Question

For the dissociation reaction, $H _{2}( g ) \rightarrow 2 H ( g ) \quad \Delta H =162 Kcal$, heat of atomisation of $H$ is

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

Answer 1

Answer: (A)

81 Kcal

Answer 2

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