Chemical Thermodynamics Question 227
Question: The enthalpy of neutralisation of $ NH_4OH $ with $ HCl $ is $ -51.46kJmo{{l}^{-1}} $ and the enthalpy of neutralization of $ NaOH $ with $ HCl $ is $ -55.90kJmo{{l}^{-1}} $ . The enthalpy of ionisation of $ NH_4OH $ is
Options:
A) $ -107.36 kJmo{{l}^{-1}} $
B) $ -4.44 kJmo{{l}^{-1}} $
C) $ +107.36,kJ,mo{{l}^{-1}} $
D) $ +4.44kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \underset{Strong,acid}{\mathop{HCl}},\xrightarrow{{}}\underset{(Completeionisation)}{\mathop{{{H}^{+}}}},+C{{l}^{-}} $ …(i) $ NH_4OH \rightarrow \underset{Weak,base}{\mathop{NH_4^{+}}},+OH{{~}^{-}} $
$ \Delta H=x,kJmo{{l}^{-1}} $ …(ii)
$ {{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}H_2O $
$ \Delta H=-55.90kJmo{{l}^{-1}} $ …(iii)
(from neutralisation of strong acid and strong base) From equation (i), (ii) and (iii)
$ NH_4OH+HCl\xrightarrow{{}}NH_4^{-}+C{{l}^{-}}+H_2O $
$ \Delta H=-51.46kJmo{{l}^{-1}} $
$ \therefore x+( -55.90 )=-51.46 $
$ x=-51.46+55.90 $
$ =4.44kJmo{{l}^{-1}} $
$ \therefore $ Enthalpy of ionisation of $ NH_4OH $ is
$ \Rightarrow 4.44kJmo{{l}^{-1}} $
BETA
