Question

$H _{2(g)}+ O _{2(g)} \rightarrow H _{2} O _{(l)} ; \Delta H_{298 K }=-68.32\, kcal$. Heat of vaporization of water at $1$ atm and $25^{\circ} C$ is $10.52\, kcal$. The standard heat of formation (in kcal) of $1$ mole of water vapour at $25^{\circ} C$ is

• A. -78.84
• B. 78.84
• C. +57.80
• D. -57.80

Answer 1

Answer: (D)

$H _{2(g)}+\frac{1}{2} O _{2(g)} \rightarrow H _{2} O _{(l)} ; \Delta H=-68.32$ kcal

Vaporisation of water

$H _{2} O _{(l)} \rightarrow H _{2} O _{(g)} ; \Delta H=10.52$ kcal

Here $\Delta H=\Delta H_{f\left[ H _{2} O _{(g)}\right]}-\Delta H_{f\left[ H _{2} O _{(l)}\right]}$

$\Delta H_{f}\left[ H _{2} O _{(g)}\right]=10.52+(-68.32)=-57.80$ kcal