Equilibrium MCQs for NEET — Chemistry Questions with Answers

X₂(g) + 4Y₂ $\rightleftharpoons$2XY₄(g);

$∆$H<0 and $∆$n<0

.^.. The forward reaction is favored at high pressure and low temperature. (According to Le-Chatelier's principle)

N₂(g) + O₂(g) $\rightleftharpoons$2NO(g);K₁ ..... (i)

2NO(g) + O₂(g)  $\rightleftharpoons$2NO₂(g);K₂ .... (ii)

N₂(g) + 2O₂(g)$\rightleftharpoons$ 2NO₂(g);K=K₁xK₂

For NO₂(g) $\rightleftharpoons$1/2N₂(g) + O₂(g);

K= [1/K₁K₂]^1/2

Key Idea CH₃COOH (weak acid) and CH₃COONa (conjugated salt) form acidic buffer and for acidic buffer,

 pH = pK_a + log[salt]/[acid]

and [H⁺] = -antilog pH

 pH = -logK_a + log[salt]/[acid]    [._.. pK_a = -logK_a]

     = -log(1.8 x 10^-5) + log(0.20)/(0.10)

     = 4.74 + log2

     = 4.74 + 0.3010 = 5.041

Now, [H⁺] = antilog(-5.045)

                = 9.0 x 10^-6 mol/L

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_g) has a different value of K_c and K_p.

(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

(c) Key Idea Electron deficient molecules behave as Lewis acid. Among the given molecules, only diborane is electron deficient, ie, does not have complete octet. Thus, it acts as a Lewis acid. NH₃ and H₂O being electron rich molecules behave as Lewis base.

HYdrolysis of ${\mathrm{NH}}_4\mathrm{Cl}$ takes place as,

$$\begin{gathered} {\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NH}}_4\mathrm{OH}+\mathrm{HCl} \\ \mathrm{or} {{\mathrm{NH}}_4}^{+}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}^{+} \\ \mathrm{Hydrolysis} \mathrm{constant}, {\mathrm{K}}_{\mathrm{h}} \\ {\mathrm{K}}_{\mathrm{h}}=\frac{\left[{\mathrm{NH}}_4\mathrm{OH}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{{\mathrm{NH}}_4}^{+}\right]\left[{\mathrm{OH}}^{-}\right]} ...\left(\mathrm{ii}\right) \\ \mathrm{or} {\mathrm{K}}_{\mathrm{h}}=\frac{\left[{\mathrm{NH}}_4{\mathrm{OH}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{{\mathrm{NH}}_4}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}....\left(\mathrm{iii}\right) \\ \mathrm{from} \mathrm{eqs}. \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{iii}\right) \\ {\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}} [\because [{\mathrm{OH}}^{-}\left]\right[{\mathrm{H}}^{+}]={\mathrm{K}}_{\mathrm{w}}] \\ =\frac{{10}^{-14}}{1.77\times {10}^{-5}} \\ =5.65\times {10}^{-10} \end{gathered}$$

Key Idea Generally acids react with bases and bases (alkalies) react with acids.

Sodium hydroxide, NaOH being a strong alkali, never react with a basic oxide (compound). among the given options, B₂O₃ and BeO are amphoteric oxides, SiO₂ is acidic oxide and CaO is a basic oxide. Therfore, NaOH does not react with CaO.

Given, CH₃COOH $\rightleftharpoons$ CH3COO^- + H⁺

K_a = 1.5 x 10^-5 ..... (i)

HCN $\rightleftharpoons$ H⁺ + CN^-; K_a = 4.5 x 10^-10 .... (ii)

CN^- + CH₃COOH $\rightleftharpoons$ HCN + CH₃COO^-

On substraction Eq. (ii)  from Eq. (i), we get

CH₃COOH+CN^-  $\rightleftharpoons$ HCN + CH₃COO^-;

K = K_a/K_a1 = 1.5 x 10^-5 / 4.5 x 10^-10 = 10⁵/3 = 3.33 x 10⁴

While adding two equations, dissociation constants are multiplied and when subtracting the equations, dissociation constants are divided.

Alternative 

CH₃COOH $\rightleftharpoons$  CH₃COO^-+H⁺;Ka = 1.5 x 10^-5

K_a = [CH₃COO^-][H⁺] / [CH₃COOH] = 1.5 x 10^-5 ..... (i)

HCN  $\rightleftharpoons$ H⁺ + CN^-; Ka = 4.5 x 10^-10

K_{a =} [H⁺][CN^-] / [HCN] = 4.5x10^-10 ..... (ii)

CN^- + CH₃COOH $\rightleftharpoons$ HCN + CH₃COOH^-

K_c = [HCN][CH₃COO^-] /[CN^-][CH₃COOH] .... (iii)

From Eqs. (i), (ii) and (iii)

K_c = 1.5 x 10^-5 / 4.5 x 10^-10 = 3.33 x 10⁴

Key Idea: The concentration of solids taken to be unity.

Fe(OH)³ (s) $\rightleftharpoons$ Fe³⁺(aq) + 3OH^-(aq)

.^.. K = [Fe³⁺][OH^-]³

Hence, if OH^- ion concentration is decreased by 1/4 times, then equilibrium concentration of Fe³⁺ will increase by 64 times.