Q.Cyanamide ($NH_2CN$) is completely burnt in excess oxygen in a bomb calorimeter, $\Delta U$ was found to be $-742.4$ kJ mol$^{-1}$, calculate the enthalpy change of the reaction at 298K. $NH_2CN(s) + \tfrac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)$, $\Delta H = ?$

Enthalpy, $H=U+PV$, is a state function reflecting a system's capacity to do mechanical work and release heat. Its central usefulness: at CONSTANT PRESSURE, the heat involved in a process exactly equals its change in enthalpy, $\Delta H=q_p$. So $\Delta H$ positive signals an endothermic process (heat absorbed), $\Delta H$ negative signals exothermic (heat released) -- a sign convention used everywhere else in the chapter.

Deriving the relation to internal energy: applying $H=U+PV$ to an initial and final state at constant pressure and subtracting gives $\Delta H=\Delta U+P\Delta V$; substituting the first law ($\Delta U=q+w$, $w=-P\Delta V$) confirms $\Delta H=q_p$ directly. For a gas-phase reaction specifically, applying the ideal gas law to reactant and product gas volumes separately and subtracting gives $P\Delta V=\Delta n_{(g)}RT$, so the chemist's working f …