For the reaction $H_{2(g)} + C_{2}H_{4(g)} \rightarrow C_{2}H_{6(g)}$,the enthalpy change is ....... $Kcal \, mol^{-1}$. Given bond energies: $H-H = 103$,$C-H = 99$,$C-C = 80$,and $C=C = 145 \, Kcal \, mol^{-1}$.

The enthalpies of combustion of $C$(graphite) and $C$(diamond) are $-393.8 \ kJ \ mol^{-1}$ and $-395.3 \ kJ \ mol^{-1}$ respectively. The enthalpy of conversion of $C$(graphite) to $C$(diamond) is

Find the value of $x$ in $\text{kJ}$ using the following equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} : \Delta H = 131 \ \text{kJ}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} : \Delta H = -282 \ \text{kJ}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} : \Delta H = -242 \ \text{kJ}$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} : \Delta H = x \ \text{kJ}$

The bond dissociation enthalpy of $H_{2(g)}$ and $N_{2(g)}$ are $436 \ kJ \ mol^{-1}$ and $940 \ kJ \ mol^{-1}$ respectively,and the enthalpy of formation of $NH_{3(g)}$ is $-45 \ kJ \ mol^{-1}$. The enthalpy of atomisation of $NH_{3(g)}$ is ..... $kJ \ mol^{-1}$.

Which of the following is an endothermic reaction?

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