Question

The heat of neutralisation of $ HCl $ by $ NaOH $ is $ -55.9\,kJ/mol, $ the energy of dissociation of $ HCN $ is:

• A. 43.8 kJ
• B. -43.8 kJ
• C. -68 kJ
• D. 68 kJ

Answer 1

Answer: (A)

Given $H ^{+}+ OH ^{-} \rightarrow H _{2} O$
$\Delta H=-55.9\, kJ$ ...(i)
$HCN + OH ^{-} \rightarrow H _{2} O + CN ^{-}$
$\Delta H=-121\, kJ$ ...(ii)
on reversing (i) we get
$H _{2} O \longrightarrow H ^{+}+ OH ^{-}$
$\Delta H=+55.9\, kJ$ ...(iii)
on adding (iii) to (ii) we get
$HCN \longrightarrow H ^{+}+ CN ^{-}$
$\Delta H=+43.8\, kJ$