# Show the relation between the enthalpies of the reactions: $H_2(g)+ \frac{1}{2} O_2(g) \rightarrow H_2O(g); \Delta H_1$ and $H_2(g)+\frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta H_2$

(a) $\Delta H_1 < \Delta H_2$
(b) $\Delta H_1 + \Delta H_2 = 0$
(c) $\Delta H_1 > \Delta H_2$
(d) $\Delta H_1 = \Delta H_2$

Answer: $\Delta H_1 < \Delta H_2$
Given these two equations,
$H_2(g)+ \frac{1}{2} O_2(g) \rightarrow H_2O(g); \Delta H_1$
$H_2(g)+\frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta H_2$
$|\Delta H_1| < |\Delta H_2 |$
i.e., amount of heat evolved in the first reaction will be less than that in the second reaction. This is because additional amount of heat is evolved in the conversion of $H_2O$(g) into $H_2O(l)$ in the second case.