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Academic session 2026-27, Chapter 6 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Equilibrium. This page offers a step-by-step solution to the specific question from Exercise 1, Question 8:
Reaction between N2 and O2– takes place as follows:
2N2 (g) + O2 (g) ↔ 2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Reaction between N2 and O2– takes place as follows:
2N2 (g) + O2 (g) ↔ 2N2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10–37, determine the composition of equilibrium mixture.
Let the concentration of N2O at equilibrium be x. The given reaction is:
2N2 (g) + O2 (g) ↔ 2N2O (g)
Intial Concentration: 0.482 mol 0.933 mol 0 mol
At equilibrium (0.482-x) mol (0.933 -x) mol x mol
Therefore, at equilibrium, in the 10 L vessel:
N2 = 0.482-x / 10
O2 = 0.933-x/2 / 10
N2O = x / 10
The value of equilibrium constant i.e.Kc = 2.0 × 10-37 is very small. Therefore, the amount of N2 and O2 reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2.
Then,
N2 = 0.482/10 = 0.0482 molL-1 and O2 = 0.933/10 = 0.0933 molL-1
Now,

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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 8: Reaction between N2 and O2– takes place as follows: 2N2 (g) + O2 (g) &harr....
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