$C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.05 \ k \ cal \ mol^{-1}$
$C_{(diamond)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 94.50 \ k \ cal \ mol^{-1}$
Therefore:

Calculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$,given the following reactions:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2}H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$

How much $kJ$ energy is released when $6 \ mol$ of octane is burnt in air? Given $\Delta H_f^o$ for $CO_{2(g)}$,$H_2O_{(g)}$ and $C_8H_{18(l)}$ respectively are $-490$,$-240$,and $+160 \ kJ/mol$.

The enthalpy change for the reaction of $50.00 \ mL$ of ethylene with $50.00 \ mL$ of $H_2$ at $1.5 \ atm$ pressure is $\Delta H = -0.31 \ kJ$. The value of $\Delta E$ will be (in $kJ$)

Given below are two statements:
Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$
Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$
The correct answer is

The enthalpies of combustion of carbon and carbon monoxide are $-393.5 \, kJ \, mol^{-1}$ and $-283 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of carbon monoxide per mole is $....... \, kJ \, mol^{-1}$. (in $.5$)