Equilibrium practice · Medium

Nitric oxide reacts with bromine and give nitrosyl bromide as per the reaction given below.

$2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOBr}(\mathrm{g})$

when 0.087 mol of NO and 0.0437 mol of $\mathrm{Br}_2$ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide:

At $25^\circ C$, the equilibrium ${K_1},{K_2}$ and ${K_3}$ of three reactions are :

${\mathrm{N}_2} + 3{\mathrm{H}_2}\;\leftrightharpoons\;2\mathrm{N}{\mathrm{H}_3};\;{K_1}$

${\mathrm{N}_2} + {\mathrm{O}_2}\;\leftrightharpoons\;2\mathrm{NO};\;{K_2}$

${\mathrm{H}_2} + \frac{1}{2}{\mathrm{O}_2}\;\leftrightharpoons\;{\mathrm{H}_2}\mathrm{O};\;{K_3}$

The equilibrium constants for the oxidation of $\mathrm{N}{\mathrm{H}_3}$ by oxygen to give $\mathrm{NO}$ is :

$0.365\;g$ of $\mathrm{HCl}$ gas was passed through $100\;\mathrm{c}{\mathrm{m}^3}$ of $0.2\;M\;\mathrm{NaOH}$ solution. The $pH$ of the resulting solution would be

For the equilibrium $\alpha$-(D) glucose $\rightleftharpoons \beta$-(D) glucose $K_c=18$ The percentage of $\alpha$-(D)-glucose at equilibrium is

In the following equilibrium amount of $\mathrm{CO}(g)$ can be decreased by

$2 \mathrm{CO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g)$

Statement-I: $\mathrm{ClO}_4^{-}$is the weakest acid.

Statement-II: In $\mathrm{ClO}_4$; ion chlorine is $\mathrm{sp}^3$ hybridized.

The dissociation constants of acids $\mathrm{HA}, \mathrm{HB}, \mathrm{HC}$ and $\mathrm{HD}$ are $2.6 \times 10^{-3}, 5.3 \times 10^{-9}, 1.1 \times 10^{-2}, 7.5 \times 10^{-5}$ respectively. The weakest acid among these acids is

Molar conductance of $0.01 \mathrm{~M}$ weak monoprotic acid $\text{(HA)}$ is $40 \mathrm{~mho} \mathrm{~cm}^2 \mathrm{~mol}^{-1}.$ $\mathrm{p}^{\mathrm{H}}$ of the solution is (Limiting molar conductance of $\text{HA}$ is $400 \mathrm{~mho} \mathrm{~cm}^2 \mathrm{~mol}^{-1})$

What will be the value of ${P^H}$ of $0.01\,\,\,{\text{mol}}\,\,\,{\text{d}}{{\text{m}}^{ - 3}}\,\,C{H_3}COOH\,\,\,\,\,\left( {{K_a} = 1.74 \times {{10}^{ - 5}}} \right)?$

Statement I : When a 10 L flask contains 4 moles each of $PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + C{l_2}(g)\,\,({K_C} = 5)$ equilibrium proceeds in backward side.
Statement II : Reaction quotient, ${Q_C} > {K_C}$

${10^{ - 6}}\;M\;\mathrm{HCl}$ is diluted to 100 times. Its $pH$ is:

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

What will be the resultant ${pH}$ when $150 \mathrm{~mL}$ of an aqueous solution of ${HCl} \, ({pH}=2.0)$ is mixed with $350 \mathrm{~mL}$ of an aqueous solution of ${NaOH} \, ({pH}=12.0)?$

The $pOH$ of a solution obtained by mixing $25 \mathrm{~mL}$ of $0.2 \mathrm{~M} \mathrm{~HCl}$ with $50 \mathrm{~mL}$ of $0.25 \mathrm{~M} \mathrm{~NaOH}$ is $\left(Kw=1.0 \times 10^{-14}\right)$

(A): A very dilute acidic solution of $\mathrm{Cd}^{2+}$ and $\mathrm{Ni}^{2+}$ gives yellow precipitate of $\mathrm{CdS}$ on passing hydrogen sulphide.
(R): Solubility product of CdS is more than that of NiS.

pH of ${10^{ - 2}}\mathrm{M}\,\,\,Ca{(OH)_2}$ aqueous solution is nearly

In the decomposition reaction of ammonia: $2 \mathrm{NH}_3(\mathrm{g}) \rightleftharpoons \mathrm{N}_2(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g}). 2$ moles of $\mathrm{NH}_3$ are introduced in the vessel of 1 litre. At equilibrium, 1 mole $\mathrm{NH}_3$ was left, the value of $\mathrm{K_c}$ will be:

Dissociation constants of two acids HA and HB are $4 \times {10^{ - 10}}$ and $1.8 \times {10^{ - 5}}$ resp. Whose ${P^H}$ value will be higher for a given molarity

pH of solution can be expressed as

Consider the reaction equilibrium

$\begin{array}{ll} \text { Ice } \rightleftharpoons & \text { Water }-\,{x} \, {kcal} \\ \left(\begin{array}{l} \text { Greater } \\ \text { volume } \end{array}\right) & \left(\begin{array}{l} \text { Lesser } \\ \text { volume } \end{array}\right) \end{array}$

The favorable conditions for forward reaction are

Equilibrium constant for the following equilibrium,

$3 \mathrm{ClO}^{-}(a q) \rightleftharpoons 2 \mathrm{Cl}^{-}(a q)+\mathrm{ClO}_3^{-}(a q)$

is $10 \times 10^{28}$ at a temperature at which $R T=2500 \mathrm{~J} \mathrm{~mol}^{-1}$.
Gibbs free energy change $\left(\Delta G^{\circ}\right)$ is

The value of $K_{s p}$ of different salts $A B, A B_2, A_2 B$ and $A_2 B_3$ are $4 \times 10^{-10}, 8 \times 10^{-15}, 2 \times 10^{-14}$, $5 \times 10^{-18}$ respectively. The most soluble salt is

The equilibrium constant in a reversible reaction at a given temperature

The equilibrium constant for ${A_2}\left( g \right)\leftrightharpoons\,2A\left( g \right)$ at 400 K and 600 K are $1 \times {10^{ - 8}}\,\,\mathrm{ and }\,\,1 \times {10^{ - 2}}$ respectively. The reaction is

A weak acid, ${HA}$, has a ${K}_{{a}}$ of $1.00 \times 10^{-5}$. If 0.100 mole of this acid dissolved in one litre of water, the percentage of acid dissociated at equilbrium is closest to

Ammonia under a pressure of $15\mathrm{ }atm$ at $27^\circ C$ is heated to $347^\circ C$ in a closed vessel in the presence of catalyst. Under the conditions, $\mathrm{N}{\mathrm{H}_3}$ is partially decomposed according to the equation, $2\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}$. The vessel is such that the volume remains effectively constant whereas pressure increases to $50\mathrm{ }atm$. Calculate the percentage of $\mathrm{N}{\mathrm{H}_3}$ actually decomposed

Addition of sodium acetate to $0.1\;M$ acetic acid will cause

$1 \mathrm{~mol}$ of helium is added to the dissociation equilibrium $\mathrm{N}_2 \mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{g})-x \mathrm{~kJ}$ set up in a cylinder fitted with a piston in such a way that the piston is moved outwards to keep the total equilibrium pressure constant

For preparing a buffer solution of $\mathrm{pH}=7.0$, which buffer system you will choose?

1M solution of $C{H_3}COOH$ should be diluted to…….. times so that pH is doubled.