Chemistry Class 11 NCERT Solutions: Chapter 7 Equilibrium Part 3

Chemistry Class 11 NCERT Solutions Chapter 7 Equilibrium Part 3

Q:8. Reaction between Equation takes place as follows:

Equation

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

Answer:

Let the concentration of Equation at equilibrium be x.

The given reaction is:

Equation

Therefore, at equilibrium, in the 10 L vessel:

Equation

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

Then,

Equation

Now,

Equation

Equation

Equation

Equation

Equation

Equation

Equation

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

Equation

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

Answer:

The given reaction is:

Equation

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

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

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

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

[NO]=0.087 mol Equation = 0.0437 mol

Therefore, the amount of No present at equilibrium is:

Equation

Equation

And, the amount of Br present at equilibrium is:

Equation

Equation

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

What is Equation at this temperature?

Answer:

For the given reaction,

Equation

Equation

Equation

Equation

We know that,

Equation

Equation

Equation

Equation

Equation

Equation

Equation

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