The enthalpy of combustion of propane,graphite and dihydrogen at $298 \ K$ are: $-2220.0 \ kJ \ mol^{-1}$,$-393.5 \ kJ \ mol^{-1}$ and $-285.8 \ kJ \ mol^{-1}$ respectively. The magnitude of the enthalpy of formation of propane $(C_{3}H_{8})$ is ......... $kJ \ mol^{-1}$. (Nearest integer)

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

What will be the enthalpy of formation of benzene in $kJ$ for the reaction: $6C_{(s)} + 3H_{2(g)} \rightarrow C_6H_{6(l)}$? Given that the enthalpy of combustion of benzene is $-3268 \ kJ$, the enthalpy of formation of $CO_{2(g)}$ is $-393.5 \ kJ$, and the enthalpy of formation of $H_2O_{(l)}$ is $-285.8 \ kJ$.

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

From the given data at $298 \, K$:
$\Delta H_f^o [CO_2, g] = -394 \, kJ/mol$
$\Delta H_f^o [H_2O, l] = -286 \, kJ/mol$
$\Delta H_f^o [propene, g] = 20 \, kJ/mol$
$cyclopropane (g) \to propene (g)$; $\Delta H^o_{isomerisation} = -33 \, kJ/mol$.
Calculate $\Delta H^o_{combustion} [cyclopropane, g]$.
$...... \, kJ/mol$

Given:
$(i) \, C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = x \, kJ \, mol^{-1}$
$(ii) \, C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\ominus = y \, kJ \, mol^{-1}$
$(iii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = z \, kJ \, mol^{-1}$
Based on the above thermochemical equations,find out which one of the following algebraic relationships is correct?