Equilibrium practice · Mixed difficulty

At a given temperature the ${K_c}$ for the reaction, $\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)$ is $2.4 \times {10^{ - 3}}$. At the same temperature, the ${K_c}$ for the reaction

$\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)$ is :

Solution of $0.1 \mathrm{~N} {~NH}_4 {OH}$ and $0.1 \mathrm{~N} {~NH}_4 {Cl}$ has pH $9.25.$ Then find out ${pK}_{{b}}$ of ${NH}_4 {OH}$

Sulphide ion $\left(\mathrm{S}^{2-}\right)$ reacts with solid sulphur forming $\mathrm{S}_2^{2-}$ and $\mathrm{S}_3^{2-}$ with formation constants 12 and 132 . Formation constant of $\mathrm{S}_3^{2-}$ from sulphur and $\mathrm{S}_2^{2-}$ is

For which reaction Kp � Kc ?

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

For the reaction, $2A\left( g \right)\;\leftrightharpoons\;3C\left( g \right) + D\left( s \right)$ the value of ${K_c}$ will be equal to

Find the value of $\mathrm{K}_{\mathrm{p}}$ for dissociation of:

$2 \mathrm{HI} \rightleftharpoons \mathrm{H}_2+\mathrm{I}_2$,

if the equilibrium concentration of $\mathrm{HI}, \mathrm{H}_2$ and $\mathrm{I}_2$ are $3.53,0.4789$ and $0.4789 \mathrm{~mole ~litre}^{-1}$ respectively.

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

Phenolphthalein can’t be used in the titration of

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:

For which reaction ${K_p}$ is less than ${K_c}$?

What is ${P^H}$ of 0.02 M solution of ammonium chloride at ${25^\mathrm{o}}C\,\,\,\,\left( {{K_b}\,\,\mathrm{ of }\,\,N{H_3} = 1.8 \times {{10}^{ - 5}}} \right)$

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

(A): Acetic acid is a weak acid.

(R): Its conjugate base is weak.

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

What is the minimum $\mathrm{pH}$ required to prevent the precipitation of $\mathrm{ZnS}$ in a solution that is $0.01 \mathrm{M} \,\, \mathrm{ZnCl}_2$ and saturated with $0.10 \mathrm{M} \,\, \mathrm{H}_2 \mathrm{~S}$ ?
[Given $\mathrm{K}_{\mathrm{sp}}=10^{-21}, \mathrm{~K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2}=10^{-20}$ ]

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

A solution of a weak monobasic acid has $\wedge_M$ for $0.1 N$ solution and $\wedge_{\infty}$ equal to $60 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ and $400 \mathrm{~S}$ $\mathrm{cm}^2 \mathrm{~mol}^{-1}$ respectively. The $\mathrm{pH}$ of this solution and $K_a$ of acid is :

The pH of the solution

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

If $C{O_2}$ is made to escape from the system $C{O_2} + {H_2}O\leftrightharpoons{H_2}C{O_3}\leftrightharpoons{H^ + } + HCO_3^ - ,$ then for the system

For the following reactions, equilibrium constants are given: $S(s) + {O_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}S{O_2}(g);{K_1} = {10^{52}}$
$2S(s) + 3{O_2}(g)\leftrightharpoons2S{O_3}(g);{K_2} = {10^{129}}$
The equilibrium constant for the reaction, $2SO(g) + {O_2}(g)\leftrightharpoons2S{O_3}(g)$ 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: