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 . . . .

The enthalpy of formation of ethane $(C_2H_6)$ from ethylene $(C_2H_4)$ by the addition of hydrogen,where the bond energies of $C-H$,$C-C$,$C=C$,and $H-H$ are $414 \ kJ/mol$,$347 \ kJ/mol$,$615 \ kJ/mol$,and $435 \ kJ/mol$ respectively,is $........$ $kJ/mol$.

From the following,the heat of neutralization is maximum in:

Which is the best definition of “heat of neutralization”?

At $25^{\circ} C$, the enthalpy of the following processes are given:
$H_{2(g)} + O_{2(g)} \rightarrow 2 OH_{(g)} \quad \Delta H^{\circ} = 78 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} \quad \Delta H^{\circ} = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)} \quad \Delta H^{\circ} = 436 \ kJ \ mol^{-1}$
$\frac{1}{2} O_{2(g)} \rightarrow O_{(g)} \quad \Delta H^{\circ} = 249 \ kJ \ mol^{-1}$
What would be the value of $X$ for the following reaction? (Nearest integer)
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} \quad \Delta H^{\circ} = X \ kJ \ mol^{-1}$

The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$. Then: