At 298 K , the enthalpy change ( in kJ ) for the reaction given below is

$$ \mathrm{CH}_4(g)+\mathrm{O}_2(g) \longrightarrow \mathrm{C}(s)+2 \mathrm{H}_2 \mathrm{O}(l) $$

$$ \begin{aligned} Given:\mathrm{H}_2(g)+\frac{1}{2} \mathrm{O}_2(g) & \longrightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta H^{\ominus}=-286 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g) ; \Delta H^{\ominus}=-394 \mathrm{~kJ} \\ \mathrm{CH}_4(g)+2 \mathrm{O}_2(g) & \longrightarrow \mathrm{CO}_2(g)+2 \mathrm{H}_2 \mathrm{O}(l) \Delta H^{\ominus}=-890 \mathrm{~kJ}\end{aligned} $$

For the reaction $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$, the correct relation between degree of dissociation $(\alpha)$ of $\mathrm{N}_2 \mathrm{O}_4(g)$ and equilibrium constant, $K_p$ is ( $p=$ total pressure of mixture)

Oxidation state of hydrogen in compound $X$ is -1 and in compound $Y$ is $+1 . X$ and $Y$ are respectively

$$ \text { Match the following } $$

The correct answer is