Calculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$ given the following:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$

What is the enthalpy of formation of $NH_3$ if the bond enthalpies are $(N \equiv N) = 941 \ kJ/mol$,$(H-H) = 436 \ kJ/mol$,and $(N-H) = 389 \ kJ/mol$?

The enthalpy change of which reaction corresponds to $\Delta H_f^o$ for $Na_2CO_{3(s)}$ at $298 \ K$?

Nitroglycerine $(MW = 227.1)$ detonates according to the following equation:
$2 C_{3}H_{5}(NO_{3})_{3(l)} \longrightarrow 3 N_{2(g)} + \frac{1}{2} O_{2(g)} + 6 CO_{2(g)} + 5 H_{2}O_{(g)}$
The standard molar enthalpies of formation,$\Delta H_{f}^{\circ}$ for all the compounds are given below:
$\Delta H_{f}^{\circ}[C_{3}H_{5}(NO_{3})_{3}] = -364 \, kJ/mol$
$\Delta H_{f}^{\circ}[CO_{2(g)}] = -393.5 \, kJ/mol$
$\Delta H_{f}^{\circ}[H_{2}O_{(g)}] = -241.8 \, kJ/mol$
$\Delta H_{f}^{\circ}[N_{2(g)}] = 0 \, kJ/mol$
$\Delta H_{f}^{\circ}[O_{2(g)}] = 0 \, kJ/mol$
The enthalpy change when $10 \, g$ of nitroglycerine is detonated is $..... \, kJ$.

Formation of $NO_{2(g)}$ from $N_{2(g)}$ and $O_{2(g)}$ is an endothermic process. Which of the following is true for this reaction?

The heat of combustion of $CH_{4(g)}$,$C_{(graphite)}$ and $H_{2(g)}$ are $-20 \ kcal$,$-40 \ kcal$ and $-10 \ kcal$ respectively. The heat of formation of methane is.......$kcal$.