If the enthalpies of combustion of benzene $(l)$,carbon $(s)$,and hydrogen $(g)$ are $Q_1$,$Q_2$,and $Q_3$ respectively,what will be the enthalpy of formation of benzene?

Diborane is formed from the elements as shown in equation $(i)$:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.

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 $:-$

In which of the following neutralisation reactions,the heat of neutralisation will be highest?

Hess's law of constant heat summation includes:

Enthalpy of formation of $CO_{(g)}$ and $CO_{2(g)}$ are $-110 \ kJ \ mol^{-1}$ and $-393 \ kJ \ mol^{-1}$ respectively. The enthalpy of combustion of $CO$ (in $kJ \ mol^{-1}$) is: