Embibe Experts Solutions for Chapter: Equilibrium, Exercise 2: Exercise 2

Which would decrease the $\mathrm{pH}$ of $25 {\mathrm{cm}}^3$ of a $0.01 \mathrm{M}$ solution of hydrochloric acid

The equilibrium constant for the reaction: ${\mathrm{H}}_{2 }\left(\mathrm{g}\right)+{\mathrm{I}}_2 \left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{HI} \left(\mathrm{g}\right)$ is $64$. If the volume of the container is reduced to one fourth of its original volume, the value of the equilibrium constant will be:

At $1000 \mathrm{K}$, the value of ${\mathrm{K}}_{\mathrm{p}}$ for the reaction:
$\mathrm{A} \left(\mathrm{g}\right)+2\mathrm{B} \left(\mathrm{g}\right)\rightleftharpoons 3\mathrm{C} \left(\mathrm{g}\right)+\mathrm{D} \left(\mathrm{g}\right)$ is $0.05$ atmosphere. The value of ${\mathrm{K}}_{\mathrm{C}}$ in terms of $\mathrm{R}$ would be:

'$\mathrm{a}$' moles of ${\mathrm{PCl}}_5$, undergoes, thermal dissociation as : ${\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3+{\mathrm{Cl}}_2$, the mole fraction of ${\mathrm{PCl}}_3$ at equilibrium is $0.25$ and the total pressure is $2.0$ atmosphere. The partial pressure of ${\mathrm{Cl}}_2$ at equilibrium is:

The extent of dissociation of ${\mathrm{PCl}}_5$ at a certain temperature and one atm pressure, is $20\%$. Calculate the pressure at which this substance is half dissociated at the same temperature.

In a container equilibrium
${\mathrm{N}}_2{\mathrm{O}}_4 \left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2 \left(\mathrm{g}\right)$
is attained at $25°\mathrm{C}$. The total equilibrium pressure in container is $380$ $\mathrm{torr}$. If equilibrium constant of above equilibrium is $0.667$ $\mathrm{atm}$, then degree of dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ at this temperature will be:

Vapour density of ${\mathrm{PCl}}_5$ is $104.16$, but when heated to $230°\mathrm{C}$, its vapour density is reduced to $62$. The degree of dissociation of ${\mathrm{PCl}}_5$ at this temperature will be:

In the reaction $\mathrm{C} \left(\mathrm{s}\right)+{\mathrm{CO}}_2 \left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{CO} \left(\mathrm{g}\right)$, the equilibrium pressure is $12 \mathrm{atm}.$ If $50 \%$ of ${\mathrm{CO}}_2$ reacts, ${\mathrm{K}}_{\mathrm{p}}$ will be: