Question

If for $H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(g), $
$ \Delta H_{1}$ is the enthalpy of reaction and for
$H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$
$ \Delta H_{2}$ is enthalpy of reaction, then

• A. $ \Delta {{H}_{1}}>\Delta {{H}_{2}} $
• B. $ \Delta {{H}_{1}}=\Delta {{H}_{2}} $
• C. $ \Delta {{H}_{1}}<\Delta {{H}_{2}} $
• D. $ \Delta {{H}_{1}}+\Delta {{H}_{2}}=0 $

Answer 1

Answer: (C)

In the process, $H _{2}(g)+\frac{1}{2} O _{2}(g) \longrightarrow H _{2} O (l)$, heat is evolved because state is changing from gaseous to liquid, ie, the process is exothermic.
Hence,$
\Delta H_{1}<\Delta H_{2}$