Question

The heat of neutralization of HCl by NaOH is -55.9 kJ/mol. If the heat of neutralization of HCN by NaOH is -12.1 kJ/ mol.

The energy of dissociation of HCN is :

• A. -43.8 kJ
• B. 43.8 kJ
• C. 68 kJ
• D. -68kJ

Answer 1

The correct option is B 43.8 kJ

$HCl+NaOH⟶NaCl+H_{2}O△H=-55.9KJ/molNaCl+H_{2}O⟶NaOH+HCl△H=55.9KJ/mol--------\left(1\right)HCN+NaOH⟶NaCH+H_{2}O△H=-12.1KJ/mol-------\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$.

$HCN+NaCl⟶NaCN+HCl⟶△H=55.9-12.1=43.8KJ/mol$

$\therefore$ Energy of dissociation for one mole of $HCN$ is $43.8KJ$

Hence, the correct option is B.