Enthalpy of formation of methane is $-75 \ kJ / mol$. What is the enthalpy change for formation of $24 \ g$ of methane (in $kJ$)?

The molar enthalpies of combustion of $C_2H_{2(g)},$ $C$ (graphite) and $H_{2(g)}$ are $-1300,$ $-394$ and $-286 \ kJ \ mol^{-1},$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is.......$kJ \ mol^{-1}$

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).

When $1 \, \text{mol}$ of anhydrous salt $AB$ is dissolved in water,$21.0 \, J \, \text{mol}^{-1}$ of heat is released. The enthalpy of hydration of $AB$ is $-29.4 \, J \, \text{mol}^{-1}$. What is the enthalpy of solution of the hydrated salt $AB \cdot 2H_2O_{(s)}$ in $J \, \text{mol}^{-1}$ (in $.4$)?

Find the bond enthalpy of $N-H$ bond in ammonia by using the change in enthalpy for the reaction given below: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta H = -23 \ kcal$. Given bond energies: $N \equiv N = 226 \ kcal/mol$,$H-H = 103 \ kcal/mol$.

Calculate $\Delta_r H$ $(kJ \ mol^{-1})$ of the following reaction:
$C_2H_5OH_{(l)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$

Molecule	$\Delta_f H^0 (kJ \ mol^{-1})$
$C_2H_5OH_{(l)}$	$-280$
$CO_{2(g)}$	$-400$
$H_2O_{(l)}$	$-290$