Application of Equilibrium Constant

When two reactants, A and B are mixed to give products C and D, the reaction quotient $\mathrm{Q}$, at the initial stage of the reaction :–

When two reactants A and B are mixed to give products C and D, the reaction quotient Q at the initial stage of the reaction is best described by-

The vapour density of a sample ${\mathrm{N}}_2{{\mathrm{O}}_4}_{\left(\mathrm{g}\right)}$ is $40$ at $10 \mathrm{atm}$. Calculate the ${\mathrm{K}}_{\mathrm{P}}$ for the reaction.

The degree of dissociation of acetic acid in a $0.1 \mathrm{N}$ solution is $1.32\times {10}^{-2}$. At what concentration of nitrous acid will its degree of dissociation be the same as that of acetic acid?

${\mathrm{K}}_{\mathrm{a}} \mathrm{for} {\mathrm{HNO}}_2 = 4 \times {10}^{-4}$

(Report your answer by multiplying the value with 10)

At what concentration of the solution will the degree of dissociation of nitrous acid be $0.2$?

 (${\mathrm{K}}_{\mathrm{a}}$ for ${\mathrm{HNO}}_2$ is $4\times {10}^{-4}$)

Report the value by multiplying with 1000 and rounding off to one significant figure.

Calculate the percentage degree of ionisation of $0.4 \mathrm{M}$ acetic acid in water. Dissociation constant of acetic acid is $1.8\times {10}^{-5}$.

Give your answer by multiplying the value with 1000.

$\mathrm{HI}$ is introduced into three identical $500 \mathrm{mL}$ bulbs at $350°\mathrm{C}$. Each bulb is opened at different time intervals and analysed for ${\mathrm{I}}_2$ by titrating with $0.015 \mathrm{M}$ hypo solution.

$\left({\mathrm{I}}_2+2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3\to {\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}\right)$

 ${\mathrm{K}}_{\mathrm{c}}$ for $2\mathrm{HI}\rightleftharpoons {\mathrm{H}}_2+{\mathrm{I}}_2$ at $350°\mathrm{C}$ is $\frac{\mathrm{y}}{2}\times {10}^{-2}$

Express the $\mathrm{y}$ value after rounding off up to nearest integer.

$5$ moles of ${\mathrm{S}\mathrm{O}}_2$ and $5$ moles of ${\mathrm{O}}_2$ react in a closed vessel. At equilibrium $60\mathrm{\%}$ of the ${\mathrm{S}\mathrm{O}}_2$ is consumed. The total number of gaseous moles $(\mathrm{S}{\mathrm{O}}_2, {\mathrm{O}}_2\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{S}{\mathrm{O}}_3)$ in the vessel is:

$0.6$ mole of ${\mathrm{P}\mathrm{C}\mathrm{l}}_5,\mathrm{ }0.3$ mole of ${\mathrm{P}\mathrm{C}\mathrm{l}}_3$ and $0.5$ mole of ${\mathrm{C}\mathrm{l}}_2$ are taken in a $1\mathrm{L}$ flask to obtain the following equilibrium:
$\mathrm{P}\mathrm{C}{\mathrm{l}}_{5\left(\mathrm{g}\right)}\leftrightharpoons \mathrm{P}\mathrm{C}{\mathrm{l}}_{3\left(\mathrm{g}\right)}+\mathrm{C}{\mathrm{l}}_{2\left(\mathrm{g}\right)}$
If the equilibrium constant ${\mathrm{K}}_{\mathrm{c}}$ for the reaction is $0.2$. Predict the direction of the reaction.

$2\mathrm{A}+\mathrm{B}\rightleftharpoons 3\mathrm{C}, {\mathrm{K}}_{\mathrm{C}}=49$

For the above reaction, $3$ $\mathrm{L}$ vessel contains $2, 1$ and $3$ moles of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. Then, what will happen to the reaction?

The reaction quotient $\left(\mathrm{Q}\right)$ predicts:

For a reaction $\mathrm{A}\to$ products, the value of equilibrium constant $\text{'}\mathrm{K}\text{'}$, when the reaction reaches completion, would be close to which among the following?

For the reaction at $27°\mathrm{C}$

${\mathrm{MgCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{MgO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2; {\mathrm{K}}_{\mathrm{P}}=100 \mathrm{atm}$

If $42 \mathrm{g}$ of  ${\mathrm{MgCO}}_3$ is taken in $20 \mathrm{L}$ vessel then calculate percentage dissociation of ${\mathrm{MgCO}}_3$. 

[Take : $\mathrm{R}=0.08 \mathrm{L} \mathrm{atm} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$]

For the reaction; $\mathrm{A}+\mathrm{B}\rightleftharpoons 3\mathrm{C}$, at $25°\mathrm{C}$, a $3$ litre vessel contains $1,2,4$ moles of $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ respectively. If ${\mathrm{K}}_{\mathrm{c}}$, for the reaction is $10$, the reaction will proceed in:

The equilibrium constant for a gaseous reaction is ${\mathrm{K}}_{\mathrm{C}}=\frac{\left[\mathrm{HI}\right]}{\sqrt{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}}$. The correct balanced equation for expression is :

${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$ may be converted to $\mathrm{Fe}$ by the reaction :-

${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{Fe}\left(s\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

for which ${\mathrm{K}}_{\mathrm{C}}=8$ at temp. $720°\mathrm{C}$
What percentage of the ${\mathrm{H}}_2$ remains unreacted after the reaction has come to equilibrium ?

At $500 \mathrm{K}$, 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 $25$. If $\frac{1}{2} \mathrm{mol}/\mathrm{L}$ of $\mathrm{HI}$ is present at equilibrium, what are the concentrations of ${\mathrm{H}}_2$ and ${\mathrm{I}}_2$, assuming that we started by taking $\mathrm{HI}$ and reached the equilibrium at $500 \mathrm{K}$ ?

${\mathrm{A}}_{\left(\mathrm{g}\right)}+3{\mathrm{B}}_{\left(\mathrm{g}\right)}\rightleftharpoons 2{\mathrm{C}}_{\left(\mathrm{g}\right)}+{\mathrm{D}}_{\left(\mathrm{g}\right)}$ Initial moles of $\mathrm{A}$ is twice that of $\mathrm{B}$. If at equilibrium, moles of $\mathrm{B}\&\mathrm{C}$ are equal, then $\%$ of $\mathrm{B}$ reacted will be :

$\mathrm{A}\left(\mathrm{g}\right)+3 \mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$ Initial moles of $\mathrm{A}$ is twice that of $\mathrm{B}$. If at equilibrium, moles of $\mathrm{B} \& \mathrm{C}$ are equal, then $\%$ of $\mathrm{B}$ reacted will be :