A) \[\text{-107}\text{.36}\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]
B) \[\text{-4}\text{.44}\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]
C) \[\text{+107}\text{.36}\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]
D) \[\text{+4}\text{.44}\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]
Correct Answer: D
Solution :
\[\underset{Strong\,acid}{\mathop{HCl\xrightarrow[{}]{{}}}}\,\underset{(Complete\,ionisation)}{\mathop{{{H}^{+}}+C{{l}^{-}}}}\,\] ...(i) \[N{{H}_{4}}OHNH_{4}^{+}+O{{H}^{-}}\] Weak base \[\Delta H=xkJ\,mo{{l}^{-1}}\] ?(ii) \[{{H}^{+}}+O{{H}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}O\] \[\underset{\begin{smallmatrix} (from\,neutralisation\,of \\ strong\,acid\,and\,strong\,base) \end{smallmatrix}}{\mathop{\Delta H=-55.90\,kJ\,mo{{l}^{-1}}}}\,\] ...(iii) From equation (i), (ii) and (iii) \[\Delta H=-51.46kJ\,mo{{l}^{-1}}\] \[\therefore \]\[x+(-55.90)=-51.46\] \[x=-51.46+55.90\] \[=4.44\,kJ\,mo{{l}^{-1}}\] \[\therefore \] Enthalpy of ionisation of \[N{{H}_{4}}OH=4.44kJ\,mo{{l}^{-1}}\]
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