Equilibrium MCQs for NEET — Chemistry Questions with Answers

(a) BaCl₂ is a salt of strong acid HCl and strong base Ba(OH)₂. So its aqueous solution is neutral with pH 7. All other salts give acidic solution due to cationic hydrolysis, so their pH is less than 7. Thus, pH value is highest for the solution of BaCl₂.

(c) H₂SO₄ is strong acid having pH<7. NaNO₂ on hydrolysis gives alkaline solution of pH>7. NaCl is neutral and H₂S is weak acid.

(b) $\mathrm{HI}\left(\mathrm{g}\right) \stackrel{ }{\rightleftharpoons }\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right) + \frac{1}{2}{\mathrm{I}}_2\left(\mathrm{g}\right)$

$$\begin{gathered} \mathrm{K}=\frac{{\left[{\mathrm{I}}_2\right]}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\left[{\mathrm{H}}_2\right]}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\left[\mathrm{HI}\right]} .........\left(\mathrm{i}\right) \\ {\mathrm{H}}_2\left(\mathrm{g}\right) + {\mathrm{I}}_2\left(\mathrm{g}\right) \stackrel{ }{\leftrightharpoons }2\mathrm{HI}\left(\mathrm{g}\right) \\ \mathrm{K}\text{'}= \frac{{\displaystyle {\left[\mathrm{HI}\right]}^2}}{\left[{\mathrm{H}}_2\right]{\displaystyle \left[{\mathrm{I}}_2\right]}} .........\left(\mathrm{ii}\right) \\ \mathrm{From} \mathrm{Eqs}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{K} \times \sqrt{\mathrm{K}\text{'}}= 1 \\ \mathrm{K}\text{'}=\frac{1}{{\mathrm{K}}^2}=\frac{1}{8^2}=\frac{1}{64} \end{gathered}$$

For the reaction , ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

Moles at t = 0       1      3         0

Moles at t=${\mathrm{t}}_{\mathrm{eq}}$   (1-$\alpha$)   (1-3$\alpha$) 2$\alpha$

By stoichiometry of the reaction

Initially, total number of moles, ${\mathrm{n}}_1=1+3=4$

At equillibrium,

Degree of dissociation, $\alpha$ = 20

Moles of ${\mathrm{N}}_2$ at equilibrium = 1-0.2=0.8

Moles of ${\mathrm{H}}_2$ at equilibrium = 3-3(0.2)=2.4

Moles of ${\mathrm{NH}}_3$ at equilibrium = 2(0.2)=0.4

Total moles at equilibium, ${\mathrm{n}}_2=0.8+2.4+0.4=3.6$

Using ideal as equation

PV=nRT

As this reaction is carried out in a closed apparatus, the volume of  the system remains constant and it is given that the temperature is also, same hence temperature and volume remains constant

Initial conditions, P=100 atm

n=4 moles

Equation becomes ${\mathrm{P}}_1\mathrm{V}={\mathrm{n}}_1\mathrm{RT}....\left(1\right)$

Final conditions, P=? atm

n=3.6 moels 

Equation becomes ${\mathrm{P}}_2\mathrm{V}={\mathrm{n}}_2\mathrm{RT}....\left(2\right)$

Dividing equation 1 by equation 2, we get

$\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{n_1}{n_2}$

Putting values in the above equation

$$\begin{gathered} \frac{100}{{\mathrm{P}}_2}=\frac{4}{3.6} \\ {P}_2=90 atm \end{gathered}$$

(b) On removal of CO₂ (one of the reaction), reaction will proceed in backward direction.

(a) Salts of strong acid and strong base do not undergo hydrolysis.

(a) A pair constituent with HNO₂ and NaNO₂ because HNO₂ is weak acid and NaNO₂ is a salt of weak (HNO₂) with strong base (NaOH). Hence, it is an example of acidic buffer solution.

(d) Key Idea The reaction for which the number of moles of gaseous products(n_p) is not equal to the number of moles of gaseous reactants (n_R), has different value of K_c and K_p.

From the equation, ${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}\times (\mathrm{RT}{)}^{∆{\mathrm{n}}_{\mathrm{g}}}$

where, $\left[∆{\mathrm{n}}_{\mathrm{g}}\mathrm{gaseous} = {\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{R}}\right]$$$

(a) n_p = n_R = 2, thus, K_p=K_c

(b) n_p = n_R = 2, thus, K_p=K_c

(c) n_p = n_R = 2, thus, K_p=K_c

(d) n_p =2, n_R = 1, thus, K_p$\ne$$$K_c

(b) K_{1 = $\frac{[\mathrm{NO}{]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}$ ;}   

      _{$K_{\mathit{2}}$= $\frac{\left[\mathrm{NO}\right]}{\left[{\mathrm{N}}_2{]}^{1/2}\right[{\mathrm{O}}_2{]}^{1/2}}$;}

        _{.^..  $K_{\mathit{1}}\mathit{=}{K}_{\mathit{2}}^{\mathit{2}}$}

(a) Ostwald's dilution law is valid only for weak electrolytes.