$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:

$S_{(g)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)} + 2x \ kcal$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow SO_{3(g)} + y \ kcal$
The heat of formation of $SO_{2(g)}$ is given by :

Calculate the heat of combustion (in $kJ$) of methane from the following data:
$(i)$ $C_{\text{(graphite)}} + 2H_{2(g)} \rightarrow CH_{4(g)} \quad \Delta H = -74.8 \ kJ$
(ii) $C_{\text{(graphite)}} + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
(iii) $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(l)} \quad \Delta H = -286.2 \ kJ$

The heats of formation of $CO_2$ and $H_2O$ are $-97 \, kcal$ and $-68 \, kcal$ respectively. The heat of combustion of benzene is $-783 \, kcal$. What is the heat of formation of benzene in $kcal$?

The enthalpy of combustion of ${C_6H_6}_{(l)}$ is $-3250 \, kJ/mol$. When $0.39 \, g$ of benzene is burnt in excess of oxygen in an open vessel,the amount of heat evolved is:

The correct relationship between heat of fusion $(\Delta H_{fus})$,heat of vaporization $(\Delta H_{vap})$,and heat of sublimation $(\Delta H_{sub})$ is: