MCQ
Highest covalent character is found in
- A$Ca{F_2}$
- B$CaC{l_2}$
- C$CaB{r_2}$
- ✓$Ca{I_2}$
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(Given : Molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ Ethyl alcohol : $46$,
water : $18$)
$C_{(graphite)} +O_2(g) \rightarrow CO_2(g)\,;$
$\Delta _rH^o=-395.5 \, kJ\,mol^{-1}$
$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O(l)\,;$ $\Delta _rH^o =-285.8\, kJ\, mol^{-1}$
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)\,;$
$\Delta _rH^o = + 890.3\, kJ\, mol^{-1}$
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 $\Delta_{r} H^{\circ}$ ........... $kJ \,mol^{-1}$