The enthalpy of neutralisation of HCl by NaOH IS `-55.9 kJ ` and that of HCN by NaOH is `-12.1 kJ mol^(-1)`. The enthalpy of ionisation of HCN is

Options

A. `-43.8 KJ`
B. `43.8 KJ`
C. 68 KJ
D. `-68` KJ

Answer

B

Explanation

`HCl+NaOH rarr NaCl + H_(2)O`
OR
`H^(+)+OH^(-)rarr H_(2)O , Delta H = -55.9 KJ mol^(-1)` …(i)
`HCN+NaOHrarr NaCN+H_(2)O`
OR
`HCN+OH^(-)rarr CN^(-)+H_(2)O , Delta H=-121.1 KJ mol^(-1)` …(ii)
`HCN rarr H^(+)+CN^(-) , Delta H = ?`
eq(iii) = eq(ii) - eq(i)
`Delta H_(3)=-12.1-(-55.9)=43.8 KJ`