Question

In which of the following pair of reactions first reaction is spontaneous while second reaction is non spontaneous?

• A. (i) ${}^{-}SH+H_{2}O\to H_{2}S+O{H}^{-}$
  (ii) $N{H}_2^{-}+H_{2}O\to N{H}_3+O{H}^{-}$
• B. (i) ${}^{-}OR+H_{2}S{O}_4\to ROH+HS{O}_4^{-}$
  (ii) $R^{-}+N{H}_3\to RH+N{H}_2^{-}$
• C. (i) $C{l}^{-}+HF\to HCl+F^{-}$
  (ii) ${}^{-}OH+HCl\to H_{2}O+C{l}^{-}$
• D. (i) ${}^{-}OH+HBr\to H_{2}O+B{r}^{-}$
  (ii) $R{O}^{-}+N{H}_3\to ROH+N{H}_2^{-}$

Answer 1

The correct option is D (i) ${}^{-}OH+HBr\to H_{2}O+B{r}^{-}$

(ii) $R{O}^{-}+N{H}_3\to ROH+N{H}_2^{-}$
(A) (i) $\underset{base}{{}^{⊝}SH}+\underset{acid}{H_{2}O}⟶H_{2}S+O{H}^{⊝}$ (ii) $\underset{base}{N{H}_2^{⊝}}+\underset{acid}{H_{2}O}⟶N{H}_3+O{H}^{⊝}$

Here both reactions are spontaneous. As $S{H}^{⊝}$ and $N{H}_2^{⊝}$ are highly basic.

(B) (i) $\underset{base}{{}^{⊝}OR}+\underset{acid}{H_{2}S{O}_4}⟶ROH+HS{O}_4^{⊝}$ (ii) $\underset{base}{R^{⊝}}+\underset{acid}{N{H}_3}⟶RH+N{H}_2^{⊝}$

Both reactions are spontaneous $R^{⊝}$ and $O{R}^{⊝}$ are highly basic.

(C) (i) $\underset{base}{C{l}^{-}}+\underset{acid}{HF}⟶HCl+F^{⊝}$ (ii) $\underset{base}{{}^{⊝}OH}+\underset{acid}{HCl}⟶H_{2}O+C{l}^{⊝}$

Here also both reactions are spontaneous.

(D) (i) $\underset{base}{O{H}^{⊝}}+\underset{acid}{HBr}⟶H_{2}O+B{r}^{⊝}$ (ii) $\underset{base}{R{O}^{⊝}}+\underset{acid}{N{H}_3}\nrightarrow ROH+N{H}_2^{⊝}$

(i) reaction is spontaneous but (ii) is not spontaneous. Because basicity of $R{P}^{⊝}$ is not high enough to abstract proton from weak acid $N{H}_3$ .