Calculate the $N-N$ bond energy in $N_2H_4$ from the given bond enthalpy data.
$\varepsilon_{N-H} = 393 \ kJ/mol$
$\varepsilon_{H-H} = 436 \ kJ/mol$
$\Delta H_{vap}[N_2H_{4(l)}] = 18 \ kJ/mol$
$N_2H_{4(l)} + H_{2(g)} \to 2NH_{3(g)} : \Delta H = -142 \ kJ/mol$
....... $kJ/mol$

From the following,the heat of neutralization is maximum in:

From the given reaction,$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H = -92.6 \ kJ$,the enthalpy of formation of $NH_3$ is (in $kJ$):

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:

Heats of combustion $(\Delta H^o)$ for $C_{(s)}$,$H_{2(g)}$ and $CH_{4(g)}$ are $-94$,$-68$ and $-213 \ kcal/mol$ respectively. The value of $\Delta H^o$ for the reaction,$C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ is $..... \ kcal$.

Two reactions are given below:
$2 Fe_{(s)} + \frac{3}{2} O_{2_{(g)}} \rightarrow Fe_2 O_{3_{(s)}}, \Delta H^{o} = -822 \ kJ/mol$
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow CO_{(g)}, \Delta H^{o} = -110 \ kJ/mol$
Then,the enthalpy change for the following reaction is:
$3 C_{(s)} + Fe_2 O_{3_{(s)}} \rightarrow 2 Fe_{(s)} + 3 CO_{(g)}$