Chemistry MCQs for NEET — Practice Questions with Answers

       A(s) + 2B(aq) $\underset{}{\overset{}{\to }}$ 2C(s) + D(aq)

Initial     1         $\frac{1}{3}$               0         0

At eq.    1 – x   $\frac{1}{3}-2\text{\hspace{0.17em}x}$         2x        x

                      $\approx$a

x $\approx$ 1/3

10²⁰ = $\frac{\frac{1}{3}}{{[\text{B]}}^2}$

10²⁰ = $\frac{\frac{1}{3}}{{\text{a}}^2}$

a² = $\frac{1}{3\times {10}^{20}}$ = $\frac{{10}^{-20}}{3}$

a = $\frac{{10}^{-10}}{\sqrt{3}}\approx$ 4 × 10^–11 M

ΔG° = – RT In K = – 2.3 × 2000 × 8.3 log K

$\frac{534.52\text{\hspace{0.17em}×\hspace{0.17em}}{10}^3}{2.3\text{\hspace{0.17em}×\hspace{0.17em}}8.3\text{\hspace{0.17em}×\hspace{0.17em}}2000}=\log \text{\hspace{0.17em}K}$

K = 10¹⁴

2O₃(g) $\stackrel{}{\rightleftharpoons }$ 3O₂(g)          K = 10¹⁴

So ${\text{P}}_{{\text{O}}_{\text{2}}}$ <<< ${\text{P}}_{{\text{O}}_{\text{2}}}$

So ${\text{P}}_{{\text{O}}_{\text{2}}}$ + ${\text{P}}_{{\text{O}}_{\text{2}}}$ = 8

${\text{P}}_{{\text{O}}_{\text{2}}}$ = 8 atm

K = 10¹⁴ = $\frac{{\left({\mathrm{P}}_{{\mathrm{O}}_{\mathrm{2}}}\right)}^3}{{\left({\mathrm{P}}_{{\mathrm{O}}_{\mathrm{3}}}\right)}^2}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{{\left(8\right)}^3}{{\mathrm{P}}_{{\mathrm{O}}_{\mathrm{3}}}}$

${\text{P}}_{{\text{O}}_{\text{3}}}$ = 22.62 × 10^–7 atm.

     NH₂COONH₄(s) $\stackrel{}{\rightleftharpoons }$ 2NH₃(g) + CO₂(g)

Initial                                    2P           P’

K_p = ${\left({\mathrm{P}}_{{\mathrm{NH}}_{\mathrm{3}}}\right)}^{\mathrm{2}}\mathrm{\hspace{0.17em}}\left({\mathrm{P}}_{{\mathrm{CO}}_{\mathrm{2}}}\right)$

K_p = (2P)² (P) …… (i)

P_T(initial) = 3P

         NH₂COONH₄(s) $\stackrel{}{\rightleftharpoons }$ 2NH₃(g) + CO₂(g)

Final                                     3P           P’

K_p = (3p)² (p’) …… (ii)

From eq. (i) and (ii)

(2P)² = (3P)² (P')

P’ = $\frac{\text{4P}}{\text{9}}$

$\frac{{\mathrm{p}}_{\mathrm{T}}\left(\mathrm{New}\right)}{{\mathrm{p}}_{\mathrm{T}}\left(\mathrm{Old}\right)}=\frac{3\mathrm{P}\hspace{0.17em}+\hspace{0.17em}\mathrm{P}\text{'}}{3\mathrm{P}}=\frac{3\mathrm{P}\hspace{0.17em}+\mathrm{\hspace{0.17em}}\frac{4\mathrm{P}}{9}}{3\mathrm{P}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{31}{27}$

If CO is added 2nd equilibrium will proceed in the backward direction and the concentration of Cl₂ will decrease. This Cl₂ will be further formed by the decomposition of PCl₅.

Formation of ammonia is an exothermic process, therefore, it is favorable at a lower temperature. But at lower temperature rate of the reaction becomes slow.

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