रसायन विज्ञान (Chemistry) NEET प्रश्न हिंदी में — उत्तर सहित MCQ अभ्यास

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

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

(b) CO₂ (एक अभिकारक) के हटने पर, अभिक्रिया पश्च दिशा में अग्रसर होगी।

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

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

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

(d) CH₃COOH (दुर्बल अम्ल) तथा CH₃COONa (संयुग्मी लवण) अम्लीय बफर बनाते हैं और अम्लीय बफर के लिए,

  pH =pK_a + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$ तथा [H⁺] = -antilog pH

  pH = -logK_a +  log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$  [^..^. pK_a = -log K_a]

  = -log(1.8 x 10-5) + log$\frac{(0.20)}{(0.10)}$

  = 4.74 + log2

  = 4.74+0.3010 =5.041

   अब,  [H⁺] = antilog(-5.045)

  = 9.0x10^-6 mol/L

(c) K₁ = $\frac{\left[{\mathrm{XeOF}}_4\right][\mathrm{HF}{]}^2}{\left[{\mathrm{XeF}}_6\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$  ...(i)

  K₂ = $\frac{\left[{\mathrm{XeOF}}_4\right]\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]}{\left[{\mathrm{XeO}}_4\right]\left[{\mathrm{XeF}}_6\right]}$  ....(ii)

  समी. (ii)/(i) से हमें प्राप्त होता है K₂/K₁ = $\frac{\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{XeO}}_4\right][\mathrm{HF}{]}^2}$ = K_c

(b) दुर्बल अम्ल तथा दुर्बल क्षार के लवणों का pH इस संबंध से निकाला जाता है:

  [H⁺]=$\sqrt{K_H}$=$\sqrt{\frac{K_w}{K_a.K_b}}$

$$\begin{gathered} \mathit{\left(}\mathit{c}\mathit{\right)} \mathrm{In} \mathrm{equation}, \mathrm{X} \stackrel{ }{\rightleftharpoons } \mathrm{Y} + \mathrm{Z} \\ \mathrm{Initial} \mathrm{moles} 1 0 0 \\ \mathrm{At} \mathrm{equil}. (1-\mathrm{\alpha }) \mathrm{\alpha } \mathrm{\alpha } \\ \mathrm{where}, \mathrm{\alpha }=\mathrm{degree} \mathrm{of} \mathrm{dissociation} \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{moles} \\ =1-\mathrm{\alpha }+\mathrm{\alpha }+\mathrm{\alpha } = (1+\mathrm{\alpha }) \\ {\mathrm{P}}_{\mathrm{x}}=\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{P}}_1 \\ {\mathrm{P}}_{\mathrm{y}}=\left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{P}}_1 \\ {\mathrm{P}}_{\mathrm{z}}=\frac{\mathrm{\alpha }}{(1-\mathrm{\alpha })}{\mathrm{P}}_1 \\ {\mathrm{K}}_{{\mathrm{P}}_1}=\frac{\left[{\mathrm{P}}_{\mathrm{y}}\right]\left[{\mathrm{P}}_{\mathrm{z}}\right]}{\left[{\mathrm{P}}_{\mathrm{x}}\right]}=\frac{\left({\displaystyle \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right){\mathrm{P}}_1 \times \left({\displaystyle \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right){\mathrm{P}}_1}{\left({\displaystyle \frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right){\mathrm{P}}_1} \\ =\frac{{\left({\displaystyle \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)}^2{\mathrm{P}}_1}{\left({\displaystyle \frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)} ............\left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{For} \mathrm{equation}, \mathrm{A} \stackrel{ }{\rightleftharpoons } 2\mathrm{B} \\ \mathrm{Initial} \mathrm{moles} 1 0 \\ \mathrm{At} \mathrm{equil}. (1-\mathrm{\alpha }) 2\mathrm{\alpha } \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{at} \mathrm{equilibrium} = (1+\mathrm{\alpha }) \\ {\mathrm{P}}_{\mathrm{B} }=\left(\frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}\right) {\mathrm{P}}_2 \\ {\mathrm{P}}_{\mathrm{A} }=\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right) {\mathrm{P}}_2 \\ {\mathrm{K}}_{{\mathrm{P}}_2}=\frac{{\left[{\mathrm{P}}_{\mathrm{B}}\right]}^2}{\left[{\mathrm{P}}_{\mathrm{A}}\right]}=\frac{\left[\left({\displaystyle \frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)\right]{\mathrm{P}}_2}{\left({\displaystyle \frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)} ...\left(\mathrm{ii}\right) \\ \mathrm{Eq}. \left(\mathrm{i}\right) \mathrm{divide} \mathrm{by} \mathrm{Eq}. \left(\mathrm{ii}\right) \\ \frac{{\mathrm{K}}_{{\mathrm{P}}_1}}{{\mathrm{K}}_{{\mathrm{P}}_2}} = \frac{{\mathrm{\alpha }}^2\times {\mathrm{P}}_1}{4{\mathrm{\alpha }}^2\times {\mathrm{P}}_2} \\ \frac{9}{1}=\frac{{\mathrm{P}}_1}{4{\mathrm{P}}_2} \\ \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{36}{1}=36:1 \end{gathered}$$

(a) A^pB^q $\rightleftharpoons$pA⁺ + qB^- ; माना विलेयता s mol/लीटर है, अतः

  p.s  q.s

  K_sp = [A⁺]^p[B^-]^q = (ps)^p.(qs)^q 

  = p^pq^q(s)^p+q

(b) N₂ का 1 मोल H₂ के 3 मोलों से अभिक्रिया करता है, अतः

    N₂ + 3H₂ $\rightleftharpoons$ 2NH₃ ; (a-x) > (a-3x)

    a  a

(a-x)    (a-3x)    2x

(b)  NH₄HS(s)  $\rightleftharpoons$NH₃ (g) + H₂S(g)

     साम्य पर दाब  P  P

  .^.. साम्य पर कुल दाब = 2P =1.12 atm

  P =1.12/2 atm

  ^..^. K_p = $P{\mathit{\text{'}}}_{N{H}_{\mathit{3}}}\mathit{ }\mathrm{x}\mathit{ }P{\mathit{\text{'}}}_{H_{\mathit{2}}S}$ 

  K_p = (1.12/2) x (1.12/2) = 0.3136 atm²