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 $kJ$):

For the reaction $R \rightarrow P$,the following potential energy diagram is given. What will be the enthalpy change $(\Delta H)$ for the given reaction (in $kJ$)?

What will be the $C-H$ bond enthalpy if:
$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)};$ $\Delta H = -890 \, kJ$
$CO_{2(g)} \rightarrow C_{(graphite)} + O_{2(g)};$ $\Delta H = 393 \, kJ$
$2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)};$ $\Delta H = 571 \, kJ$
$2H_{2(g)} \rightarrow 4H_{(g)};$ $\Delta H = 871 \, kJ$
$C_{(graphite)} \rightarrow C_{(g)};$ $\Delta H = 716 \, kJ$

The heat of formation of water is $260 \ kJ$. How much $H_2O$ is decomposed by $130 \ kJ$ of heat (in $mol$)?

Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by:

The calorific value of glucose is ...... $KJ/g$. Given: $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O ; \Delta H = -2900 \, KJ/mol$.