Chemistry Class 11 NCERT Solutions: Chapter 7 Equilibrium Part 3

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Chemistry Class 11 NCERT Solutions Chapter 7 Equilibrium Part 3

Q:8. Reaction between takes place as follows:

If a mixture of 0.482 mol of mol of is placed in a 10 L reaction vessel and allowed to form at a temperature for which determine the composition of equilibrium mixture.


Let the concentration of at equilibrium be x.

The given reaction is:

Therefore, at equilibrium, in the 10 L vessel:

The value of equilibrium constant i.e., is very small. Therefore, the amount of reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of



Q: 9. Nitric oxide reacts with and gives Nitrosyl bromide as per reaction given below:

When 0.087 mol of NO and 0.0437 mol of are mixed in a closed container at constant temperature, 0.0518 mol of is obtained at equilibrium. Calculate equilibrium amount of NO and


The given reaction is:

Now, 2 mol of are formed from 2 mol of NO. Therefore, 0.0518 mol of are formed from 0.0518 mol of NO.

Again, 2 mol of are formed from 1 mol of Br.

Therefore, 0.0518 mol of are formed from mol of Br, or 0.0259 mol of No.

The amount of NO and Br present initially is as follows:

[NO]=0.087 mol = 0.0437 mol

Therefore, the amount of No present at equilibrium is:

And, the amount of Br present at equilibrium is:

Q: 10. At 450 bar for the given reaction at equilibrium

What is at this temperature?


For the given reaction,

We know that,