Enthalpy of formation of $CO_{2(g)}$,$H_2O_{(l)}$ and $C_6H_{12}O_{6(s)}$ are $-393$,$-286$ and $-1170 \ kJ \ mol^{-1}$ respectively. The quantity of heat liberated when $18 \ g$ of $C_6H_{12}O_{6(s)}$ is burnt completely in oxygen is (in $kJ$)

What is the value of $\Delta S_{\text{total}}$ for the following reaction at $300 \ K$:
$Fe_{2}O_{3(s)} + 3CO_{(g)} \longrightarrow 2Fe_{(s)} + 3CO_{2(g)}$
$\Delta H^{\circ} = -25 \ kJ, \Delta S^{\circ} = 15 \ J \ K^{-1}$ (in $J \ K^{-1}$)

Two liters of $N_2$ gas at $0 \, ^\circ C$ and $5 \, atm$ pressure undergoes isothermal expansion against a constant external pressure of $1 \, atm$ until the pressure of the gas becomes $1 \, atm$. If the gas is ideal,what is the work done in the expansion in $J$?

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

The enthalpies of formation of gaseous $N_2O$ and $NO$ at $298 \ K$ are $82.0$ and $90.0 \ kJ \ mol^{-1}$ respectively. The enthalpy change of the reaction $N_2O_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow 2 NO_{(g)}$ is

Water vapor is an ideal gas. Calculate the internal energy change $(\Delta U)$ when $1 \, \text{mol}$ of water is vaporized at $1 \, \text{bar}$ pressure and $100 \, ^{\circ}C$. (Given: Molar enthalpy of vaporization of water at $1 \, \text{bar}$ and $373 \, K = 41 \, kJ \, \text{mol}^{-1}$ and $R = 8.3 \, J \, \text{mol}^{-1} \, K^{-1}$)