Equilibrium practice · Mixed difficulty

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.

Ionisation of weak acid can be calculated by the formula

In this reaction water acts as
I. Bronsted acid
II. Bronsted base
III. Amphoteric oxide

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

What is the molar solubility of $Al{\left( {OH} \right)_3}$ in 0.2 M NaOH solution? Given that, solubility product of $Al{\left( {OH} \right)_3} = 2.4 \times {10^{ - 24}}$

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 $40^{\circ} \mathrm{C}?$

Given, $p K_b$ values
$\begin{array}{cccccccccccccccccccc} {{\text{Species}}}&{p{K_b}} \\ {N{H_3}}&{4.74} \\ {{A^ - }}&{4.00} \\ {{B^ - }}&{5.00} \\ {{C^ - }}&{6.00} \end{array}$
queous solutions of $0.01 \mathrm{M}$ salts of $\mathrm{HX}$

$(X=A, B, C)$ and $\mathrm{NH}_4 \mathrm{OH}$ have $\mathrm{pH}$ in the order

The strongest base among the following is:

$\mathrm{I}_2+\mathrm{I}^{-} \rightleftharpoons \mathrm{I}_3^{-}$This reaction is set-up in aqueous medium. We start with 1 mole of $\mathrm{I}_2$ and 0.5 mole of $\mathrm{I}^{-}$in $1 \mathrm{~L}$ flask. After equilibrium is reached, excess of $\mathrm{AgNO}_3$ gave 0.25 mole of yellow precipitate. equilibrium constant 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

The equilibrium $\mathrm{NH}_4 \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_3(\mathrm{g})+\mathrm{H}_2 \mathrm{S}(\mathrm{g})$ is achieved at the equilibrium pressure of '$\mathrm{X}$' bar at $T \mathrm{~K}.$ The value of $\Delta_{\mathrm{r}} G^{\circ}$ for the reaction is

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

A $50 \mathrm{~ml}$ of mixture of $\mathrm{He}, \mathrm{CO}$ and $\mathrm{C}_2 \mathrm{H}_6$ was exploded in $120 \mathrm{~ml}$ of oxygen. On cooling, the gases occupied $142 \mathrm{~ml}$. After treatment with $\mathrm{KOH}$ the volume reduced by $26 \mathrm{~ml}$ and again on treatment with alkaline pyrogallol the volume further reduced, then wrong one 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

The solubility of metal sulphide in saturated solution of $\mathrm{H}_2 \mathrm{S}$ (concentration $=0.1 \mathrm{~M}$) can be represented as follows.

$\mathrm{MS}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{M}^{2+}(\mathrm{aq})+\mathrm{H}_2 \mathrm{S}(\mathrm{aq}) ; K_{\mathrm{eq}}=\frac{\left[\mathrm{M}^{2+}\right]\left[\mathrm{H}_2 \mathrm{S}\right]}{\left[\mathrm{H}^{+}\right]^2}$

The values of $K_{\mathrm{eq}}$ for the metal sulphides, $\text{MnS, } \mathrm{ZnS}, \mathrm{CoS}$ and $\mathrm{PbS}$ are $3 \times 10^{10}, 3 \times 10^{-2}, 3$ and $3\times 10^{-7},$ respectively. If the concentration of each metal ion in a saturated solution of $\mathrm{H}_2 \mathrm{S}$ is $0.01 \, \mathrm{M},$ then which metal sulphide(s) will precipitate at $\left[\mathrm{H}^{+}\right]=1.0 \mathrm{~M}?$

It you have saturated solution of $Ca{F_2}$ then

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

The pH of solution is increased from 3 to 6. Its $\mathrm{H}^{+}$ ion conc. will be:

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.

When a bottle of cold drink is opened the gas comes out with a fizzle due to:

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,

The solubility of ${AgCl}$ in water at $298 \mathrm{~K}$ is $1.06 \times 10^{-5} \text{ mol L}^{-1}.$ The solubility product of ${AgCl}$ at the same temperature was found to be ${p} \times 10^{-q}$. The values of ${p}$ and ${q}$ are respectively