The heat of neutralization of a strong acid by a strong base is nearly equal to

Three thermochemical equations are given below:
$(i) C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\circ = x \ kJ \ mol^{-1}$
$(ii) C_{(graphite)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\circ = y \ kJ \ mol^{-1}$
$(iii) CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\circ = z \ kJ \ mol^{-1}$
Based on the above equations,find out which of the relationship given below is correct.

Find the value of $x$ in $\text{kJ}$ using the following equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} : \Delta H = 131 \ \text{kJ}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} : \Delta H = -282 \ \text{kJ}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} : \Delta H = -242 \ \text{kJ}$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} : \Delta H = x \ \text{kJ}$

What is enthalpy of reaction?

Given that bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively and $\Delta H_f$ for $HCl$ is $-90 \ kJ \ mol^{-1},$ the bond enthalpy of $HCl$ is ............... $kJ \ mol^{-1}$.

The heat of atomisation of methane and ethane are $x \ kJ \ mol^{-1}$ and $y \ kJ \ mol^{-1}$ respectively. The longest wavelength $(\lambda)$ of light capable of breaking the $C-C$ bond can be expressed in $SI$ unit as :