Consider the reaction at $300 \ K$:
$C_6H_{6(l)} + \frac{15}{2}O_{2(g)} \longrightarrow 6CO_{2(g)} + 3H_2O_{(l)} ; \Delta H = -3271 \ kJ$
What is $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^\circ C$? .....$kJ$

Which of the following relations are correct?
$(A)$ $\Delta U = q + p \Delta V$
$(B)$ $\Delta G = \Delta H - T \Delta S$
$(C)$ $\Delta S = \frac{q_{rev}}{T}$
$(D)$ $\Delta H = \Delta U - \Delta nRT$
Choose the most appropriate answer from the options given below:

Assuming the water vapour to be a perfect gas,calculate the internal energy change when $1 \ mol$ of water at $100^{\circ} C$ and $1 \ bar$ pressure is converted to ice at $0^{\circ} C$. Given the enthalpy of fusion of ice is $6.00 \ kJ \ mol^{-1}$ and heat capacity of water is $4.2 \ J \ g^{-1} {\circ} C^{-1}$.

The enthalpy of combustion of carbon monoxide at $17 ^\circ C$ and constant volume is $-283.3 \ kJ$. Calculate the enthalpy of combustion at constant pressure in $kJ$.

The heat evolved during the combination of $24 \, g$ of $C$ and $128 \, g$ of $S$ according to the reaction $C + 2S \to CS_2$ with $\Delta H = 22.0 \, kcal$ is:

The standard molar heat of formation of ethane,$CO_2$,and water $(l)$ are respectively $-21.1$,$-94.1$,and $-68.3 \ kcal$. The standard molar heat of combustion of ethane will be $kcal$.