Seema Saini Solutions for Chapter: Chemical Equilibrium, Exercise 4: DPP 6.4

${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$
$1$ $\text{mole}$ ${\mathrm{N}}_2$ and $3 \text{mole}$ ${\mathrm{H}}_2$ are present at start in $1 \mathrm{L}$ flask. At equilibrium, ${\mathrm{NH}}_3$ formed required $100 \mathrm{mL}$ of $5 \mathrm{M}$ $\mathrm{HCl}$ for neutralization. Hence, ${\mathrm{K}}_{\mathrm{C}}$ is

For equilibrium, $2\mathrm{NOBr} \left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO} \left(\mathrm{g}\right)+{\mathrm{Br}}_2 \left(\mathrm{g}\right)$. Calculate the ratio $\frac{{\mathrm{K}}_{\mathrm{P}}}{\mathrm{P}},$ where $\mathrm{P}$ is the total pressure and ${\mathrm{P}}_{{\mathrm{Br}}_2}=\frac{\mathrm{P}}{9}$ at a certain temperature.

For the reaction: $2\mathrm{HI} \left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2 \left(\mathrm{g}\right)+{\mathrm{I}}_2 \left(\mathrm{g}\right),$ the degree of dissociation $\left(\mathrm{\alpha }\right)$ of $\mathrm{HI} \left(\mathrm{g}\right)$ is related to equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$, by the expression

At $727°\mathrm{C}$ and $1.23 \mathrm{atm}$ of total equilibrium pressure, ${\mathrm{SO}}_3$ is partially dissociated into ${\mathrm{SO}}_2$ and ${\mathrm{O}}_2$, according to the following reaction: ${\mathrm{SO}}_3 \left(\mathrm{g}\right)\rightleftharpoons {\mathrm{SO}}_2 \left(\mathrm{g}\right)+1/2{\mathrm{O}}_2 \left(\mathrm{g}\right).$ The density of the equilibrium mixture is $0.9 \mathrm{gm}/\mathrm{litre}$. The degree of dissociation is

The formation constant of $\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6^{2+}$ is $6\times {10}^8$ at $25°\mathrm{C}$. If $50 \mathrm{mL}$ of $2.0 \mathrm{M} {\mathrm{NH}}_3$ is added to $50 \mathrm{mL}$ of $0.20 \mathrm{M}$ solution of ${\mathrm{Ni}}^{2+}$, then the concentration of ${\mathrm{Ni}}^{2+}$ ion will be nearly equal to:

The vapour density of ${\mathrm{N}}_2{\mathrm{O}}_4$ at a certain temperature is $30$. What is the $\%$ dissociation of ${\mathrm{N}}_2{\mathrm{O}}_4$ at this temperature?

The equation $\mathrm{\alpha }=\frac{\mathrm{D}-\mathrm{d}}{(\mathrm{n}-1)\mathrm{d}}$ is correctly matched for:

$138 \mathrm{g}$ of ${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)$ is placed in a $8.2 \mathrm{L}$ container at $300 \mathrm{K}$. The equilibrium vapour density of mixture was found to be $30.67$. Then,$\left(\mathrm{R}=0.082 \mathrm{L}-\mathrm{atm} {\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\right)$