If pressure is applied to the following equilibrium, liquid vapours the boiling point of liquid

Q: If pressure is applied to the following equilibrium, liquid vapours the boiling point of liquid

(A) will increase

(B) will decrease

(C) may increase or decrease

(D) will not change

Solution: Boiling point of a liquid is the temperature at which vapour pressure became equal to atm pressure. If the pressure is applied to the above equilibrium the reaction will go to the backward direction, i.e. vapour pressure decrease hence the boiling point increase.

Ans: (A)

For the reaction $ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) $, the forward reaction at constant temperature

Q: For the reaction $ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) $, the forward reaction at constant temperature is favoured by

(A) Introducing an inert gas at constant volume

(B) Introducing Cl2 gas at constant pressure

(C) Introducing an inert gas at constant pressure

(D) Increasing the volume of the container

Solution: Introducing the inert gas at constant pressure will increase the volume of container and when the volume of container is increased equilibrium shift in the direction where no. of moles are increasing i.e. toward forward direction.

Ans: (C) and (D)

The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases…

Q: The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the final temperature is

(A) 80

(B) 60

(C) 40

(D) 70

Solution:

$ N_2O_4 \rightleftharpoons 2NO_2 $
$1 \quad \quad \quad \quad 0 $ (at initial)
$1-\alpha \quad \quad \quad \quad 2\alpha $ (at equilibrium)

$\large \frac{V.D_{initial}}{V.D_{final}} = \frac{n_{initial}}{n_{final}} $

$\large \frac{46}{25.4} = \frac{25.4}{46}= \frac{1+\alpha}{1} $

1.8 = 1 + α

α = 0.8 or 80 %