For $NaCl_{(s)}$,the enthalpy of solution is $4 \ kJ \ mol^{-1}$ and the lattice enthalpy is $790 \ kJ \ mol^{-1}$. What is the hydration enthalpy of $NaCl$?

Find the value of $Q$ from the following equations:
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$

$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$
$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$
$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$
The heat of formation of $KOH$ is (in $kcal$):

Given that $:$
$2 C_{(s)} + 2 O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} ; \Delta H = -787 \ kJ$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_2 O_{(l)} ; \Delta H = -286 \ kJ$
$C_2 H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} + H_2 O_{(l)} ; \Delta H = -1301 \ kJ$
The heat of formation of acetylene will be $:-$

The heat of reaction does not depend upon

The heat evolved in the combustion of benzene is given by the equation $C_6H_{6(l)} + 7.5 O_{2(g)} \to 3H_2O_{(l)} + 6CO_{2(g)}$,$\Delta H = -781.0 \ kcal \ mol^{-1}$. Which of the following quantities of heat energy will be evolved when $39 \ g$ of benzene is burnt in an open container?