Chemistry MCQs for NEET — Practice Questions with Answers

(b)    Given :  $\frac{{\mathit{K}}_{{\mathit{P}}_{\mathit{1}}}}{{\mathit{K}}_{{\mathit{P}}_{\mathit{2}}}}$ = 1/9     and ${\alpha }_{\mathit{x}}\mathit{=}{\alpha }_{\mathit{z}}$

        $$\begin{gathered} \mathrm{X} \rightleftharpoons 2\mathrm{Y} \\ 1 0 \\ 1-\mathrm{\alpha } 2\mathrm{\alpha } \end{gathered}$$      .^..     $K_{P_{\mathit{1}}}\mathit{ }\mathit{=}\mathit{ }\frac{\mathit{(}\mathit{2}{\alpha }_x{\mathit{)}}^{\mathit{2}}}{\mathit{(}\mathit{1}\mathit{-}{\alpha }_x\mathit{)}}\mathit{ }\mathrm{x}\mathit{ }{\left[\frac{P_{\mathit{1}}}{\mathit{1}\mathit{+}\mathit{ }{\alpha }_x}\right]}^{\mathit{1}}$        

         $$\begin{gathered} \mathrm{Z} \rightleftharpoons \mathrm{P} + \mathrm{Q} \\ 1 0 \\ 1-\mathrm{\alpha } 2\mathrm{\alpha } \end{gathered}$$    .^..      $K_{P_2}\mathit{ }\mathit{=}\mathit{ }\frac{\mathit{(}{\alpha }_z{\mathit{)}}^{\mathit{2}}}{\mathit{(}\mathit{1}\mathit{-}{\alpha }_z\mathit{)}}\mathit{ }\mathrm{x}\mathit{ }{\left[\frac{P_2}{\mathit{1}\mathit{+}\mathit{ }{\alpha }_z}\right]}^{\mathit{1}}$                                                                            

                          .^..           P₁/P₂ = $\frac{{\mathit{K}}_{{\mathit{P}}_{\mathit{1}}}}{{\mathit{K}}_{{\mathit{P}}_{\mathit{2}}}}$x 1/4 = 1/9 x 1/4 = 1/36

For $A\left(\mathrm{s}\right)\stackrel{ }{\rightleftharpoons }2B\left(\mathrm{g}\right) + 3C\left(\mathrm{g}\right).$

$$\begin{gathered} \therefore {\mathrm{K}}_{\mathrm{c}}={\left[\mathrm{C}\right]}^3{\left[\mathrm{B}\right]}^2; \mathrm{if} \left[\mathrm{C}\right] \mathrm{becomes} \mathrm{twice}, \\ \mathrm{Then} \mathrm{let} \mathrm{conc}. \mathrm{of} \mathrm{B} \mathrm{becomes} \mathrm{B}\text{'} \mathrm{then} \\ {\mathrm{K}}_{\mathrm{c}}={\left[2\mathrm{C}\right]}^3{\left[\mathrm{B}\text{'}\right]}^2 \\ \mathrm{or} {\left[\mathrm{C}\right]}^3.{\left[\mathrm{B}\right]}^2={\left[2\mathrm{C}\right]}^3{\left[\mathrm{B}\text{'}\right]}^2 \\ \therefore \frac{\left[\mathrm{B}\text{'}\right]}{\left[\mathrm{B}\right]}=\frac{1}{8}=\frac{1}{2\sqrt{2}} \end{gathered}$$

(c) $\mathit{∆}n$=1-1-1/2 = -1/2

      .^..    K_p =K_c(RT)^-1/2

(b) HA $\rightleftharpoons$H⁺ + A^-

     At equlibrium, [H⁺]=[A^-]

         ^{K_a = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$ =  $\frac{[{\mathrm{H}}^{+}{]}^2}{\left[\mathrm{HA}\right]}$}

             ^{[H+] =$\sqrt{K_{\mathit{a}}\left[\mathrm{HA}\right]}$ = $\sqrt{1\mathrm{x}{10}^{-5} \mathrm{x}0.1}$=$\sqrt{1\mathrm{x}{10}^{-6}}$ =  ${10}^{-3}$x 1}

             ^{$\mathrm{\alpha }=\frac{\mathrm{Actual} \mathrm{ionisation}}{\mathrm{Molar} \mathrm{ionisation}}$ = ${10}^{-3}$/0.1 = ${10}^{-2}$}

        ^{% of acid dissociated = ${10}^{-2}$ x 100 = 1.00%}

(d) Heat will be used to melt ice.

(a) K_p =$\frac{({\mathrm{p}}_{{\mathrm{SO}}_3}{)}^2}{\left({\mathrm{p}}_{{\mathrm{SO}}_2}{)}^2\right({\mathrm{p}}_{{\mathrm{O}}_2})}$ =$\frac{(0.331{)}^2}{(0.662{)}^2(0.101)}$=2.5

       Now,     K_p =$\frac{({\mathrm{p}}_{{\mathrm{SO}}_3}{)}^2}{\left({\mathrm{p}}_{{\mathrm{SO}}_2}{)}^2\right({\mathrm{p}}_{{\mathrm{O}}_2})}$

                ${\mathrm{p}}_{{\mathrm{SO}}_3}={\mathrm{p}}_{{\mathrm{SO}}_2}$ 

           Then, ${\mathrm{p}}_{{\mathrm{O}}_2}$ = 1/K_p  =1/0.25 = 0.4 atm

(d)

$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH} \stackrel{ }{\rightleftharpoons }{\mathrm{CH}}_3{\mathrm{COO}}^{-} + {\mathrm{H}}^{+} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]} \\ \mathrm{Given} \mathrm{that}, \\ \left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]=\left[{\mathrm{H}}^{+}\right]=3.4 \mathrm{X} {10}^{-4} \mathrm{M} \\ {\mathrm{K}}_{\mathrm{a} }\mathrm{for} {\mathrm{CH}}_3\mathrm{COOH} = 1.7 \mathrm{X} {10}^{-5} \\ {\mathrm{CH}}_3\mathrm{COOH} \mathrm{is} \mathrm{week} \mathrm{acid}, \mathrm{so} \mathrm{in} \mathrm{it} \left[{\mathrm{CH}}_3\mathrm{COOH}\right] \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{initial} \mathrm{concentration}. \mathrm{Hence}, \\ 1.7 \times {10}^{-5}=\frac{\left(3.4 \times {10}^{-4}\right)\left(3.4 \times {10}^{-4}\right)}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]} \\ \left[{\mathrm{CH}}_3\mathrm{COOH}\right]=\frac{\left(3.4 \times {10}^{-4}\right)\left(3.4 \times {10}^{-4}\right)}{1.7 \times {10}^{-5}}=6.8 \times {10}^{-3} \mathrm{M} \end{gathered}$$$$

(b) [CO] =0.1/0.5

     [Cl₂] =0.1/0.5

     [COCl₂] = 0.3/0.5

      .^..  K_c = $\frac{0.3/0.5}{\frac{0.1}{0.5}\mathrm{x}\frac{0.1}{0.5}}$ = 15

(a) CH₃-CH₃ is neutral or least acidic and thus, its conjugate base should be strongest.

(c)    PCl₅ (g) $\rightleftharpoons$PCl₃(g) +Cl₂ (g)

        Mole

  At eq. (a-x)        x            x       

Addition of inert gas at constant V has no effect on reactions having $\mathit{∆}n$ = 0 or $\mathit{∆}n$$\ne$0. But addition of inert gas at constant P has an effect on reactions having $\mathit{∆}n$$\ne$0, and no effect if $\mathit{∆}n$=0. The given reaction has $\mathit{∆}n$$\ne$0 and thus, only choice (c) is correct. Also, the effect may be drawn as: On addition of inert gas at constant P, volume increases. To have K_c constant, X must increase.