HCN_((aq)) + NaOH_((aq)) to NaCN_((aq)) + H_2O_((l)) , DetlaH = -12.13 kJ. Calculate heat of ionisation of HCN.

HCN_((aq)) + NaOH_((aq)) to NaCN_((aq)) + H_2O_((l)) , DetlaH = -12.13

1.	HCN_((aq)) + NaOH_((aq)) to NaCN_((aq)) + H_2O_((l)) , DetlaH = -12.13 kJ. Calculate heat of ionisation of HCN.
Answer» SOLUTION :`H_((aq))^(+) + OH_((aq))^(-) to H_2O_((l)) , Delta H = -57.3 KJ` ……(1) `HCN_((aq)) + OH_((aq))^(-) to H_2O_((l)) + CN_((aq))^(-)`, `DELTAH = -12.13 kJ`.....(2) Equation (2) – equation (1) gives , `HCN_((aq0) to H_((aq))^(-) + CN_((aq))^(-) , DeltaH = +45.17 kJ` Heat of ionisation of `HCN = +45.17 kJ mol^(-1)`.