The heat evolved in the combustion of benzene is given by the equation $C_6H_{6(l)} + 7.5 O_{2(g)} \to 3H_2O_{(l)} + 6CO_{2(g)}$,$\Delta H = -781.0 \ kcal \ mol^{-1}$. Which of the following quantities of heat energy will be evolved when $39 \ g$ of benzene is burnt in an open container?

For the reaction $N_2 + 3 X_2 \longrightarrow 2 NX_3$,where $X = F, Cl$ (the average bond energies are $F-F = 155 \ kJ \ mol^{-1}$,$N-F = 272 \ kJ \ mol^{-1}$,$Cl-Cl = 242 \ kJ \ mol^{-1}$,$N-Cl = 200 \ kJ \ mol^{-1}$ and $N \equiv N = 941 \ kJ \ mol^{-1}$),the heats of formation of $NF_3$ and $NCl_3$ in $kJ \ mol^{-1}$,respectively,are closest to

Calculate the enthalpy change for the following reaction:
$H_2C=CH_{2(g)} + H_{2(g)} \longrightarrow H_3C-CH_{3(g)}$
[The bond energies of $C-H, C-C, C=C$ and $H-H$ are $414, 347, 615$ and $435 \ kJ/mol$ respectively.] (in $kJ$)

Consider the following reactions:
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is

Calculate the enthalpy of formation of carbon monoxide $(CO).$ Given: $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -393.3 \ kJ \ mol^{-1}$ and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -282.2 \ kJ \ mol^{-1}.$

What will be the amount of heat evolved by burning $0.4 \ mol$ of methane? (Given heats of formation of $CH_4$,$CO_2$,and $H_2O$ are $-75$,$-400$,and $-240 \ kJ \ mol^{-1}$ respectively) ..... $kJ$