Question

The enthalpy of reaction for HCN(aq) + OH^-(aq) -> CN^-(aq) + H₂O(l) is -12.1 kJ/mol. Assuming that the correct enthalpy of reaction of the experiment is -54.0 kJ/mol, calculate the enthalpy change for the ionization of HCN:

HCN(aq) -> H⁺(aq) + CN^-(aq)

Be sure to report your answer to the correct number of significant figures.

ΔH = ____ kJ/mol

Answer 1

consider the given reaction

HCN + OH- ---> CN- + H20 (l)

enthalpy of this reaction is -12.1 kJ/mol

but

the correct enthalpy of reaction of the experiment is -54 kJ/mol

So

the missing energy is used for the ionization of HCN

HCN --> H+ + CN-

so

enthalpy change for ionization of HCN = enthalpy of the reaction - correct enthalpy of the reaction

so

enthalpy change for ionization of HCN = -12.1 - (-54)

enthalpy change for ionization of HCN = 41.9

so

the enthalpy change for ionization of HCN is 41.9 kJ/mol

so

the answer is dH = 41.9 kJ/mol