All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
JEE Exam  >  JEE Questions  >  Equilibrium constant for the given reaction i... Start Learning for Free
SaveJoin the Discussion on the App
Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer?
Clear all your JEE doubts with EduRev
Most Upvoted Answer
Equilibrium constant for the given reaction is C = 1020 at temperature...

Given Data:
- Equilibrium constant (K) = 1020
- Temperature (T) = 300 K
- Initial moles of A = 1 mole
- Initial moles of B = 1/3 mole

Calculating Equilibrium Concentrations:
- Let x moles of B react to form C and D
- Therefore, at equilibrium, moles of B = 1/3 - 2x
- As per the stoichiometry of the reaction, moles of C = 2x
- Applying equilibrium constant expression:
K = [C]^2[D] / [A][B]^2
1020 = (2x)^2(1/3 - 2x) / (1)(1/3)^2
1020 = 4x^2(1/3 - 2x) / 1/9
1020 = 36x^2(1 - 6x)
1020 = 36x^2 - 216x^3
216x^3 - 36x^2 + 1020 = 0
Solving this cubic equation, we get x ≈ 0.00000000004

Calculating Final Concentration of B:
- Moles of B at equilibrium = 1/3 - 2(0.00000000004) ≈ 1/3
- Concentration of B = Moles of B / Total volume = (1/3) / (1 + 1/3) ≈ 1/4

Therefore, the equilibrium concentration of B starting with a mixture of 1 mole of A and 1/3 mole of B at 300 K is approximately 1/4, which can be represented as -4 x 10^-11 in scientific notation. Hence, the correct answer is option 'A'.
Explore Courses for JEE exam

Similar JEE Doubts

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumq.Above homogeneous reaction is carried out in a 2 litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If Kc for the reaction is then equilibrium concentration of C is

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.When C,H5OH and CH3COOH are mixed in equivalent proportion equilibrium is reached when 2/3 of acid and alcohol are used. How much ester will be present when 2 moles of acid were to react with 2 moles of alcohol? .

The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.Initial concentration of A is twice the initial concentration of `B. At equilibrium concentration of A and a are same then equilibrium constant for the reaction is

Top Courses for JEEView all

Top Courses for JEE

Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer?
Question Description
Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer?.
Solutions for Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Equilibrium constant for the given reaction is C = 1020 at temperature 300 K A(s) +2B(aq) 2C (s) + D(aq.)K = 1020 The equilibrium conc. of B starting with mixture of 1 mole of A and 1/3 mote/titre of B at 300 K is a)-4 x 10-11 b)-2 x 10-10 c)-2 x 10-11 d)-10-10 Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses

Suggested Free Tests

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup