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9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)?
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9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establis...
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9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establis...
Given Data
- Mass of N2O4 = 9.2 g
- Molar mass of N2O4 = 92 g/mol
- Volume of the vessel = 1 L
- Percentage of N2O4 dissociated = 50%
Calculating Moles of N2O4
- Moles of N2O4 = Mass / Molar mass
= 9.2 g / 92 g/mol
= 0.1 moles
Initial Concentration of N2O4
- Initial concentration (C0) = Moles / Volume
= 0.1 moles / 1 L
= 0.1 M
Dissociation at Equilibrium
- Since 50% of N2O4 is dissociated:
Moles dissociated = 0.1 moles × 50% = 0.05 moles
- Remaining moles of N2O4 = 0.1 - 0.05 = 0.05 moles
- Moles of NO formed = 2 × 0.05 = 0.1 moles (from stoichiometry of the reaction)
Equilibrium Concentrations
- Concentration of N2O4 at equilibrium = 0.05 moles / 1 L = 0.05 M
- Concentration of NO at equilibrium = 0.1 moles / 1 L = 0.1 M
Calculating the Equilibrium Constant (Kc)
- The equilibrium expression for the reaction:
N2O4(g) ⇌ 2NO(g)
Kc = [NO]^2 / [N2O4]
- Substitute equilibrium concentrations:
Kc = (0.1)^2 / (0.05)
= 0.01 / 0.05
= 0.2
Final Result
- The equilibrium constant (Kc) for the reaction at this state is 0.2.
- Mass of N2O4 = 9.2 g
- Molar mass of N2O4 = 92 g/mol
- Volume of the vessel = 1 L
- Percentage of N2O4 dissociated = 50%
Calculating Moles of N2O4
- Moles of N2O4 = Mass / Molar mass
= 9.2 g / 92 g/mol
= 0.1 moles
Initial Concentration of N2O4
- Initial concentration (C0) = Moles / Volume
= 0.1 moles / 1 L
= 0.1 M
Dissociation at Equilibrium
- Since 50% of N2O4 is dissociated:
Moles dissociated = 0.1 moles × 50% = 0.05 moles
- Remaining moles of N2O4 = 0.1 - 0.05 = 0.05 moles
- Moles of NO formed = 2 × 0.05 = 0.1 moles (from stoichiometry of the reaction)
Equilibrium Concentrations
- Concentration of N2O4 at equilibrium = 0.05 moles / 1 L = 0.05 M
- Concentration of NO at equilibrium = 0.1 moles / 1 L = 0.1 M
Calculating the Equilibrium Constant (Kc)
- The equilibrium expression for the reaction:
N2O4(g) ⇌ 2NO(g)
Kc = [NO]^2 / [N2O4]
- Substitute equilibrium concentrations:
Kc = (0.1)^2 / (0.05)
= 0.01 / 0.05
= 0.2
Final Result
- The equilibrium constant (Kc) for the reaction at this state is 0.2.
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9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)?
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9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)?.
9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 9.2 g N2O4, is heated in a 1L vessel tillequilibrium state is establishedN.04) - 2NO,(0)In equilibrium state 50% N.O. wasdissociated, equilibrium constant will be(mol, wt. of N.04 92)?.
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