Equilibrium MCQs for NEET — Chemistry Questions with Answers

HgO(s) $\stackrel{}{\rightleftharpoons }$ Hg (g) + $\frac{1}{2}$O₂ (g)

   1                     0        0

t =t , (1-x)          X          X/2

K_p = P_Hg(g) × P(O₂)^1/2

Total moles at equilibrium = $\frac{3\mathrm{x}}{\mathrm{2}}$

P_Hg = $\frac{\mathrm{x}}{3\mathrm{x}\hspace{0.17em}/\hspace{0.17em}2}\mathrm{\hspace{0.17em}}=\hspace{0.17em}\mathrm{P}\hspace{0.17em}=\mathrm{\hspace{0.17em}}\frac{\mathrm{2}}{\mathrm{3}}\mathrm{P}$

PO₂ = $\frac{\mathrm{x}/2}{3\mathrm{x}/2}\mathrm{\hspace{0.17em}}\mathrm{P}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{1}{3}\mathrm{P}$

K_p = $\frac{2}{3}\mathrm{P}{\left(\frac{{\displaystyle 1}}{{\displaystyle 3}}\mathrm{P}\right)}^{1/2}=\frac{\mathrm{2}}{{\mathrm{3}}^{3/2}}\hspace{0.17em}{\mathrm{P}}^{3/2}$

  Br₂(l) + Cl₂(g) $\stackrel{}{\rightleftharpoons }$ 2BrCl(g)

       t = 0        1                0

                  (1 – x)           2x

K_p = $\frac{({\mathrm{P}}_{\mathrm{BrCl}}{\mathrm{)}}^{\mathrm{2}}}{{\mathrm{P}}_{{\mathrm{Cl}}_{\mathrm{2}}}}$ = 1 so, ${\mathrm{P}}_{{\mathrm{Cl}}_{\mathrm{2}}}$ = ${{\text{(P}}_{\mathrm{BrCl}})}^2$ = 0.01 atm

them at equilibrium, $\frac{{\mathrm{n}}_{\mathrm{BrCl}}}{{\mathrm{n}}_{{\mathrm{Cl}}_{\mathrm{2}}}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{0.1}{0.01}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}10\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{2\mathrm{x}}{1\mathrm{\hspace{0.17em}}-\hspace{0.17em}\mathrm{x}}$

so,10 – 10x = 2xorx = $\frac{10}{12}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{5}{6}$ moles

Moles of Br(l) required for maintaining vapour pressure of 0.1 atm

= 2 × $\frac{5}{6}$ moles = $\frac{10}{6}$ moles = moles of BrCl(g).

Moles required for taking part in reaction = moles of Cl₂ used up = $\frac{5}{6}$moles.

K_w = 55.5 × 3.6 × 10^–15 × 0.1 = 2 × 10^–14

Hence temperature must be > 25°C

HA $\stackrel{}{\rightleftharpoons }$ H⁺ + A^– ; K₁ = $\frac{\left[{\mathrm{H}}^{\mathrm{+}}\right][{\mathrm{A}}^{-}\mathrm{]}}{\left[\mathrm{HA}\right]}$

HB $\stackrel{}{\rightleftharpoons }$ H⁺ + B^– ; K₂ = $\frac{\left[{\mathrm{H}}^{\mathrm{+}}\right][{\mathrm{B}}^{\mathrm{-}}\mathrm{]}}{\left[\mathrm{HB}\right]}$

By mass balance,

[HA]_initial = [HA]_eq + [A^–] = C₁

[HB]_initial = [HB]_eq + [B^–] = C₂

By charge balance,

[H⁺] = $\frac{{\mathrm{K}}_1\mathrm{\hspace{0.17em}}\left[\mathrm{HA}\right]}{\left[{\mathrm{H}}^{+}\right]}\mathrm{\hspace{0.17em}}+\mathrm{\hspace{0.17em}}\frac{{\mathrm{K}}_2\mathrm{\hspace{0.17em}}\left[\mathrm{HB}\right]}{\left[{\mathrm{H}}^{+}\right]}$

[H⁺]² = K₁ [HA] + K₂ [HB] = K₁ {C₁ – [A^–]} + K₂ {C₂ – [B^–[}

If K₁, K₂ are very less then

[H⁺] = $\sqrt{{\mathrm{K}}_1\hspace{0.17em}{\mathrm{C}}_1\mathrm{\hspace{0.17em}}+\hspace{0.17em}{\mathrm{K}}_2\hspace{0.17em}{\mathrm{C}}_2\mathrm{\hspace{0.17em}}+\mathrm{\hspace{0.17em}}....\hspace{0.17em}{\mathrm{K}}_{\mathrm{n}}\hspace{0.17em}{\mathrm{C}}_{\mathrm{n}}}$ = $\sqrt{\sum _{i\hspace{0.17em}=\hspace{0.17em}1}^n{K}_i{C}_i}$ for ‘n’ number of weak acids

pH = pKa₁ = 2.34.

pH = pKa₂ = 9.6.

pH = $\frac{{\mathrm{pKa}}_{\mathrm{1}}\hspace{0.17em}+\hspace{0.17em}{\mathrm{pKa}}_{\mathrm{2}}}{\mathrm{2}}$ = 5.97. 

Mg(aq) + 2HCl(aq) → MgCl₂ (aq) + H₂

$\frac{1.458}{24.3}$

Millimoles of HCl = 3 × 80 = 240 mM

Moles of HCl after reaction = 240 – 60 × 2 = 120

New Molarity = $\frac{120}{80}$ = 1.5 M

pH = – log[H⁺] = – log 1.5 = – 0.176

Change is pH = – 0.176 – (– 0.477) = 0.3

pH of NaHCO₃ solution = 9

Now H+ + HCO₃^– → H₂CO₃

∴ no. of mmole of HCl remaining = 1 – 0.1 = 0.9 mmole

∴ pH = – log (9 × 10^–3) = – 2 log 3 + 3

In pure water:    C₆H₅COOAg $\stackrel{}{\rightleftharpoons }$ C₆H₅COO^– + Ag⁺

                                  (s – x)                 s

                             C₆H₅COO^– + H₂O $\stackrel{}{\rightleftharpoons }$ C₆H₅COOH + OH^–

                                        (s – x)                    x            x

s (s – x) = 2.5 × 10–13 ..……….

$\frac{{\mathrm{x}}^{\mathrm{2}}}{(\mathrm{s}\hspace{0.17em}-\hspace{0.17em}\mathrm{x})}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{{10}^{-14}}{(6.46\mathrm{\hspace{0.17em}}\times \mathrm{\hspace{0.17em}}{10}^{-5})}$ .………. (2)

Calculate s.

In buffer:  C₆H₅COOAg $\stackrel{}{\rightleftharpoons }$ C₆H₅COO^– + Ag⁺

                    (s’ – x’)                s’

C₆H₅COO^– + H⁺ $\stackrel{}{\rightleftharpoons }$ C₆H₅COOH

S’ (s’ – x’) = 2.5 × 10^–13.………. (3)

$\frac{(\mathrm{s}\text{'}\hspace{0.17em}-\hspace{0.17em}\mathrm{x}\text{'})\hspace{0.17em}\times \mathrm{\hspace{0.17em}}{10}^{-3.19}}{\mathrm{x}\text{'}}$ = 6.46 × 10^–5 .………. (4)

solve for s’.

Calculate $\frac{\mathrm{s}\text{'}}{\mathrm{s}}\mathrm{.}$

pH = $\frac{{\mathrm{pK}}_{{\mathrm{a}}_{\mathrm{1}}} + {\mathrm{pK}}_{{\mathrm{a}}_{\mathrm{2}}}}{2}$ = $\frac{2.28\hspace{0.17em}+\hspace{0.17em}9.2}{2}$ = 5.74