When equal volumes of $1 \ M \ HCl$ and $1 \ M \ H_2SO_4$ are separately neutralized by an excess volume of $1 \ M \ NaOH$ solution,$x \ J$ and $y \ J$ of heat are liberated,respectively. The value of $y / x$ is . . . .

Which among the following represents the reaction of formation of the product?

Carbon and carbon monoxide burn in oxygen to form carbon dioxide according to the following reactions:
$C(s) + O_2(g) \to CO_2(g)$; $\Delta H = -394 \ kJ \ mol^{-1}$
$2CO(g) + O_2(g) \to 2CO_2(g)$; $\Delta H = -569 \ kJ \ mol^{-1}$
The heat of formation of $1 \ mol$ of carbon monoxide is thus ..... $kJ \ mol^{-1}$.

Consider the following data:
$(i)$ $2Al(s) + 6HCl(aq) \to Al_2Cl_6(aq) + 3H_2(g) + 1200 \text{ kJ/mol}$
(ii) $H_2(g) + Cl_2(g) \to 2HCl(g) + 164 \text{ kJ/mol}$
(iii) $HCl(g) + aq \to HCl(aq) + 83 \text{ kJ/mol}$
(iv) $Al_2Cl_6(s) + aq \to Al_2Cl_6(aq) + 663 \text{ kJ/mol}$
The enthalpy of formation of anhydrous solid $Al_2Cl_6$ is:

For the reaction,$3 C_2 H_{2(g)} \longrightarrow C_6 H_{6(g)}$,calculate the standard enthalpy change. The values of $\Delta H_f$ for $C_2 H_2$ and $C_6 H_6$ respectively are $250 \ kJ \ mol^{-1}$ and $90 \ kJ \ mol^{-1}$.

For the reaction $2 H_2 + O_2 \rightarrow 2 H_2 O$,$\Delta H = -571 \ kJ$. Bond energy of $H-H = 435 \ kJ$ and $O=O = 498 \ kJ$. Then the average bond energy of $O-H$ bond will be: