For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter,$\Delta U$ and $w$ correspond to

If the standard enthalpy change $\left(\Delta_{r} H^{\theta}\right)$ for a certain reaction at $298 \ K$ and constant pressure is $-1860 \ kJ \ mol^{-1}$,and the standard entropy change $\left(\Delta_{\text{sys}} S^{\theta}\right)$ of the same reaction is $-550 \ J \ K^{-1} \ mol^{-1}$,which one of the following statements is correct?

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 molar heat of formation of $NH_4NO_{3(s)}$ is $-367.54 \ kJ \ mol^{-1}$ and those of $N_2O_{(g)}$ and $H_2O_{(l)}$ are $+81.46 \ kJ \ mol^{-1}$ and $-285.78 \ kJ \ mol^{-1}$ respectively at $25 \ ^oC$ and $1.0 \ atm$ pressure. Calculate $\Delta U$ at $25 \ ^oC$ for the reaction:
$NH_4NO_{3(s)} \rightarrow N_2O_{(g)} + 2H_2O_{(l)}$ (in $kJ$)

The heat liberated when $1.89 \ g$ of benzoic acid is burnt in a bomb calorimeter at $25^{\circ} C$ increases the temperature of $18.94 \ kg$ of water by $0.632^{\circ} C$. If the specific heat of water at $25^{\circ} C$ is $0.998 \ cal / (g^{\circ} C)$,then find the heat of combustion of benzoic acid.

In the reaction at $300 \ K$:
$C_6H_6(\ell) + \frac{15}{2}O_{2(g)} \to 6CO_{2(g)} + 3H_2O(\ell) \quad \Delta H = -3271 \ kJ$
What is the value of $\Delta U$ for the combustion of $1.5 \ mol$ of benzene at $27 \ ^oC$? $..... \ kJ$