Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)

Standard enthalpy of formation of water is $-286 \ kJ \ mol^{-1}$. When $1800 \ mg$ of water is formed from its constituent elements in their standard states,the amount of energy liberated is: (in $kJ$)

Enthalpy change of the reaction $4H_{(g)} \to 2H_{2(g)}$ is $-869.6\, kJ$. The bond dissociation energy of $H-H$ bond is .....$kJ$

At standard conditions,if the change in the enthalpy for the following reaction is $-109 \; kJ \ mol^{-1}$:
$H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$
Given that bond energy of $H_2$ and $Br_2$ is $435 \; kJ \ mol^{-1}$ and $192 \; kJ \ mol^{-1}$,respectively,what is the bond energy (in $kJ \ mol^{-1}$) of $HBr$?

$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$
$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$
$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$
The heat of formation of $KOH$ is (in $kcal$):

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).