Equilibrium practice · Medium

Given the equilibrium constant values,
I. $\mathrm{N}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{N}_2 \mathrm{O}(g) ; K_c=2.7 \times 10^{-18}$
II. $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g); K_c=4.6 \times 10^{-3}$
III. $\frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g) ; K_c=4.1 \times 10^{-9}$
Thus, for the reaction, $2 \mathrm{~N}_2 \mathrm{O}(g)+3 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{~N}_2 \mathrm{O}_4(g), K_c$ is

Statement I: The $\mathrm{pH}$ of $10^{-7} \mathrm{M}-\mathrm{NaOH}$ solution is in between 7.0 to 7.3 at $25^{\circ} \mathrm{C}.$

Statement II: Due to common ion effect, ionization of water is reduced.

For the reaction, $\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)$, the standard free energy is $\Delta G^{\circ} > 0$. The equilibrium constant $\left(K_C\right)$ would be

Assertion (A): The dynamic nature of chemical equilibrium can be demonstrated in the synthesis of ammonia by Haber's process.

Reason (R): In a series of experiments, Haber started with known amounts of dinitrogen and dihydrogen maintained at high temperature and pressure and at regular intervals determinated the amount of ammonia present and concentration of unreacted dihydrogen and dinitrogen.

Calculate the $\mathrm{pH}$ of solution made by Three solution having $\mathrm{pH}=2,4$ and 6 .

The range of pH in which methyl orange works as indicator:

The equilibrium constant for the reaction, $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$ is $K_1$ and the equilibrium constant for the reaction, $\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$ is $K_2$ both at the same temperature. The value of $K_1$ and $K_2$ are related as:

Consider the reaction, $C{H_3}COOH\mathrm{ } + \mathrm{ }{C_2}{H_5}OH\;\;\leftrightharpoons\;\;C{H_3}COO{C_2}{H_5} + \mathrm{ }{H_2}O.$ One mole each of $C{H_3}COOH$ and ${C_2}{H_5}OH$ are heated in the presence of little concentrated ${H_2}S{O_4}.$ On equilibrium:

Lewis base among the following is

For the reaction ${BaO}_2({~s}) \rightleftharpoons {BaO}({s})+{O}_2({~g}) ; \Delta H=+{ve}$.
In an equilibrium condition, pressure of ${O}_2$ depends on

When 3 moles of ethyl alcohol are mixed with 3 moles of acetic acid, 2 moles of ester are formed at equilibrium. According to the equation,

$\mathrm{CH}_3 \mathrm{COOH}(l)+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)+\mathrm{H}_2 \mathrm{O}(l)$
The value of the equilibrium constant for the reaction is

The pH of the solution

$5\;mL\;\frac{\mathrm{M}}{5}\;\mathrm{HCl}\; + \;10\;\mathrm{mL}\;\mathrm{ of}\;\frac{\mathrm{M}}{{10}}\;\mathrm{NaOH}\;\mathrm{is}$

Statement I : On opening a sealed soda water bottle, the dissolved $\mathrm{CO}_2$ escapes out

Statement II : Gas escapes to reach the new equilibrium condition of lower pressure

The ${K_{sp}}$ for bismuth sulphide $(B{i_2}{S_3})$ is $1.08 \times {10^{ - 73}}.$ The solubility of $B{i_2}{S_3}$ in $mol\,{L^{ - 1}}$ at 298 K is

Ammoniacal solution of $\mathrm{Ag}^{+}+$Acidified solution of $\mathrm{N}_2 \mathrm{H}_4 \rightarrow$ ppt of (X)

The equilibrium constant for the reaction,${\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right)$ is $4 \times {10^{ - 4}}$ at 2000 K. In presence of a catalyst the equilibrium is attained ten times faster. Therefore, the equilibrium constant, in present of the catalyst, at 2000 K is:

What will be the final $\mathrm{pH}$ if $0.2 \mathrm{~mol}$ of $\mathrm{NaOH}$ is added to $1 \mathrm{~L}$ of buffer solution consisting of $0.4 \mathrm{~M} \mathrm{~CH}_3 \mathrm{COOH}$ and $0.3 \mathrm{~M} \mathrm{~CH}_3 \mathrm{COONa}?$ $\left(\mathrm{pK}_{\mathrm{a}} \text{ of } \mathrm{CH}_3 \mathrm{COOH}=4.75\right)$

The pH indicators are

The $\mathrm{pH}$ value of $10^{-7} \mathrm{~M}$ solution ${HCl}$ is

The pH of $1\,M\,C{H_3}COONa$ solution in water will be nearly $\left( {p{K_a}\,\,for\,C{H_3}COOH = 4.74} \right)$

A $0.01 \mathrm{M}$ ammonia solution is $5 \%$ ionized, its $\mathrm{pH}$ will be

Assertion :- The value of $K$ increases when concentration of the reactants are increased.

Reason :- With increases of concentration of reactants the equilibrium shifts in forward direction.

The equilibrium ${SOCl}_2(\mathrm{g}) \rightleftharpoons {SO}_2(\mathrm{g})+{Cl}_2(\mathrm{g})$ is attained at $25^{\circ} \mathrm{C}$ in a closed rigid container and helium gas is introduced. Which of the following statements is correct?

An equilibrium mixture of the reaction, $2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right)$ contains 0.120 mole of $\mathrm{N}{\mathrm{O}_2}$, 0.080 mole of $\mathrm{NO}$ and 0.640 mole of ${\mathrm{O}_2}$ in a 4 litre flask at constant temperature. The value of ${K_c}$ for the reaction at this temperature is:

For reaction $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$
Some amount of A, B, C and D are added for reaction, then reaction quotient ' $\mathrm{Q}$ '

The dissociation constant of $\mathrm{N}{\mathrm{H}_4}\mathrm{OH}$ is $1.8 \times {10^{ - 5}}.$ The hydrolysis constant of $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ would be:

STATEMENT I : $\mathrm{pH}$ of $10^{-8} \mathrm{M}$ $\mathrm{HCl}$ lies between 6 and 7.

STATEMENT II : For very dilute solutions of acids, $\mathrm{H}^{+}$ion contribution form water is also taken into consideration.

The solubility product constant $\mathrm{K}_{\text {sp }}$ of $\mathrm{Mg}(\mathrm{OH})_2$ is $9.0 \times 10^{-12}$. If a solution is $0.010 \mathrm{M}$ with respect to $\mathrm{Mg}^{2+}$ ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of $\mathrm{Mg}(\mathrm{OH})_2$.

Calculate ${pH}$ of $0.1 \, \mathrm{M}$ solution of ${NH}_4 {Cl}?$ $\left({K}_{{b}}\right.$ for ${NH}_3$ is $2.5 \times 10^{-6}$)

Given ${N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right);\,\,{K_1}$
${N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right);\,\,{K_2}$
${H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons {H_2}O\left( g \right);\,\,{K_3}$
The equilibrium constant for $2N{H_3}\left( g \right) + \frac{5}{2}{O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right) + 3{H_2}O\left( g \right)$ will be