Under similar conditions,the enthalpy of freezing is exactly opposite to:

Calculate the heat of formation for propene $(C_3H_6)$ using the following thermochemical equations:
$(i) C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H_1 = -94.05 \ k.cal/mole$
$(ii) H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}; \Delta H_2 = -68.32 \ k.cal/mole$
$(iii) C_3H_{6(g)} + \frac{9}{2} O_{2(g)} \to 3 CO_{2(g)} + 3 H_2O_{(l)}; \Delta H_3 = -499.7 \ k.cal/mole$
(Note: The original question provided propane combustion data; assuming propene combustion data for consistency).

If the enthalpy of atomisation for $Br_{2(l)}$ is $x \ kJ/mol$ and the bond enthalpy for $Br_{2(g)}$ is $y \ kJ/mol$,what is the relation between them?

In an exothermic reaction,$\Delta H$ is

If enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.

Calculate the heat required to convert $9 \ g$ of liquid water to water vapor using the following equations:
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(g)} \quad \Delta H = -57 \ kCal$
$H_{2(g)} + 1/2 O_{2(g)} \longrightarrow H_2O_{(l)} \quad \Delta H = -68.3 \ kCal$ (in $kCal$)