Chemistry - Eudiometry Question with Solution | TestHub

(A) ${\mathrm{N}}_2+{\mathrm{H}}_2⟶{\mathrm{N}}_2{\mathrm{H}}_2(\mathrm{l}),\Delta {\mathrm{H}}_1=-30$ unit

20 lit. $40$ lit.

${\mathrm{N}}_2+2{\mathrm{H}}_2⟶{\mathrm{N}}_2{\mathrm{H}}_4(\mathrm{g}),\Delta {\mathrm{H}}_2=+30$ unit

${\mathrm{N}}_2{\mathrm{H}}_2(\mathrm{l})+{\mathrm{H}}_2(\mathrm{g})⟶{\mathrm{N}}_2{\mathrm{H}}_4(\mathrm{g})$

40 lit. 20 lit.

20 lit. x 20 lit.

Net energy change $=20\times (-30)+20\times (+30)$

$$\text{ ⇒ }0$$

(B) ${\mathrm{N}}_2+{\mathrm{H}}_2⟶{\mathrm{N}}_2{\mathrm{H}}_2(\mathrm{l})$

$\begin{array}{lll}5.6& 5.6& 0\end{array}$

Contraction = 11.2 lit

(C) 11.2 amol 15.2 amol

${\mathrm{N}}_2+{\mathrm{H}}_2\to {\mathrm{N}}_2{\mathrm{H}}_2(\mathrm{l})$

4 a mol 11.2 a mol

${\mathrm{N}}_2{\mathrm{H}}_2(\mathrm{l})+$ ${\mathrm{H}}_2$ ⟶ ${\mathrm{N}}_2{\mathrm{H}}_4(\mathrm{g})$

11.2 a mol      4 a mol        4 a mol                                                

  4 lit vol.         

Total contraction in vol. = 22.4 lit.    

(D)   10     22.4             

N₂ + 2H₂—→ N₂H₄ (g)             

  > 2.4 lit.       10 lit.         

Volume contraction = 20 lit.