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18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it .......
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18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the e...
Given:
- Initial amount of N2O4 = 18.4 g
- Volume of the closed vessel = 1 L
- Equilibrium reaction: N2O4 ⇌ 2NO2
- At equilibrium, 50% of N2O4 is dissociated
To find:
The value of the equilibrium constant, Kc
Solution:
Step 1: Calculate the moles of N2O4:
The molar mass of N2O4 = 92 g/mol
Moles of N2O4 = Mass / Molar mass
Moles of N2O4 = 18.4 g / 92 g/mol
Moles of N2O4 = 0.2 mol
Step 2: Calculate the moles of NO2:
Since the reaction is 1:2, the moles of NO2 formed will be twice the moles of N2O4 dissociated.
Moles of NO2 = 2 * Moles of N2O4 dissociated
Moles of NO2 = 2 * (0.5 * 0.2 mol) (50% of N2O4 is dissociated)
Moles of NO2 = 0.2 mol
Step 3: Write the expression for the equilibrium constant, Kc:
Kc = [NO2]^2 / [N2O4]
Kc = (0.2 mol/L)^2 / (0.2 mol/L)
Kc = 0.2 mol/L
Step 4: Interpret the value of Kc:
The value of Kc is 0.2 mol/L. Since the concentration of both NO2 and N2O4 is the same, the value of Kc is equal to the concentration of NO2 squared.
Conclusion:
The value of the equilibrium constant, Kc, for the reaction N2O4 ⇌ 2NO2, is 0.2 mol/L.
- Initial amount of N2O4 = 18.4 g
- Volume of the closed vessel = 1 L
- Equilibrium reaction: N2O4 ⇌ 2NO2
- At equilibrium, 50% of N2O4 is dissociated
To find:
The value of the equilibrium constant, Kc
Solution:
Step 1: Calculate the moles of N2O4:
The molar mass of N2O4 = 92 g/mol
Moles of N2O4 = Mass / Molar mass
Moles of N2O4 = 18.4 g / 92 g/mol
Moles of N2O4 = 0.2 mol
Step 2: Calculate the moles of NO2:
Since the reaction is 1:2, the moles of NO2 formed will be twice the moles of N2O4 dissociated.
Moles of NO2 = 2 * Moles of N2O4 dissociated
Moles of NO2 = 2 * (0.5 * 0.2 mol) (50% of N2O4 is dissociated)
Moles of NO2 = 0.2 mol
Step 3: Write the expression for the equilibrium constant, Kc:
Kc = [NO2]^2 / [N2O4]
Kc = (0.2 mol/L)^2 / (0.2 mol/L)
Kc = 0.2 mol/L
Step 4: Interpret the value of Kc:
The value of Kc is 0.2 mol/L. Since the concentration of both NO2 and N2O4 is the same, the value of Kc is equal to the concentration of NO2 squared.
Conclusion:
The value of the equilibrium constant, Kc, for the reaction N2O4 ⇌ 2NO2, is 0.2 mol/L.
Community Answer
18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the e...
0.4
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18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it .......
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18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ....... for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ....... covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ........
18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ....... for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ....... covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ........
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18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ......., a detailed solution for 18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ....... has been provided alongside types of 18.4 g of N2O4 is taken in a 1 lit closed vessel and heated till the equilibrium is reached N2O4==2NO2 .at equilibrium it is found that 50% of N2O4 is dissociated .what wil be the value of Kc ??? please any one help me by solving it ....... theory, EduRev gives you an
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