Equilibrium practice · Medium

Statement-I: $\mathrm{HNO}_3$ is not a Bronsted acid in $\mathrm{CHCl}_3$.

Statement-II: $\mathrm{CHCl}_3$ is not an example for protophilic solvent.

${O}_2$ is kept in a vessel at $300 \mathrm{~K}$. Following equilibrium is established $3 {O}_2(\mathrm{g}) \rightleftharpoons 2 {O}_3(\mathrm{g})$. At equilibrium partial pressure of ${O}_2=10^5$ bar & partial pressure of ${O}_3=10^{-10}$ bar. Calculate $\Delta {G}^{\circ}$ (in kcal) of the above reaction. [Given: ${R}=2 \mathrm{~cal ~K}^{-1} \mathrm{~mol}^{-1} ; \log _{{e}} 10=2.3$]

$0.1\;\mathrm{millimole}$ of $\mathrm{CdS}{\mathrm{O}_4}$ are present in $10\;\mathrm{mL}$ acid solution of $0.08\;N\;\mathrm{HCl}\mathrm{.}$ Now ${\mathrm{H}_2}\mathrm{S}$ is passed to precipitate all the $\mathrm{C}{\mathrm{d}^{2 + }}$ ions. The $\mathrm{pH}$ of the solution after filtering off precipitate, boiling of ${\mathrm{H}_2}\mathrm{S}$ and making the solution $100\;\mathrm{mL}$ by adding ${\mathrm{H}_2}O$ is:

If $\mathrm{\alpha }$ is the degree of ionisation, $C$ the concentration of a weak electrolyte and ${{K_a}}$ the acid ionisation constant then the correct relationship between $\mathrm{\alpha },C\;and\;{K_a}$ is

For the reaction, $F{e_2}N(s) + \frac{3}{2}{H_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}2Fe(s) + N{H_3}(g)$

At the state of dynamic equilibrium, for solute $+$ solvent $\leftrightharpoons$ solution.

The highest pH value is of:

$S{O_2}\left( s \right) + 2{H_2}\left( g \right) \rightleftharpoons Sn\left( s \right) + 2{H_2}O\left( g \right)$ in equilibrium mixture 40% water vapour is present by volume. Then find ${K_p}$ if total pressure at equilibrium is 5 atm.

4 mole of $A$ are mixed with 4 mol of $B$, when 2 mole of $C$ are formed at equilibrium according to the reaction $A + B\,\leftrightharpoons\,C + D$, Then equilibrium constant is

STATEMENT-1 : For the reaction $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$ if the volume of vessel is reduced to half of its original volume, equilibrium concentration of all gases will be doubled.

STATEMENT-2 : According to Le-Chatelier's principle, reaction shifts in a direction that tends to minimized the effect of the stress.

The ionic product of water at $25^{\circ} \mathrm{C}$ is $10^{-14}$. Its ionic product at $90^{\circ} \mathrm{C}$ will be

At $40^{\circ} \mathrm{C},$ the density of heavy water is $1.02 \mathrm{~g} / \mathrm{ml}$ and its ionic product is $5.1 \times 10^{-15}.$ Which of the following if the only incorrect information regarding heavy water at $40^{\circ} \mathrm{C}?$

The strongest base among the following is:

A: At higher temperature, $\mathrm{K}_{\mathrm{w}}$ of water remains unaltered.
R: $\mathrm{K_w}$ is a constant.

Solubility product of $\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2}$ at ordinary temperature is $1.96 \times {10^{ - 11}}$. pH of a saturated solution of $\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2}$ will be

At $500 \mathrm{~K}$, equilibrium constant, $K_C$, for the following reaction is 5. $\frac{1}{2} \mathrm{H}_2(g)+\frac{1}{2} \mathrm{I}_2(g) \rightleftharpoons \mathrm{HI}(g)$ What would be the equilibrium constant $K_C$ for the reaction? $2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_2(g)+\mathrm{I}_2(g)$

The interfering radicals interfere in the test of usual inorganic analysis after II group analysis due to:

$0.04\;\mathrm{g}$ of pure $\mathrm{NaOH}$ is dissolved in $10\;\mathrm{litre}$ of distilled water. The $\mathrm{pH}$ of the solution is:

A solution of $Cu{(N{O_3})_2}$ in water is acidic due to

$\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{OH}^{-}$In the reaction water acts as

I. Bronsted acid

II. Bronsted Base

III. Amphoteric oxide

$HX$ is a weak acid ${K_a} = {10^{ - 5}}$. It forms a salt $NaX$ ($0.1\mathrm{ }M$ on reacting with caustic soda. The degree of hydrolysis of $NaX$ is

Assertion (A): The equilibrium (given below) attained in a closed vessel remains unaltered by the addition of $\mathrm{CaCO}_{3(\mathrm{s})}.$

$\mathrm{CaCO}_{3(\mathrm{s})} \rightleftharpoons \mathrm{CaO}_{(\mathrm{s})}+\mathrm{CO}_2 \uparrow$

Reason (R): The active mass of a solid is a constant and independent of its mass and is always taken to be unity.

For the decomposition of ${CaCO}_3({s})$ at $1080 \mathrm{~K}$, equilibrium constant is equal to $8.87$ atmospheres. Equilibrium concentration of carbondioxide is equal to

For the equilibrium,

${Ag}^{+}+2 {NH}_3 \rightleftharpoons {Ag}\left({NH}_3\right)_2^{+} ; K_1=1.8 \times 10^7$
$\text {and } {Ag}^{+}+{Cl}^{-} \rightleftharpoons {AgCl} ; K_2=5.6 \times 10^9$

Hence, for the equilibrium,

${AgCl}+2 {NH}_3 \rightleftharpoons {Ag}\left({NH}_3\right)_2^{+}+{Cl}^{-}$

equilibrium constant is

For the reaction $2 {H}_2+2 {NO} \rightleftharpoons {N}_2+2 {H}_2 {O}$ the following mechanism has been suggested :

$\begin{aligned} & 2 {NO} \rightleftharpoons {N}_2 {O}_2 \text { [equilibrium constant }=K_1 \text { (fast)] } \\ & {N}_2 {O}_2+{H}_2 \xrightarrow{{K}_2} {~N}_2 {O}+{H}_2 {O} \text { (slow) } \\ & {N}_2 {O}+{H}_2 \xrightarrow{{K}_3} {~N}_2+{H}_2 {O} \text { (fast) } \end{aligned}$

According to following mechanism is correctly matched.

The equilibrium,

${\mathrm{P}_4}\left( s \right) + 6\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;4\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right)$ attained by mixing equal moles of ${\mathrm{P}_4}$ and $\mathrm{C}{\mathrm{l}_2}$ in a evacuated vessel. Then, at equilibrium,

Statement I: In the dissociation of $\mathrm{PCl}_5$ at constant pressure and temperature addition of helium at equilibrium increases the dissociation of $\mathrm{PCl}_5$.

Statement II : Helium removes $\mathrm{Cl}_2$ from the field of action.

Which of the following affects the value of equilibrium constant?

What will be the resultant $\mathrm{pH}$ when $200 \mathrm{~ml}$ of an aqueous solution of ${HCl}(\mathrm{pH}=2.0)$ is mixed with $300 \mathrm{~ml}$ of an aqueous solution of ${NaOH}$ $(\mathrm{pH}=12.0)?$

Assertion (A): pH of $x \mathrm{~M~HCl}$ is less than pH of $x \mathrm{~M} \mathrm{~CH}_3 \mathrm{COOH}.$

Reason (R): The degree of ionization of HCl and $\mathrm{CH}_3 \mathrm{COOH}$ are equal at infinite dilution.