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:

Image of N2 and O2 equilibrium

Image of N2 and O2

Image of N2 and O2 equilibrium

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.

Answer:

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

Then,

Now,

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

Answer:

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

Structure of Sulfur Dioxide and Sulfur Trioxide

Structure of Sulfur Dioxide and Sulfur Trioxide

Structure of Sulfur Dioxide and Sulfur Trioxide

What is at this temperature?

Answer:

For the given reaction,

We know that,