Question

For the dissociation reaction, $H_2(g)\to 2H(g)\Delta H=162Kcal$, heat of atomisation of $H$ is

• A. 81 Kcal
• B. 162 Kcal
• C. 208 Kcal
• D. 218 Kcal

Answer 1

Answer: (A)

$\Delta H=\Delta H_{(\text{product })}-\Delta H_{(\text{reactant })}$
$162=2\times \Delta H_H-\Delta H_{H_2}$
$\Delta H_H=\frac{162}{2}\left(\because \Delta H_{H_2}=0\right)$
$\Delta H_H=81Kcal$