Question

The heat of neutralization of HClby NaOH is 55.9 kJ/molefthe heat of neutralization of HCN by NaOH is
- 12.1 kJ/mole, the heat of ionization of HCN is
(A) -438 kJ / mole
(B) 43.8 kJ/mole
(C) 68 kJ/mole
(D). -68 kJ / mole​

Answer 1

Answer:

Ans is b)43.8kJ/mole for your question

Answer 2

Answer:

Explanation:

HCl+NaOH⟶NaCl+H  

​  

O△H=−55.9KJ/mol

NaCl+H  

​  

O⟶NaOH+HCl△H=55.9KJ/mole

HCN+NaOH⟶NaCH+H  

​  

O△H=−12.1KJ/mo)e

Adding (1) and (2).

HCN+NaCl⟶NaCN+HCl⟶△H=55.9−12.1

=43.8KJ/mol

∴ Energy of dissociation for one mole of HCN is 43.8KJ

Hence, the correct option is