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Question: The heat of neutralization of NaOH is \( - 55.9\;KJ\;mo{l^{ - 1}}\).If the heat of neutralization of...

The heat of neutralization of NaOH is −55.9  KJ  mol−1 - 55.9\;KJ\;mo{l^{ - 1}}.If the heat of neutralization of HCN by NaOH is −12.1  KJ  mol−1 - 12.1\;KJ\;mo{l^{ - 1}}
The energy of dissociation of HCN is:
(A) −43.8 - 43.8 KJKJ
(B) 43.843.8 KJKJ
(C) 6868 KJKJ
(D) −68 - 68 KJKJ

Explanation

Solution

In the case of the reaction of strong acids with strong bases the net reaction is the neutralisation of the base by acid to form salt and water. The ionisation reaction is nothing but the reversal of the above reaction. So the heat of ionisation or energy of dissociation is equal to the sum of both enthalpies.

Complete answer:
We know that the heat of neutralization is the amount of heat evolved when one equivalent of acid is neutralised by one equivalent of a base in a fairly dilute solution. Neutralisation reactions are always exothermic and the value of change in enthalpy is negative. Here we have been given the heat of neutralisation of sodium hydroxide by hydrochloric acid. The following reaction will be as follows
HCl+NaOH→NaCl+H2OHCl + NaOH \to NaCl + {H_2}O ΔH=−55.9  KJ  mol−1\Delta H = - 55.9\;KJ\;mo{l^{ - 1}}.
Here the enthalpy of neutralisation is given to be −55.9  KJ  mol−1 - 55.9\;KJ\;mo{l^{ - 1}}
Similarly, for the ionisation of sodium chloride, the reaction will be as follows
NaCl+H2O→NaOH+HClNaCl + {H_2}O \to NaOH + HCl ΔH=55.9  KJ  mol−1\Delta H = 55.9\;KJ\;mo{l^{ - 1}}----(1)
Now we have been given the neutralisation of hydrogen cyanide by sodium hydroxide to be −12.1  KJ  mol−1 - 12.1\;KJ\;mo{l^{ - 1}}
The reaction of the following will be as follows
HCN+NaOH→NaCN+H2OHCN + NaOH \to NaCN + {H_2}O ΔH=−12.1  KJ  mol−1\Delta H = - 12.1\;KJ\;mo{l^{ - 1}}----(2)
The enthalpy of neutralisation of hydrogen cyanide is given to be −12.1  KJ  mol−1 - 12.1\;KJ\;mo{l^{ - 1}}
Adding both the equations we get,
HCN+NaCl→NaCN+HClHCN + NaCl \to NaCN + HCl
⇒ΔH=55.9−12.1\Rightarrow \Delta H = 55.9 - 12.1
⇒ΔH=43.8  KJ  mol−1\Rightarrow \Delta H = 43.8\;KJ\;mo{l^{ - 1}}
∴\therefore Energy of dissociation for one mole of HCNHCN is 43.843.8 KJKJ

**Hence the correct answer is option B.

Note:**
In case of neutralisation of a weak acid or a weak base against a strong acid or strong base respectively a part of evolved heat is used up in ionising the weak acid or base. Bond dissociation energy is a positive quantity as energy is being provided to break the bond.