Equilibrium MCQs for NEET — Chemistry Questions with Answers

(B)  ${\mathrm{CN}}^{-}$ is more basic than ${\mathrm{F}}^{-} {\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HCN}\right)<{\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HF}\right)$

       ${\mathrm{CN}}^{-}$ is more basic than ${\mathrm{CH}}_3{\mathrm{COO}}^{-} {\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HCN}\right)<{\mathrm{K}}_{\mathrm{a}}\left({\mathrm{CH}}_3\mathrm{COOH}\right)$

(B) Ampholyte ion can undergo hydrolysis as well as ionization.

$∆G=∆G^{\circ }+RT \ln Q$

If reaction quotient = 1

$\therefore ∆G=∆G^{\circ }$

(d) ${\mathrm{H}}_2{\mathrm{O}}_2 \stackrel{ }{\rightleftharpoons }{\mathrm{H}}^{+} + {\mathrm{HO}}_2^{-}$

(A). Because it is most soluble in water and solubility decreases as

  $BeS{O}_4$ > $MgS{O}_4$ >$CaS{O}_4$ > $BaS{O}_4$

Extent of back bonding decreases from BF₃ to BI₃. So, lewis acidic character increases from BF₃ to BI₃. Hence, BF₃ is the weakest lewis acid.

(a) The required equation of oxidation of NH₃ by oxygen to give NO is:

$$\begin{gathered} 4{\mathrm{NH}}_3 +\hspace{0.17em}5{\mathrm{O}}_2 \underset{800°\mathrm{C}}{\overset{\mathrm{Pt} \left(\mathrm{gauze}\right)}{\to }}4\mathrm{NO} +\hspace{0.17em}6{\mathrm{H}}_2\mathrm{O} \\ \mathrm{For} \mathrm{this} \mathrm{K}=\frac{{\left[\mathrm{NO}\right]}^4{\left[{\mathrm{H}}_2\mathrm{O}\right]}^6}{{\left[{\mathrm{NH}}_3\right]}^4\left[{\mathrm{O}}_{2}5\right]} \\ \mathrm{For} \mathrm{the} \mathrm{equation} \left(\mathrm{I}\right) {\mathrm{K}}_1=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3} \\ \mathrm{For} \mathrm{the} \mathrm{equation} \left(\mathrm{II}\right) {\mathrm{K}}_2=\frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3} \\ \mathrm{For} \mathrm{the} \mathrm{equation} \left(\mathrm{III}\right) {\mathrm{K}}_3=\frac{\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{H}}_2\right]{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \mathrm{For} \mathrm{getting} \mathrm{the} \mathrm{K} \mathrm{we} \mathrm{must} \mathrm{do} \\ \mathrm{K}{ }_1^2=\frac{{\left[{\mathrm{NH}}_3\right]}^4}{{\left[{\mathrm{N}}_2\right]}^2{\left[{\mathrm{H}}_2\right]}^6}, {\mathrm{K}}_2^2=\frac{{\left[\mathrm{NO}\right]}^4}{{\left[{\mathrm{N}}_2\right]}^2{\left[{\mathrm{O}}_2\right]}^2} \\ {\mathrm{K}}_3^6=\frac{{\left[{\mathrm{H}}_2\mathrm{O}\right]}^6}{{\left[{\mathrm{H}}_2\right]}^6{\left[{\mathrm{O}}_2\right]}^{{\displaystyle \raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=3}}} \\ \mathrm{K}=\frac{{\mathrm{K}}_2^2 \times {\mathrm{K}}_3^6}{{\mathrm{K}}_1^2} \mathrm{substituting} \mathrm{the} \mathrm{value} \mathrm{we} \mathrm{get}, \\ \mathrm{K}=\frac{{\left[\mathrm{NO}\right]}^4{\left[{\mathrm{H}}_2\mathrm{O}\right]}^6}{{\left[{\mathrm{NH}}_3\right]}^4{\left[{\mathrm{O}}_2\right]}^5}\mathrm{so} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{K} \mathrm{in} \mathrm{terms} \mathrm{of} \\ {\mathrm{K}}_1, {\mathrm{K}}_2 \mathrm{and} {\mathrm{K}}_3 \mathrm{is} \\ \mathrm{K}=\frac{{\mathrm{K}}_2{\mathrm{K}}_3^3}{{\mathrm{K}}_1} \end{gathered}$$

(A) $P_4{O}_{10}+12N{H}_3+6H_{2}O\to 4{\left(N{H}_4\right)}_{3}P{O}_4$