
- A

- ✓

- C

- D








Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$STATEMENT-2$: The colour of the compound formed in the reaction of aniline with $\mathrm{NaNO}_2 / \mathrm{HCl}$ at $0^{\circ} \mathrm{C}$ followed by coupling with $\beta$-naphthol is due to the extended conjugation.
Reason : The reaction between nitrogen and oxygen requires high temperature.
$R - C \equiv C - R\xrightarrow{{Pd/BaS{O_4}}}A$
$R - C \equiv C - R\xrightarrow{{Na/Liq.\,NH_3}}B$
$C_{(graphite)} +O_2(g) \rightarrow CO_2(g)\,;$
$\Delta _rH^o=-395.5 \, kJ\,mol^{-1}$
$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O(l)\,;$ $\Delta _rH^o =-285.8\, kJ\, mol^{-1}$
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)\,;$
$\Delta _rH^o = + 890.3\, kJ\, mol^{-1}$
Based on the above thermochemical equations, the value of, $\Delta _rH^o $ at $298\, K$ for the reaction
$C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g) $ will be for $\Delta_{r} H^{\circ}$ ........... $kJ \,mol^{-1}$
