron atom in boron trifluoride.

This reaction is best explained by Lewis acid-base theory. According to this theory, an acid is defined as a substance that can accept an electron pair, while a base is a substance that can donate an electron pair. In the given reaction, boron trifluoride acts as the Lewis acid by accepting the lone pair of electrons from ammonia, which acts as the Lewis base.

The hybridization of boron (BB) and nitrogen (NN) in the reactants can be determined by considering the electron configuration and bonding in the molecules:

  1. In boron trifluoride (BF3BF3), boron has an electron configuration of 1s22s22p11s22s22p1. Boron forms three bonds with fluorine atoms, resulting in sp^2 hybridization for boron. The three hybridized orbitals on boron overlap with the p orbitals of the three fluorine atoms to form three sigma bonds.

  2. In ammonia (NH3NH3), nitrogen has an electron configuration of 1s22s22p31s22s22p3. Nitrogen forms three bonds with hydrogen atoms, resulting in sp^3 hybridization for nitrogen. The three hybridized orbitals on nitrogen overlap with the s orbitals of the three hydrogen atoms to form three sigma bonds.

Therefore, in the reactants, the hybridization of boron is sp^2 and the hybridization of nitrogen is sp^3.

 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

To calculate the volume of water required to dissolve 0.1 g0.1g of lead (II) chloride (PbCl2PbCl2) to get a saturated solution, we first need to determine the number of moles of PbCl2PbCl2 dissolved. Then, we can use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+... read more

To calculate the volume of water required to dissolve 0.1 g0.1g of lead (II) chloride (PbCl2PbCl2) to get a saturated solution, we first need to determine the number of moles of PbCl2PbCl2 dissolved. Then, we can use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+ ions in the saturated solution, which will allow us to calculate the volume of water required.

First, let's calculate the number of moles of PbCl2PbCl2 dissolved:

  1. Calculate the molar mass of PbCl2PbCl2: Molar mass of Pb=207 g/molMolar mass of Pb=207g/mol Molar mass of Cl=35.453 g/molMolar mass of Cl=35.453g/mol Molar mass of ( \text{PbCl}_2 = \text{Molar mass of Pb} + 2 \times \text{Molar mass of Cl} ] =207+2×35.453=207+70.906=277.906 g/mol=207+2×35.453=207+70.906=277.906g/mol

  2. Calculate the number of moles of PbCl2PbCl2: Number of moles=MassMolar mass=0.1 g277.906 g/molNumber of moles=Molar massMass=277.906g/mol0.1g Number of moles≈3.598×10−4 molNumber of moles≈3.598×10−4mol

Now, let's use the solubility product constant (KspKsp) to find the concentration of Pb2+Pb2+ ions in the saturated solution:

PbCl2→Pb2++2Cl−PbCl2→Pb2++2Cl

Given Ksp=3.2×10−8Ksp=3.2×10−8, we can set up an ice table:

PbCl2→Pb2++2Cl−InitialChange+x+2xEquilibrium3.598×10−42×3.598×10−4PbCl2InitialChangeEquilibrium→Pb2+++x3.598×10−42Cl+2x2×3.598×10−4

Substituting the equilibrium concentrations into the KspKsp expression:

Ksp=[Pb2+]×[Cl−]2Ksp=[Pb2+]×[Cl]2

3.2×10−8=(3.598×10−4)×(2×3.598×10−4)23.2×10−8=(3.598×10−4)×(2×3.598×10−4)2

Solve for [Pb2+][Pb2+]:

[Pb2+]=3.2×10−8(2×3.598×10−4)2[Pb2+]=(2×3.598×10−4)23.2×10−8

[Pb2+]≈2.222×10−2 M[Pb2+]≈2.222×10−2M

Now, we can use the concentration to calculate the volume of water required to dissolve 0.1 g0.1g of PbCl2PbCl2:

[Pb2+]=Amount of substanceVolume of solution[Pb2+]=Volume of solutionAmount of substance

Volume of solution=Amount of substance[Pb2+]Volume of solution=[Pb2+]Amount of substance

Volume of solution=3.598×10−4 mol2.222×10−2 MVolume of solution=2.222×10−2M3.598×10−4mol

Volume of solution≈0.0162 LVolume of solution≈0.0162L

Therefore, approximately 0.0162 L0.0162L of water is required to dissolve 0.1 g0.1g of lead (II) chloride to get a saturated solution.

 
 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

To find the solubility of Al(OH)3Al(OH)3 in grams per liter (g/L), we first need to determine the concentration of Al3+Al3+ ions in the saturated solution. Then, we can use stoichiometry to find the concentration of Al(OH)3Al(OH)3 and finally convert it to grams per liter. Given: Solubility product... read more

To find the solubility of Al(OH)3Al(OH)3 in grams per liter (g/L), we first need to determine the concentration of Al3+Al3+ ions in the saturated solution. Then, we can use stoichiometry to find the concentration of Al(OH)3Al(OH)3 and finally convert it to grams per liter.

Given:

  • Solubility product (KspKsp) of Al(OH)3Al(OH)3 = 2.7×10−112.7×10−11
  • Atomic mass of Al (AlAl) = 27 u
  1. Calculate the concentration of Al3+Al3+ ions:

The dissolution of Al(OH)3Al(OH)3 proceeds as follows:

Al(OH)3⇌Al3++3OH−Al(OH)3⇌Al3++3OH

Let xx be the solubility of Al(OH)3Al(OH)3 in moles per liter (M). Then, the concentration of Al3+Al3+ ions will also be xx M.

Using the solubility product expression:

Ksp=[Al3+]×[OH−]3Ksp=[Al3+]×[OH]3

Given that [OH−]=3x[OH]=3x, we can substitute these values into the expression:

2.7×10−11=x×(3x)32.7×10−11=x×(3x)3

Solving for xx:

2.7×10−11=27x42.7×10−11=27x4

x4=2.7×10−1127x4=272.7×10−11

x4=1×10−12x4=1×10−12

x=1×10−124x=41×10−12

x=0.01 Mx=0.01M

  1. Calculate the molar mass of Al(OH)3Al(OH)3:

Molar mass of Al(OH)3=Molar mass of Al+3×Molar mass of O+3×Molar mass of HMolar mass of Al(OH)3=Molar mass of Al+3×Molar mass of O+3×Molar mass of H

=27+3×16+3×1=27+3×16+3×1

=27+48+3=27+48+3

=78 g/mol=78g/mol

  1. Calculate the solubility in grams per liter (g/L):

The solubility of Al(OH)3Al(OH)3 in grams per liter (g/L) is the molar mass of Al(OH)3Al(OH)3 multiplied by its molar solubility:

Solubility=Molar solubility×Molar massSolubility=Molar solubility×Molar mass

Solubility=0.01 M×78 g/molSolubility=0.01M×78g/mol

Solubility=0.78 g/LSolubility=0.78g/L

  1. Calculate the pH of the solution:

Since Al(OH)3Al(OH)3 is a sparingly soluble salt, we can assume that it dissociates completely in solution. Therefore, the concentration of Al3+Al3+ ions is equal to the solubility, which is 0.01 M0.01M.

To find the pH of the solution, we need to calculate the concentration of H+H+ ions. Since Al3+Al3+ ions are neutral, for each mole of Al3+Al3+ ions, three moles of H+H+ ions are produced.

[H+]=3×0.01 M=0.03 M[H+]=3×0.01M=0.03M

Now, we can use the formula for pH:

pH=−log⁡[H+]pH=−log[H+]

pH=−log⁡(0.03)pH=−log(0.03)

pH≈−log⁡(3×10−2)pH≈−log(3×10−2)

pH≈−(−1.5229)pH≈−(−1.5229)

pH≈1.5229pH≈1.5229

Therefore, the solubility of Al(OH)3Al(OH)3 in g/L is 0.78 g/L0.78g/L, and the pH of the solution is approximately 1.521.52.

 
 
 
read less
Answers 1 Comments
Dislike Bookmark

Answered on 10 Apr Learn CBSE/Class 11/Science/Chemistry/Equilibrium

Sadika

Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively. When equal volumes of two solutions are mixed, the resulting pH can be calculated using the formula for dilution: pH=12(pHA+pHB)pH=21(pHA+pHB) pH=12(6+4)=12(10)=5pH=21(6+4)=21(10)=5 ... read more

 

  • Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

    When equal volumes of two solutions are mixed, the resulting pH can be calculated using the formula for dilution: pH=12(pHA+pHB)pH=21(pHA+pHB) pH=12(6+4)=12(10)=5pH=21(6+4)=21(10)=5

 

read less
Answers 1 Comments
Dislike Bookmark

Top topics in Class 11 Tuition

Looking for Class 11 Tuition ?

Find Online or Offline Class 11 Tuition on UrbanPro.

Do you offer Class 11 Tuition ?

Create Free Profile »

Looking for best Class 11 Tuition ?

POST YOUR REQUIREMENT
x

Ask a Question

Please enter your Question

Please select a Tag

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more